algebra 1 ehs 2014-2015
DESCRIPTION
Traditional Common Core Algebra 1 TextTRANSCRIPT
Algebra 1 - Essex High School
Lea Ann SmithAndrew Gloag, (AndrewG)
Anne Gloag, (AnneG)
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Printed: August 4, 2014
AUTHORSLea Ann SmithAndrew Gloag, (AndrewG)Anne Gloag, (AnneG)
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Contents www.ck12.org
Contents
1 Unit 1 - Review 11.1 Variables and Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Order of Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 Introductory Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.4 Introductory Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.5 Properties of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.6 Solving Equations Using One Step . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.7 Solving Equations using Two or More Steps . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.8 Equations with Variables on Both Sides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321.9 Word Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351.10 Solving Equations for a Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381.11 Ratios, Rates and Proportions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411.12 Percent Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471.13 Multi-Step Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521.14 Absolute Value Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 561.15 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
2 Unit 2 - Linear Functions 612.1 Introduction to Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 622.2 Finding the Domain and Range of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 652.3 Graphing using Table of Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 702.4 Graphing Using Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 762.5 Slope and Rate of Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 802.6 Graphs Using Slope-Intercept Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 872.7 Writing Equations of Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 962.8 Parallel and Perpendicular Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1032.9 Graphing Linear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1072.10 Graphs of Absolute Value Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1132.11 Fitting a Line to Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1172.12 Problem Solving with Linear Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
3 Unit 3 - Systems of Linear Equations 1333.1 Solving Linear Systems by Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1343.2 Solving Linear Systems by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1433.3 Solving Linear Systems by Combination (Elimination) . . . . . . . . . . . . . . . . . . . . . . 1463.4 Special Types of Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1533.5 Applications of Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1573.6 Systems of Linear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
4 Unit 4 - One Variable Statistics 1704.1 Dot Plots, Frequency Tables, Histograms and Bar Graphs . . . . . . . . . . . . . . . . . . . . . 1714.2 Measures of Central Tendency (Mean, Median and Mode) . . . . . . . . . . . . . . . . . . . . . 181
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4.3 Box and Whisker Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1884.4 Shape, Center and Spread . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192
5 Unit 5 - Polynomials, Exponents and Radicals 2035.1 Addition and Subtraction of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2045.2 Exponent Properties Involving Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2105.3 Multiplication of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2185.4 Special Products of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2225.5 Exponent Properties Involving Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2265.6 Zero and Negative Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2335.7 Division of Polynomials by Monomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2385.8 Division of Polynomials by Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2405.9 Scientific Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2445.10 Square Roots and Irrational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2505.11 Pythagorean Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2535.12 Multiplication and Division of Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2555.13 Addition and Subtraction of Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257
6 Unit 6 - Quadratic Functions 2596.1 Monomial Factors of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2606.2 Factoring Quadratic Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2636.3 Factoring Polynomials where a does not equal 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 2696.4 Factoring the Difference of Two Squares (DOTS) . . . . . . . . . . . . . . . . . . . . . . . . . 2736.5 Solve Quadratic Equations by Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2766.6 Solve Quadratic Equations by Square Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2806.7 Solve Quadratic Equations by Completing the Square . . . . . . . . . . . . . . . . . . . . . . . 2866.8 Solving Quadratic Equations by the Quadratic Formula . . . . . . . . . . . . . . . . . . . . . . 2916.9 Graphs of Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2986.10 Applications of Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305
7 Unit 7 - Exponential Functions, Sequences and Modeling 3097.1 Exponential Growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3107.2 Exponential Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3187.3 Arithmetic Sequences and Linear Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3267.4 Geometric Sequences and Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . 3307.5 Describing the Pattern and Writing a Recursive Rule for a Sequence . . . . . . . . . . . . . . . 3357.6 Linear, Exponential, and Quadratic Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3387.7 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353
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www.ck12.org Chapter 1. Unit 1 - Review
CHAPTER 1 Unit 1 - ReviewChapter Outline
1.1 VARIABLES AND EXPRESSIONS
1.2 ORDER OF OPERATIONS
1.3 INTRODUCTORY EQUATIONS
1.4 INTRODUCTORY INEQUALITIES
1.5 PROPERTIES OF REAL NUMBERS
1.6 SOLVING EQUATIONS USING ONE STEP
1.7 SOLVING EQUATIONS USING TWO OR MORE STEPS
1.8 EQUATIONS WITH VARIABLES ON BOTH SIDES
1.9 WORD PROBLEMS
1.10 SOLVING EQUATIONS FOR A VARIABLE
1.11 RATIOS, RATES AND PROPORTIONS
1.12 PERCENT PROBLEMS
1.13 MULTI-STEP INEQUALITIES
1.14 ABSOLUTE VALUE EQUATIONS
1.15 REFERENCES
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1.1. Variables and Expressions www.ck12.org
1.1 Variables and Expressions
What You Will Learn
• Evaluate single-variable expressions with given values for the variable.• Evaluate multi-variable expressions with given values for the variables.
Evaluate Single-Variable Expressions with Given Values for the Variable
An algebraic expression is a mathematical phrase involving letters, numbers, and operation symbols.
A variable can be any letter, such as x, m, R, y, P, a, and others, that we use in an expression.
The variable represents possible values of a quantity.
Take a look at the following examples of variable expressions.
3x+10
10r
b3 +2
mx−3
Variable expressions are used to describe real-world situations when we don’t know a value or a quantity. Sometimesa value is dependent on a changing variable, for example, the total height of a stack of paper depends on the thicknessof each piece of paper.
When we work with a variable expression, we can say that we evaluate the expression. To evaluate means to“find the value of”. We find the value of an expression.
Why don’t we “solve” expressions?
To solve something in math means that it has to equal a value. An expression does not have an equal sign, thereforewe can’t solve it. Also, a variable in an expression can have several different values. Whereas in an equation, thevariable has one value that makes the equation true.
Example
Evaluate the expression 10k−44 if k = 12.
Remember that a number next to a letter means to multiply.
First, we substitute the 12 for the letter k.
10(12) – 44
Now we multiply 10 and 12, then we subtract 44.
120−4476
Our answer is 76.
Sometimes, you will also have an expression to evaluate that uses division.
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Example
Evaluate the expression x3 +2 if x is 24
First, we substitute the 24 for x.243 +2
Next, we divide twenty-four by three.
24÷3 = 8
Now we add 2.
8+2 = 10
Our answer is 10.
Practice Exercises
Now it is your turn. Evaluate the following single-variable expressions.
a. 4x−9 if x is 20b. 5y+6 if y is 9c. a
4 −8 if a is 36
Evaluate Multi-Variable Expressions with Given Values for the Variables
Algebraic expressions can have more than one variable. Look at the following examples of multi-variable expres-sions.
xy+4x
mx+b
25r+(x−7)
When we know the value of the variables, we can evaluate multi-variable expressions the same way we evaluatedsingle-variable expressions, by substituting the value for the variables in the expression and solving from left toright.
Example
Evaluate xy+ x if x = 2 and y = 4.
In this case we are only given one possible value for x and y. We know that x = 2 and y = 4, we can evaluate theexpression using the given values.
First, we can rewrite the expression by substituting the given values into the expression.
xy+ x
(2)(4)+2
We used the parentheses here to show multiplication. When two variables are next to each other it meansmultiplication. Here we used the parentheses because we needed to show multiplication between 2 and 4.
Now we can multiply first.
2×4 = 8
Next, we add two.
8+2 = 10
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1.1. Variables and Expressions www.ck12.org
Our answer is 10.
Let’s look at another example where a given value is a fraction.
Example 4
Evaluate mx+3m if x = 23 and m = 9.
First, we substitute the given values into the expression.
9(2
3
)+3(9)
Do you remember how to multiply a whole number and a fraction?
To multiply a whole number and a fraction you must first make the whole number a fraction by placing the numberover one.
9 becomes 91
Now you multiply numerator × numerator and denominator × denominator.91 ·
23
Next, we multiply the two fractions and simplify.183 = 6
Now we can substitute the 6 back into the expression.
6+3(9)
Next we multiply.
3(9) = 27
Finally, we add the numbers.
6+2733
Our answer is 33.
Practice Exercises
Evaluate the following expressions using the given values.
a. ab+7 when a is 9 and b is 8b. xy+ zx when x is 2, y is 5 and z is 7c. xy+ x when x is 4, and y is 1
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Video Overview
MEDIAClick image to the left for more content.
Homework
Directions: Evaluate each expression if the given value of r = 9.
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www.ck12.org Chapter 1. Unit 1 - Review
1. r3
2. 63− r
3. 11r
4. 2r+7
5. 3r+ r
6. 4r−2r
7. r+5r
8. 12r−1
Directions: Evaluate each expression when h =−12.
9. 70−3h
10. 6h+6
11. 4h−9
12. 11+ h4
13. 3h+h
14. 2h+5h
15. 6h−2h
Directions: Evaluate each multi-variable expression.
16. Evaluate xy+9y if x = 3 and y = 13
17. Evaluate 2x+3y if x = 25 and y = 1
4
18. Evaluate 4y+3y−2z if y = 2 and z =−4
19. Evaluate 6z−2(z+ x) if x is 3 and z is 4
20. Evaluate 8a+3b−2c if a is 5, b is 4 and c is 3
Here are the vocabulary words that are found in this lesson.
Algebraic Expressionan expression that contains numbers, variables and operations.
Variablea letter used to represent an unknown quantity.
Variable Expressionan algebraic expression that contains one or more variables.
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1.2. Order of Operations www.ck12.org
1.2 Order of Operations
What You Will Learn
• The correct order of operations and how to use them• How to use a calculator to evaluate multistep expressions
Math operations have to be done in a specific order so that the answer to an expression always comes out the same. Asaying that can help you remember the Order of Operations is PEMDAS - Please Excuse My Dear Aunt Sally.
The Order of Operations:
Whatever is found inside PARENTHESES must be done first. EXPONENTS are to be simplified next. MULTI-PLICATION and DIVISION are equally important and must be performed moving left to right. ADDITION andSUBTRACTION are also equally important and must be performed moving left to right.
Example 1: Use the Order of Operations to simplify (7−2)×4÷2−3
Solution: First, we check for parentheses. Yes, there they are and must be done first.
(7−2)×4÷2−3 = (5)×4÷2−3
Next we look for exponents (little numbers written a little above the others). No, there are no exponents so we skipto the next math verb.
Multiplication and division are equally important and must be done from left to right.
5×4÷2−3 = 20÷2−3
20÷2−3 = 10−3
Finally, addition and subtraction are equally important and must be done from left to right.
10−3 = 7 This is our answer.
Example 2: Use the Order of Operations to simplify the following expressions.
a) 3×5−7÷2
b) 3× (5−7)÷2
c) (3×5)− (7÷2)
Solutions:
a) There are no parentheses and no exponents. Go directly to multiplication and division from left to right: 3×5−7÷2 = 15−7÷2 = 15−3.5
Now subtract: 15−3.5 = 11.5
b) Parentheses must be done first: 3× (−2)÷2
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There are no exponents, so multiplication and division come next and are done left to right: 3×(−2)÷2 =−6÷2 =−3
c) Parentheses must be done first: (3×5)− (7÷2) = 15−3.5
There are no exponents, multiplication, division, or addition, so simplify:
15−3.5 = 11.5
Parentheses are used two ways. The first is to alter the Order of Operations in a given expression, such as example(b). The second way is to clarify an expression, making it easier to understand.
Some expressions contain no parentheses while others contain several sets of parentheses. Some expressions evenhave parentheses inside parentheses! When faced with nested parentheses, start at the innermost parentheses andwork outward.
Example 3: Use the Order of Operations to simplify 8− [19− (2+5)−7]
Solution: Begin with the innermost parentheses:
8− [19− (2+5)−7] = 8− [19−7−7]
Simplify according to the Order of Operations:
8− [19−7−7] = 8− [5] = 3
Evaluating Algebraic Expressions with Fraction Bars
Fraction bars count as grouping symbols for PEMDAS, and should be treated as a set of parentheses. All numeratorsand all denominators can be treated as if they have invisible parentheses. When real parentheses are also present,remember that the innermost grouping symbols should be evaluated first. If, for example, parentheses appear on anumerator, they would take precedence over the fraction bar. If the parentheses appear outside of the fraction, thenthe fraction bar takes precedence.
Example 4: Use the Order of Operations to simplify the following expressions.
a) z+34 −1 when z = 2
b)(a+2
b+4 −1)+b when a = 3 and b = 1
c) 2×(
w+(x−2z)(y+2)2 −1
)when w = 11, x = 3, y = 1 and z =−2
Solutions: Begin each expression by substituting the appropriate value for the variable:
a) (2+3)4 −1 = 5
4 −1. Rewriting 1 as a fraction, the expression becomes:
54− 4
4=
14
b) (3+2)(1+4) =
55 = 1
(1−1)+b Substituting 1 for b, the expression becomes 0+1 = 1
c) 2([11+(3−2(−2))]
[(1+2)2)]−1)= 2
((11+7)
32 −1)= 2
(189 −1
)Continue simplifying: 2
(189 −
99
)= 2
(99
)= 2(1) = 2
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1.2. Order of Operations www.ck12.org
Using a Calculator to Evaluate Algebraic Expressions
A calculator, especially a graphing calculator, is a very useful tool in evaluating algebraic expressions. The graphingcalculator follows the Order of Operations, PEMDAS. In this section, we will explain two ways of evaluatingexpressions with the graphing calculator.
Method #1: This method is the direct input method. After substituting all values for the variables, you type in theexpression, symbol for symbol, into your calculator.
Evaluate [3(x2−1)2− x4 +12]+5x3−1 when x =−3.
Substitute the value x =−3 into the expression.
[3((−3)2−1)2− (−3)4 +12]+5(−3)3−1
The potential error here is that you may forget a sign or a set of parentheses, especially if the expression is long orcomplicated. Make sure you check your input before writing your answer. An alternative is to type the expressionin by appropriate chunks – do one set of parentheses, then another, and so on.
Method #2: This method uses the STORE function of the Texas Instrument graphing calculators, such as the TI-83,TI-84, or TI-84 Plus.
First, store the value x = −3 in the calculator. Type -3 [STO] x. ( The letter x can be entered using the x- [VAR]button or [ALPHA] + [STO]). Then type in the expression in the calculator and press [ENTER].
The answer is −13.
Note: On graphing calculators there is a difference between the minus sign and the negative sign. When we storedthe value negative three, we needed to use the negative sign, which is to the left of the [ENTER] button on thecalculator. On the other hand, to perform the subtraction operation in the expression we used the minus sign. Theminus sign is right above the plus sign on the right.
You can also use a graphing calculator to evaluate expressions with more than one variable.
Evaluate the expression: 3x2−4y2+x4
(x+y)12
for x =−2,y = 1.
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Store the values of x and y. −2 [STO] x, 1 [STO] y. The letters x and y can be entered using [ALPHA] + [KEY].Input the expression in the calculator. When an expression shows the division of two expressions be sure to useparentheses: (numerator) ÷ (denominator). Press [ENTER] to obtain the answer −.88̄ or −8
9 .
Video Overview
MEDIAClick image to the left for more content.
Homework
Use the Order of Operations to simplify the following expressions.
1. 8− (19− (2+5)−7)2. 2+7×11−12÷33. (3+7)÷ (7−12)4. 2·(3+(2−1))
4−(6+2) − (3−5)5. 8 ·5+62
6. 9÷3×7−23 +77. 8+12÷6+68. (72−32)÷8
Evaluate the following expressions involving variables.
9. jkj+k when j = 6 and k = 12.
10. 2y2 when x = 1 and y = 511. 3x2 +2x+1 when x = 512. (y2− x)2 when x = 2 and y = 1
Evaluate the following expressions involving variables.
13. 4x9x2−3x+1 when x = 2
14. z2
x+y +x2
x−y when x = 1, y =−2, and z = 4.
15. 4xyzy2−x2 when x = 3, y = 2, and z = 5
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1.2. Order of Operations www.ck12.org
16. x2−z2
xz−2x(z−x) when x =−1 and z = 3
The formula to find the volume of a square pyramid is V = s2(h)3 . Evaluate the volume for the given values.
17. s = 4 inches,h = 18 inches18. s = 10 f eet,h = 50 f eet19. h = 7 meters,s = 12 meters20. h = 27 f eet,s = 13 f eet21. s = 16 cm,h = 90 cm
In 22 – 25, insert parentheses in each expression to make a true equation.
22. 5−2 ·6−4+2 = 523. 12÷4+10−3 ·3+7 = 1124. 22−32−5 ·3−6 = 3025. 12−8−4 ·5 =−8
In 26 – 29, evaluate each expression using a graphing calculator.
26. x2 +2x− xy when x = 250 and y =−12027. (xy− y4)2 when x = 0.02 and y =−0.02528. x+y−z
xy+yz+xz when x = 12 , y = 3
2 , and z =−1
29. (x+y)2
4x2−y2 when x = 3 and y =−5d
30. The formula to find the volume of a spherical object (like a ball) is V = 43(π)r
3, where r = the radius of thesphere. Determine the volume for a grapefruit with a radius of 9 cm.
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1.3 Introductory Equations
What You Will Learn
• Solve single-variable addition and subtraction equations using mental math.• Solve single-variable multiplication and division equations using mental math.
Solve Single-Variable Addition and Subtraction Equations Using Mental Math
We have been working with algebraic expressions. Remember that algebraic expressions combine numbers, vari-ables and operations together. When given the value of a variable, we can evaluate any expression.
Take the following example: 10r+11.
If r = 22, we substitute the value of the variable into the expression and evaluate.
Be careful to follow the order of operations.
10r+11 = 10(22)+11
220+11
231
The answer is 231.
An equation is a mathematical statement that two expressions are equal. The key thing to notice in an equation isthat there is an equals sign.
For example, we can say that
15+7 = 24−2
Since both sides of the equation equal 22, these equations are equivalent.
15+7 = 22 and 24−2 = 22
An equation can also include an algebraic unknown, or a variable. If you think about this it makes perfect sense. Wehave an unknown in an equation so we use a variable to represent the unknown quantity.
Take a look at some of these equations with variables in them:
15t = 45
12 = x+9
25 = 3(x−7)+1
When we have an equation with a variable in it, we can solve the equation to figure out the value of thevariable.
Remember, an equation states that two expressions are equal.
When we solve an equation with a variable, we are finding the value of the variable that makes the equation true.
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1.3. Introductory Equations www.ck12.org
Example
y+10 = 15
The equal sign tells us that y+10 and 15 have the same value.
Therefore, the value of y must be a number that, when added to 10, equals 15.
What could be the value of y?
We can ask ourselves, “What number plus ten is equal to fifteen?”
You can use mental math to determine that y = 5 because 5+10 = 15.
One advantage of working with equations is that you can always check your work.
Think of it like a balance.
This scale is not balanced. When we solve an equation, we want the scale to balance. One side will be equalto the other side.
When you think you know the value of a variable, substitute it into the equation.
If your value for the variable is the correct one, both sides will be equal.
The two expressions will balance!
The first example had addition in it, what about subtraction?
You can work on a subtraction equation in the same way. Let’s look at an example.
Example
x−5 = 10
This is an equation once again. We want to figure out what number, minus five, is equal to ten. That way both sidesof the equations will be equal and be balanced. We ask ourselves the question, “What number take away five is equalto 10?”
We can use mental math to figure out that x is equal to 15. When working with small numbers, mental math isthe quickest way to figuring out the value of the variable.
Now let’s check our answer. To do this we substitute our answer for x back into the equation and see if both sidesare equal.
15−5 = 10
10 = 10
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You can see that if we pick the correct quantity for x that the two halves of the scale will be balanced. If wepick the wrong quantity, one that is too big or too small, then the scales will tip.
Solving equations is often called BALANCING EQUATIONS for this very reason!!
Practice Exercies
Practice using mental math to solve each equation.
a. x+5 = 12b. x−3 = 18c. 6− y = 4
Solve Single-Variable Multiplication and Division Equations Using Mental Math
We have just demonstrated how to solve single-variable addition and subtraction equations. To solve variableequations using multiplication and division we follow the same procedure. First, we examine the problem anduse mental math.
Using mental math requires you to remember your multiplication tables. Remember that division is theopposite of multiplication, so when using small numbers mental math and your times tables will be your beststrategy to balancing equations.
Let’s look at an example.
Example
9p = 72
Here is a multiplication problem. Whenever you see a variable next to a number it means multiplication.Here we need to figure out, “What number times 9 is equal to 72?”
Using mental math and division, you can see that p is equal to 8.
Now let’s check our answer. We do this by substituting the value of the variable, 8, back into the original problem.
9(8) = 72
72 = 72
One side of the equation is equal to the other side, so our equation balances! Our work is complete andaccurate.
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1.3. Introductory Equations www.ck12.org
We can apply mental math when solving division problems too. Remember that a fraction bar means division,so when you see one you know that you are going to be dividing.
Examplex3 = 4
Remember that when the variable is on top of the fraction bar that we are dividing the bottom number intothis number. So we ask ourselves, “What number divided by three is equal to four?”
We use mental math, and our knowledge of the times table, to figure this out. Our missing value is 12.
Next, we can check our work. We substitute 12 back into the equation for the variable x.
123
= 4
4 = 4
One side of the equation is equal to the other side, so our equation balances! Our work is complete andaccurate.
Practice Exercises
Practice solving these equations.
a. 5y = 20b. 6g = 42c. x
7 = 49
Homework
Directions: Use mental math to solve each addition or subtraction equation.
1. x+2 = 7
2. y+5 = 10
3. a+7 = 20
4. b+8 = 13
5. z+10 = 32
6. s+14 = 26
7. 4+ y = 8
8. 6+ x = 21
9. 17+a = 23
10. 18+b = 30
11. 12+ x = 24
12. 13+ y = 18
13. 15+a = 22
14. x+17 = 24
15. y+3 = 45
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16. x−4 = 10
17. y−8 = 20
18. 5− y = 2
19. 22−a = 15
20. 18− y = 2
Directions: Use mental math to solve each multiplication or division equation.
21. 5x = 25
22. 6x = 48
23. 2y = 18
24. 3y = 21
25. 4a = 16
26. 13b = 26
27. 15a = 30
28. 15x = 45
29. x2 = 3
30. x4 = 5
31. x3 = 11
32. x5 = 12
33. x7 = 8
34. x8 = 9
35. x3 = 12
Here are the vocabulary words that are part of this lesson.
Algebraic Expressionan expression that contains a combination of numbers, variables and operations. It does not have an equalssign.
Equationa number sentence with two expressions divided by an equal sign. One quantity on one side of the equationequals the quantity on the other side of the equation.
Variable Equationan equation where a variable is used to represent an unknown quantity.
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1.4. Introductory Inequalities www.ck12.org
1.4 Introductory Inequalities
What You Will Learn
• The meaning of inequality symbols• How to graph inequalities on a number line• How to express an inequality using interval notation
The Meaning of Inequality Symbols
We are going to begin working with quantities that may or may not be equal. These are called inequalities.
Just like an equation is shown using an equal sign, an inequality is expressed through symbols too.
> means “is greater than.”
< means “is less than.”
≥ means “is greater than or equal to.”
≤ means “is less than or equal to.”
Let’s begin by taking a look at some inequalities.
2 > x
Here we have an inequality where we know that two is greater than the quantity of x. This problem has manydifferent solutions. Any number that would make this a true statement can be substituted in for the variable. Weknow that we can have negative numbers here too. We can write our answer as a set of numbers that begins with oneand goes down from there.
The answer is {1, 0 , -1...} Remember that any fraction or decimal that is smaller than two could also work.
Here is another one.
5≤ y
Here we have an inequality that involves less than or equal to. We can substitute any value in for the variable thatmakes this a true statement. This means that we can start with five and go greater.
The answer is {5, 6, 7...}
Graphing Inequalities on a Number Line
We graph inequalities on a number line to visually show the set of numbers that would make the statement a truestatement. Given that we can have less than, greater than, or less than or equal to, and greater than or equal to, wecan have four different types of graphing. Here are some hints to help you with graphing inequalities.
• Use an open circle to show that a value is not a solution for the inequality. You will use open circles to graphinequalities that include the symbols > or <.
• Use a closed circle to show that a value is a solution for the inequality. You will use closed circles to graphinequalities that include the symbols ≥ or ≤.
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Write an inequality to represent all possible values of n if n is less than 2. Then graph those solutions on a numberline.
First, translate the description above into an inequality. You can do this in the same way that you wroteequations. Just pay close attention to the words being used.
n, is less than 2.
↓ ↓ ↓n < 2
Now, graph the inequality.
Draw a number line from -5 to 5.
You know that n represents all numbers less than 2, and can be represented as n < 2. So, the solutions forthis inequality include all numbers less than 2. The number 2 is not a solution for this inequality, so draw anopen circle at 2. Then draw an arrow showing all numbers less than 2. The arrow should face left because thelesser numbers are to the left on a number line.
Write an inequality for each example.
Example A
The quantities less than or equal to 4.
Solution: x≤ 4
Example B
A number is greater than or equal to -12.
Solution: a≥−12
Example C
Two times a number is less than 7.
Solution: 2x < 7
Practice Exercise
Here is one for you to try on your own.
Write an inequality to represent all possible values of n if n is greater than or equal to -4. Then graph those solutionson a number line.
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1.4. Introductory Inequalities www.ck12.org
Solution
First, translate the description above into an inequality.
n is greater than or equal to −4.
↓ ↓ ↓n ≥ −4
Now, graph the inequality.
Draw a number line from -5 to 5.
You know that n represents all numbers greater than or equal to -4, and can be represented as n≥−4. So, thesolutions for this inequality include all numbers greater than or equal to -4. The number -4 is a solution forthis inequality, so draw a closed circle at -4. Then draw an arrow showing all numbers greater than -4. Thearrow should face right because the greater numbers are to the right on a number line.
Express an Inequality Using Interval Notation
Interval notation is another commonly used way to express an inequality. The graphs we have been drawing canbe thought of as an interval on the number line. The inequality x > 5 means x can be any number in the intervalbetween 5 and infinity. This is written in interval notation as (5,∞). If x≥ 5, the interval notation would be [5,∞).If the graph ends in an open circle, use a parenthesis, if it ends in a closed circle, use a square bracket.
Video Overview
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Homework
Directions: List at least three possible values for the variable in each inequality
1. x < 132. y > 53. x < 24. y >−3
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5. a > 126. x≤ 47. y≥ 38. b≥−39. a≤−5
10. b≥ 11
Directions: Write an inequality to describe each situation, graph it on a number line and write it in interval notation.
11. A number is less than or equal to -8.12. A number is greater than 50.13. A number is less than -4.14. A number is greater than -12.15. A number is greater than or equal to 11.
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1.5. Properties of Real Numbers www.ck12.org
1.5 Properties of Real Numbers
What You Will Learn
• Algebraic Properties of Addition• Algebraic Properties of Multiplication
Algebraic Properties of Addition
The Commutative Property of Addition: For all real numbers a and b, a+b = b+a.
To commute means to change locations, so the Commutative Property of Addition allows you to rearrange the objectsin an addition problem.
The Associative Property of Addition: For all real numbers a, b, and c, (a+b)+ c = a+(b+ c).
To associate means to group together, so the Associative Property of Addition allows you to regroup the objects inan addition problem.
The Identity Property of Addition: For any real number a, a+0 = a.
This property allows you to use the fact that the sum of any number and zero is the original value.
These properties apply to all real numbers, but in this lesson we are applying them to integers, which are just aspecial kind of real number.
Guided Practice
Simplify the following using the properties of addition:
a) 9+(1+22)
b) 4,211+0
Solution:
a) It is easier to regroup 9+1, so by applying the Associative Property of Addition, (9+1)+22 = 10+22 = 32.
b) The Additive Identity Property states the sum of a number and zero is itself; therefore, 4,211+0 = 4,211.
Algebraic Properties of Multiplication
The four mathematical properties which involve multiplication are the Commutative, Associative, MultiplicativeIdentity and Distributive Properties.
• Commutative property When two numbers are multiplied together, the product is the same regardless of theorder in which they are written:
Example 4 ·2 = 2 ·4
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We can see a geometrical interpretation of The Commutative Property of Multiplication to the right. The Areaof the shape (length × width) is the same no matter which way we draw it.
• Associative Property When three or more numbers are multiplied, the product is the same regardless of theirgrouping
Example (2 ·3) ·4 = 2 · (3 ·4)
• Multiplicative Identity Property: The product of one and any number is that number.
Example 5 ·1 = 5.
• Distributive property The multiplication of a number and the sum of two numbers is equal to the first numbertimes the second number plus the first number times the third number.
Example: 4(6+3) = 4 ·6+4 ·3
Video Overview
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Homework
1. Complete each statement using the commutative property
a) x + 3 = _____ + x
b) 4(2) = ____(4)
2. Complete each statement using the associative property
a) (3•4)(-2) = ____ [4(-2)]
b) [x + (-4)] + (-3) = ___ + [(-4) + (-3)]
3. Complete each statement using an identity property
a) 6 + 0 = ____
b) 1(5) = ____
4. Complete each statement using an inverse property
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1.5. Properties of Real Numbers www.ck12.org
a) -8 + ___ = 0
b) 23 • ____ = 1
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1.6 Solving Equations Using One Step
What You Will Learn
• Solve an equation using addition.• Solve an equation using subtraction.• Solve an equation using multiplication.• Solve an equation using division.
Introduction
Nadia is buying a new bicycle helmet. Peter watches her pay for the helmet with a $100 bill. She receives $22.00 inchange, and from only this information, Peter works out how much the helmet costs. How much was the helmet?
In math, we can solve problems like this using an equation. An equation is an algebraic expression that involves anequals sign. If we use the letter x to represent the cost of the bicycle helmet we could write the following equation.
x+22 = 100
This tells us that the value of the helmet plus the value of the change received is equal to the $100 that Nadia paid.
Peter saw the transaction from a different viewpoint. He saw Nadia receive the helmet, give the salesperson $100then he saw Nadia receive $22 change. Another way we could write the equation would be:
x = 100−22
This tells us that the value of the player is equal to the amount of money Nadia paid (100 - 22).
Mathematically, these two equations are equivalent, though it is easier to determine the cost of the bicycle helmetfrom the second equation. In this chapter, we will learn how to solve for the variable in a one variable linearequation. Linear equations are equations in which each term is either a constant or the product of a constant and asingle variable (to the first power). The term linear comes from the word line. You will see in later chapters thatlinear equations define lines when graphed.
We will start with simple problems such as the one in the last example.
Solve an Equation Using Addition
When we write an algebraic equation, equality means that whatever we do to one side of the equation, we have todo to the other side. For example, to get from the second equation in the introduction back to the first equation, wewould add a quantity of 22 to both sides:
x = 100−22
x+22 = 100−22+22 or x+22 = 100
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1.6. Solving Equations Using One Step www.ck12.org
Similarly, we can add numbers to each side of an equation to help solve for our unknown.
Example 1
Solve x−3 = 9
Solution
We need to isolate x. Change our equation so that x appears by itself on one side of the equals sign. Right now ourx has a 3 subtracted from it. To reverse this, we could add 3, but we must do this to both sides.
x−3 = 9
x−3+3 = 9+3 The +3 and −3 on the left cancel each other. We evaluate 9+3
x = 12
Example 2
Solve x−3 = 11
Solution
To isolate x we need to add 3 to both sides of the equation. This time we will add vertically.
x��−3 = 11
��+3 =+3
x = 14
Notice how this format works. One term will always cancel (in this case the three), but we need to remember tocarry the x down and evaluate the sum on the other side of the equals sign.
Example 3
Solve z−5 =−7
Solution
This time our variable is z, but don’t let that worry you. Treat this variable like any other variable.
z−5 =−7
+5 =+5
z =−2
Solve an Equation Using Subtraction
When our variable appears with a number added to it, we follow the same process, only this time to isolate thevariable we subtract a number from both sides of the equation.
Example 4
Solve x+6 = 26
Solution
To isolate x we need to subtract six from both sides.
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x��+6 = 26
��−6 =−6
x = 20
Example 5
Solve x+20 =−11
Solution
To isolate x we need to subtract 20 from both sides of the equation.
x+20 =−11
−20 =−20
x =−31
Solve an Equation Using Multiplication
Suppose you are selling pizza for $1.50 a slice and you get eight slices out of a single pizza. How much do you getfor a single pizza? It shouldn’t take you long to figure out that you get 8×$1.50= $12.00. You solve this problem bymultiplying. The following examples do the same algebraically, using the unknown variable x as the cost in dollarsof the whole pizza.
Example 6
Solve 18 · x = 1.5
Our x is being multiplied by one-eighth. We need to cancel this factor, so we multiply by the reciprocal 8. Do notforget to multiply both sides of the equation.
�8(
1�8· x)= 8(1.5)
x = 12
In general, when x is multiplied by a fraction, we multiply by the reciprocal of that fraction.
Example 7
Solve 9x5 = 5
9x5 is equivalent to 9
5 · x so x is being multiplied by 95 . To cancel, multiply by the reciprocal, 5
9 .
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1.6. Solving Equations Using One Step www.ck12.org
59
(9x5
)=
59·5
x =259
Solve an Equation Using Division
Solving by division is another way to cancel any terms that are being multiplied x. Suppose you buy five identicalcandy bars, and you are charged $3.25. How much did each candy bar cost? You might just divide $3.25 by 5. Oryou may convert to cents and divide 325 by 5. Let’s see how this problem looks in algebra.
Example 8
Solve 5x = 325 To cancel the 5 we divide both sides by 5.
�5x�5
=3255
x = 65
Video Overview of Adding/Subtracting to solve equations
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Video Overview of Multiplying/Dividing to solve equations
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Lesson Summary
• An equation in which each term is either a constant or a product of a constant and a single variable is a linearequation.
• Adding, subtracting, multiplying, or dividing both sides of an equation by the same value results in anequivalent equation.
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• To solve an equation, isolate the unknown variable on one side of the equation by applying one or morearithmetic operations to both sides.
Homework
1. Solve the following equations for x.
a. x+11 = 7b. x−1.1 = 3.2c. 7x = 21d. 4x = 1e. 5x
12 = 23
f. x+ 52 = 2
3g. x− 5
6 = 38
2. Solve the following equations for the unknown variable.
a. q−13 =−13b. z+13 = 9c. 21s = 3d. t + 1
2 = 13
e. 7 f11 = 7
11f. 3
4 =−12 · y
g. −6r =−42
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1.7. Solving Equations using Two or More Steps www.ck12.org
1.7 Solving Equations using Two or MoreSteps
What You Will Learn
• Solve a two-step equation using addition, subtraction, multiplication, and division.• Solve a multi-step equation by using the distributive property and combining like terms.• Solve a multi-step equation with fractions
Solve a Two-Step Equation
We have seen that in order to solve for an unknown variable we can "undo" the operation on the variable to get it byitself. In this chapter we will expand our ability to do that, with problems that require us to combine more than onetechnique in order to solve for our unknown.
First, we will look at equations that have require two steps to isolate the variable. An example is
3x−5 = 10
We are looking for a number that will give you an answer of 10 if you multiply it by 3 and subtract 5. It is possibleto figure that out in your head, but algebra can be very helpful as well, especially when the equations get longer andfull of fractions. Here’s how you use algebra to solve it
3x−5 = 10
+5 =+5 Add 5 to both sides.
3x = 153x3
=153
Divide both sides by 3.
x = 5
We had to do two operations in order to get x by itself. First, we used adding to "undo" the subtracted 5. Then weused dividing to "undo" the multiplied 3. Using the opposite operation to undo an equation is what algebra is allabout.
The same logic holds for all of the algebra operations, you just have to keep doing the opposite until you are leftwith x by itself.
28 = 4x+3
−3 = −3 Subtract 3 from both sides.
25 = 4x254
=4x4
Divide both sides by 4
x =254
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Combining Like Terms and the Distributive Property
If it is possible to combine like terms on one or both sides of the equation, you must do that before you start undoingany of the operations.
3x+2x+4 = 14 Combine like terms.
5x+4 = 14
−4 =−4 Subtract 4 from both sides.
5x = 105x5
=105
Divide both sides by 5.
x = 2
If the x is inside a set of parenthesis, you need to use the distributive property to get rid of them.
6(x+4) = 12 Distribute the 6.
6x+24 = 12
−24 =−24 Subtract 24 from both sides6x6
=−12
6Divide both sides by 6
x =−2
This last example will combine all the ideas from this section.
4(x+2)−3(4x−7) = 21 Use the Distributive Property
4x+8−12x+21 = 21 Combine like terms
−8x+29 = 21 Subtract 29 from each side
−29 =−29−8x−8
=−8−8
Divide both sides by -8
x = 1
Solve a Multi-Step Equation with Fractions
If an equation has a large number of fractions, it is often easier to multiply both sides of the equation by the leastcommon denominator of the fractions. This will cancel out all of the denominators and there will be no fractionsremaining.
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1.7. Solving Equations using Two or More Steps www.ck12.org
35
x+710
=−125
LCD is 10
10(35
x+7
10) = (−12
5)10 Multiply each side by 10
6x+7 =−24 Subtract 7 from each side
−7 =−76x6
=−31
6Divide both sides by 6
x =−31
6
Lesson Summary
• Some equations require more than one operation to solve.• Terms with the same variable in them (or no variable in them) are like terms. Combine like terms (adding
or subtracting them from each other) is necessary to simplify the expression and solve for the unknown. Youmay need to use the Distributive Property before you combine like terms.
Video Overview of Solving Two Step Equations
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Video Overview of Solving Equations with the Distributive Property and Combining Like Terms
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Homework
Solve the following equations for the unknown variable.
1. 4x−3 = 92. 15 =−5x+43. 5x− (3x+2) = 14. 4(x+3) = 1
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5. 45 = 9−4x6. 6x+12−3x = 77. 12+ 2
3 x = 88. 7(x−3)−4(x+1) = 159. 5
6 x− 23 = 3
410. 3− 4
7 x = 25
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1.8. Equations with Variables on Both Sides www.ck12.org
1.8 Equations with Variables on Both Sides
What You Will Learn
• Solve an equation with variables on both sides• Solve an equation with no solution or infinite solutions
Solve an Equation with Variables on Both Sides
When the variable appears on both sides of the equation, we need to manipulate our equation so that all variablesappear on one side, and only numbers remain on the other.
Example 1
Solve 3x+4 = 5x
Solution
This equation has x on both sides. However, there is only a number term on the left. We will therefore move all thex terms to the right of the equal sign leaving the number on the left.
��3x+4 = 5x Subtract 3x from both sides.
−��3x −3x
4 = 2x Divide by 242=
2x2
Solution
x = 2
Example 2
Solve 9x = 4−5x
This time we will collect like terms (x terms) on the left of the equal sign.
9x = 4���−5x
+5x ���+5x Add 5x to both sides.
14x = 4
14x = 4 Divide by 14.14x14
=414
x =27
Solution
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x = 27
If there are like terms to combine, do that before you collect the variables on one side of the equal sign
Example 3
3x+5−6x = 13x+4 Combine like terms
−3x+5 = 13x+4 Add 3x to each side
+3x = +3x
5 = 16x+4 Subtract 4 from each side
−4 = −41
16=
16x16
Divide each side by 16
116
= x
If there are parenthesis in the equation, do the distributive property first.
Example 4
−4(2x+3)+7 = 4x Use the Distributive Property−8x−12+7 = 4x Combine like terms−8x−5 = 4x Add 8x to each side+8x =+8x
−512
=12x12
Divide both sides by 12
−512
= x
Solve an Equation with No Solutions or Infinite Solutions
Now that we have variables on both sides, it is possible to set up an equation where it isn’t possible to get a numberfor x. This example will show how this can happen.
4x+5(x+1) = 3x−3(−2x+7) Use the Distributive Property
4x+5x+5 = 3x+6x−21 Combine like terms
9x+5 = 9x−21 Subtract 9x from each side
−9x =−9x
5 =−21
No matter what number is substituted in for x, the final result will be two numbers that don’t equal each other. Thismeans that the equation has no solution.
It is also possible to set up an equation where any number can work.
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1.8. Equations with Variables on Both Sides www.ck12.org
2(5x+3)−8x = 3(3x+2)−7x Use the Distributive Property
10x+6−8x = 9x+6−7x Combine like terms
2x+6 = 2x+6 Subtract 2x from each side
−2x =−2x
6 = 6
In this case, any value for x will give us two numbers that equal each other. This means the solution to the equationis all real numbers. There are an infinite number of numbers, so another way to say this is the equation has infinitesolutions.
Video Overview
MEDIAClick image to the left for more content.
Lesson Summary
• If an unknown variable appears on both sides of an equation, distribute and combine like terms as necessary.Then subtract (or add) one term to both sides to simplify the equation to have the unknown on only one side.
Homework
Solve the following equations for the unknown variable.
1. 4x+3 = 2x+92. −5x+2x+7 = 10x−133. 3(x−1) = 2(x+3)4. 8− (x+6) = 3x+65. 7(x+20) = x+56. 9(x−2) = 3x+37. 5x− (x−3) = 2x+7−9x8. 3
5 x− 110 x = 1
2 x+19. x−5 =−(5− x)
10. 9x+3x−10 = 3(3x+1)
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1.9 Word Problems
What You Will Learn
• Solve a word problem by defining variables, setting up an equation and solving
Word Problems
Word problems are challenging. It can be difficult to figure out how to read the problem and turn it into an equation.This book will outline several different types of problems and how to set them up.
The first problems we will do involve word phrases that are used to describe math operations.
Example 1
Write "A number increased by seven" as an algebraic expression.
Solution: x+4
Example 2
Write "The product of a number and three times the same number is 27" as an algebraic equation.
Solution:
x ·3x = 27
Now we will try some similar problems where we use our equation solving skills to find the number
Example 3
Write "If four is added to three times a number, the result is seven" as an equation. Then solve to find thenumber
3x+4 = 7 Subtract 4 from both sides
−4 =−43x3
=33
Divide both sides by three
x = 1
Solution: x=1, so the number being discussed is 1.
As the problems get more difficult, you will need a strategy. Try doing the following five steps when you set up aword problem
1. Read the problem
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1.9. Word Problems www.ck12.org
2. Highlight the important information3. Define a variable4. Write an equation5. Solve and check
Try that strategy on the next problem
Example 4
A piece of lumber 70 inches long is divided into 3 pieces. The longest piece must be three times the middlesize piece and the shortest piece is 10 inches shorter than the middle sized piece. Find the length of eachpiece.
First let’s think about the important information. The lumber is divided into three pieces. We don’t knowhow long each piece is, but we do know how they both compare to the middle piece. That means we shoulddefine a variable "m" for the length of the middle piece and we will write an equation using that.
The longest piece is three times the size of the middle piece, so we will call the longest piece 3m. The shortestpiece is 10 inches shorter than the middle piece, so we will call it m-10. All of the pieces add up to 70, sothis is our final equation
3m+m+m−10 = 70
Now solve this equation to find m
Solution: This means the middle sized piece is 16 inches. The longest piece is three times that, or 48 inches.The shortest piece is ten inches shorter than that, or 6 inches
Video Overview
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Homework
Read the following problems, define a variable, write an equation and solve using algebra.
1. The product of two and a number is ten.2. The sum of five times a number and ten is equal to the product of 15 and the number3. The quotient of a number and 5 is equal to the sum of six and one.4. The perimeter of a rectange is 120 inches. The length is twice the width. Find the length and width of the
rectangle.5. The Fast Track company makes toy remote-control cars, which it sells for $18 each. The production cost for
the company is $2000 per day plus $13 per race car. How many race cars must the company sell to breakeven?
6. Lauren is two years younger than Emily. If Emily is 12, how old is Lauren?7. Kathleen is twice as old as Bob. The sum of their ages is 48. Find their ages
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www.ck12.org Chapter 1. Unit 1 - Review
8. Bill received 100 more votes than Martha in a recent election. The total number of votes was 458. How manyvotes did each receive?
9. The sum of two consecutive even integers is 186. What are the integers?10. The perimeter of a triangle is 22 inches. The first side is 3 times longer than the second side. The third side
is 4 more than twice the second side. How long is each side of the triangle?
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1.10. Solving Equations for a Variable www.ck12.org
1.10 Solving Equations for a Variable
What Will You Learn
• Solve a formula for a given variable• Use that formula to evaluate a variable
Solve a Formula for a Given Variable
A formula is an equation where variables are used to describe how one thing depends on other things. For example,here is a common formula
A = LW
This formula says the area (A) of a rectangle is equal to the length (L) times the width (W). If the length of therectangle were increased, the area would increase. The area changes if the length changes, so the area depends onthe length.
It would also make sense to say that if the length increased, maybe it was because the area increased. In order tohave the length be the answer the formula gave us, we have to do some algebra to the formula.
AW
=LWW
Divide both sides by W
AW
=L��W��W
The W’s cancel out
AW
= L or L =AW
Now when you use the formula, the length is the answer you get, not the area.
Now use the formula to evaluate a variable
Example 1
The area of a rectangle is 40 sq ft and the width is 8 ft, what is the length
L =AW
L =408
L = 5 f t
Let’s look at a more complicated formula. The area of a triangle is given by
A = 12 bh
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www.ck12.org Chapter 1. Unit 1 - Review
where b is the base and h is the height. What if you wanted the base to be the variable that was by itself? Youwould do the same algebra you do to solve an equation, except some of the steps would be with variables instead ofnumbers.
21·A =
12
bh · 21
Multiply both sides by21
to clear the fraction
2A = bh
2Ah
=b�h�h
Divide both sides by h to get the b by itself
2Ah
= b or b =2Ah
Example 2
The area of a triangle is 18 square feet and the height is 4 feet, find the base
b =2Ah
b =2 ·18
4
b =364
b = 9 f t
Here’s one last example with even more steps. The perimeter of a rectangle is given by the formula
P = 2L+2W
In this formula, L stands for length and W stands for width. What if you knew the perimeter of the rectangle andthe length of the rectangle and you wanted to figure out the width? Once again, just do algebra steps to the formulauntil you get the width all by itself
P = 2L+2W
−2L =−2L Subtract 2L from each side
P−2L = 2W
P−2L2
=�2W�2
Now divide each side by 2
P−2L2
=W or W =P−2L
2
Now if someone gave you the perimeter and the length, you would be able to figure out the width.
Example 3
The perimeter of a rectangle is 24 ft and the length is 5 ft, find the width
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1.10. Solving Equations for a Variable www.ck12.org
W =P−2L
2
W =24−2 ·5
2
W =24−10
2
W =142
W = 7 f t
Homework
Solve the following formulas for the given variable
1. a. d = rt,solve for r
b. If d=10 and t=2, what is r?
2. a. P = a+b+ c, solve for b
b. If P=53, a=5 and c =16, what is b?
3. a. I = prt, solve for p
b. If I=26, r=3 and t=5, what is p?
4. a. V = 13 bh, solve for b
b. If V=24 and h=8, what is b?
5. a. C = 2πr, solve for r
b. If C=100, what is r? (π is a constant that is always equal to 3.14)
6. a. y = 2x−5, solve for x
b. If y =23, what is x?
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www.ck12.org Chapter 1. Unit 1 - Review
1.11 Ratios, Rates and Proportions
What You Will Learn
• Write and understand a ratio.• Write and understand a rate• Write and solve a proportion.• Solve proportions using cross products.
Ratio
A ratio shows the relative sizes of two or more numbers, it is best explained by looking at an example.
Example 1
The State Dining Room in the White House measures approximately 48 feet long by 36 feet wide. Compare the lengthof room to the width, and express your answer as a ratio.
Solution48 f eet36 f eet =
4836 = 4
3
Example 2
A tournament size shuffleboard table measures 30 inches wide by 14 feet long. Compare the length of the table to itswidth and express the answer as a ratio.
We could write the ratio immediately as:14 f eet
30 inches Notice that we cannot cancel the units.
Sometimes it is OK to leave the units in, but as we are comparing two lengths, it makes sense to convert all themeasurements to the same units.
Solution14 f eet
30 inches =14×12 inches
30 inches = 16830 = 28
5
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1.11. Ratios, Rates and Proportions www.ck12.org
Rates
A rate is similar to a ratio, except the numbers being compared have different units. Typical examples include speedin miles per hour or gas mileage in miles per gallon.
Example 3
A family car is being tested for fuel efficiency. It drives non-stop for 100 miles, and uses 3.2 gallons of gasoline.Write the ratio of distance traveled to fuel used as a rate.
100 miles3.2 gallons
To convert this to miles/gallon we need to divide both numerator and denominator by 3.2.
(1003.2
)miles(3.2
3.2
)gallons
=31.25 miles
1 gallon
Solution
The ratio of distance to fuel used is 31.25 miles1 gallon or 31.25 miles per gallon.
Write and Solve a Proportion
When two ratios are equal to each other, we call it a proportion.
1015
=69
This statement is a proportion. We know the statement is true because we can reduce both fractions to 23 .
Check this yourself to make sure!
We often use proportions in science and business. For example, when scaling up the size of something. We use themto solve for an unknown, so we will use algebra and label our unknown variable x. We assume that a certain ratioholds true whatever the size of the thing we are enlarging (or reducing). The next few examples demonstrate this.
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Example 4
A small fast food chain operates 60 stores and makes $1.2 million profit every year. How much profit would the chainmake if it operated 250 stores?
First, we need to write a ratio. This will be the ratio of profit to number of stores.
Ratio =$1,200,000
60 stores
We now need to determine our unknown, x which will be in dollars. It is the profit of 250 stores. Here is the ratiothat compares unknown dollars to 250 stores.
Ratio =$x
250 stores
We now write equal ratios and solve the resulting proportion.
$1,200,00060 stores
=$x
250 storesor
1,200,00060
=x
250
Note that we can drop the units – not because they are the same on the numerator and denominator, but because theyare the same on both sides of the equation.
1,200,00060
=x
250Simplify fractions.
20,000 =x
250Multiply both sides by 250.
5,000,000 = x
Solution
If the chain operated 250 stores the annual profit would be 5 million dollars.
Solve Proportions Using Cross Products
One neat way to simplify proportions is to cross multiply. Consider the following proportion.
164
=205
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1.11. Ratios, Rates and Proportions www.ck12.org
If we want to eliminate the fractions, we could multiply both sides by 4 and then multiply both sides by 5. In factwe could do both at once:
�4 ·5 ·16�4
= 4 · �5 ·20�5
5 ·16 = 4 ·20
Now comparing this to the proportion we started with, we see that the denominator from the left hand side ends upmultiplying with the numerator on the right hand side.
You can also see that the denominator from the right hand side ends up multiplying the numerator on the left handside.
In effect the two denominators have multiplied across the equal sign:
This movement of denominators is known as cross multiplying. It is extremely useful in solving proportions,especially when the unknown variable is on the denominator.
Example 5
Solve the proportion for x.43 = 9
x
Cross multiply:
x ·4 = 9 ·34x4
=274
Divide both sides by 4.
Solution
x = 6.75
Example 6
Solve the following proportion for x.
0.53
=56x
Cross multiply:
x ·0.5 = 56 ·30.5x0.5
=1680.5
Divide both sides by 0.5.
Solution:
x = 336
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Example 7
x+23
=x−1
4Cross Multiply
4(x+2) = 3(x−1)
4x+8 = 3x−3
−3x −3x
x+8 =−3
−8 =−8
x =−11
Video Overview of Ratios
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Video Overview of Rates
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Video Overview of Proportions and Cross Multiplication
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Lesson Summary
• A ratio is a way to compare two numbers, measurements or quantities by dividing one number by the otherand expressing the answer as a fraction. 2
3 ,32 miles
1.4 gallons , and x13 are all ratios.
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1.11. Ratios, Rates and Proportions www.ck12.org
• A rate compares two numbers with different units, for example miles per hour• A proportion is formed when two ratios are set equal to each other.• Cross multiplication is useful for solving equations in the form of proportions. To cross multiply, multiply
the bottom of each ratio by the top of the other ratio and set them equal. For instance, cross multiplying
results in 11 ·3 = 5 · x.
Homework
1. Write the following comparisons as ratios. Simplify fractions where possible.
a. $150 to $3b. 150 boys to 175 girlsc. 200 minutes to 1 hourd. 10 days to 2 weeks
2. Write the following ratios as a unit rate.
a. 54 hotdogs to 12 minutesb. 5000 lbs to 250 in2
c. 20 computers to 80 studentsd. 180 students to 6 teacherse. 12 meters to 4 floorsf. 18 minutes to 15 appointments
3. Solve the following proportions.
a. 136 = 5
xb. 1.25
7 = 3.6x
c. 619 = x
11d. 1
x = 0.015
e. 3004 = x
99f.
3x9
=x−5
7
g. 42x+4 = 9
5h. 4x−10
5 = 3x+83
4. A restaurant serves 100 people per day and takes $908. If the restaurant were to serve 250 people per day,what might the taking be?
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1.12 Percent Problems
What You Will Learn
• Find a percent of a number.• Use the percent equation.
Percent
A percent is simply a ratio with a base unit of 100. When we write a ratio as a fraction, the base unit is thedenominator. Whatever percentage we want to represent is the number on the numerator. For example, the followingratios and percents are equivalent.
TABLE 1.1:
Fraction Percent Fraction Percent( 50100
)50%
( 501000
)=( 0.5
100
)0.5%( 10
100
)10%
( 125
)=( 4
100
)4%( 99
100
)99%
(35
)=( 60
100
)60%(125
100
)12.5%
(1
10,000
)=(0.01
100
)0.01%
Fractions are easily converted to decimals, just as fractions with denominators of 10, 100, 1000, 10000 are convertedto decimals. When we wish to convert a percent to a decimal, we divide by 100, or simply move the decimal pointtwo units to the left.
TABLE 1.2:
Percent Decimal Percent Decimal Percent Decimal10% 0.1 0.05% .0005 0% 099% 0.99 0.25% .0025 100% 1
Find a Percent of a Number
One thing we need to do before we work with percents is to practice converting between fractions, decimals andpercentages. We will start by converting decimals to percents.
Example 1
Express 0.2 as a percent.
The word percent means “for every hundred”. Therefore, to find the percent, we want to change the decimal to afraction with a denominator of 100. For the decimal 0.2 we know the following is true:
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1.12. Percent Problems www.ck12.org
0.2 = 0.2×100×(
1100
)Since 100×
(1
100
)= 1
0.2 = 20×(
1100
)0.2 =
(20
100
)= 20%
Solution
0.2 = 20%
Another approach is to think of a decimal using it’s word name. For example, 0.35 is 35 hundredths. This means itcold be written as 35
100 or 35%.
Notice that the answer is the same as if you just moved the decimal point two places to the right. This is a morepractical way to convert decimals to percents.
Example 2
Express 0.07 as a percent.
Move the decimal point to places to the right to get that 0.07 = 7%
Example 3
Express 35% as a decimal.
35% =
(35100
)= 0.35
There is also a shortcut here, notice that we have just moved the decimal point two places to the left
When converting fractions to percents, we can substitute x100 for x%, where x is the unknown percentage we can
solve for.
Example 5
Express 35 as a percent.
We start by representing the unknown as x% or x100 .
(35
)=
x100
Cross multiply.
5x = 100 ·3 Divide both sides by 5 to solve for x.
5x = 300
x =300
5= 60
Solution(35
)= 60%
Example 6
Express 1340 as a percent.
Again, represent the unknown percent as x100 , cross-multiply, and solve for x.
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www.ck12.org Chapter 1. Unit 1 - Review
1340
=x
10040x = 1300
x =130040
= 32.5
Solution(1340
)= 32.5%
Converting percentages to simplified fractions can be done by writing the percent as a fraction with a denominatorof 100 and then simplifying
Example 7
Express 28% as a simplified fraction.
First write as a fraction, then simplify.
28% =28100
Now reduce the fraction.
(�4 ·7�4 ·25
)=
725
Solution
28% =( 7
25
)Use the Percent Equation
The percent equation is often used to solve problems.
Percent Equation:
percent100
=iso f
Example 8
What is 25% of $80
In this equation, the percent is the easier part. The left side of the equation will be
25100
The "is" is associated with x, because the question is asking "what is". The "of" is associated with the 80, becausethe question is asking "of $80". This means the final equation is:
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1.12. Percent Problems www.ck12.org
25100
=x
80Cross multiply
100x = 25 ·80
100x = 2000
x = 20
Solution
25% of $80 is $20.
Example 9
$90 is what percentage of $160.
Use the percent equation. This time the percentage is the unknown, so the x will be over 100
x100
=90
160160x = 100 ·90
160x = 9000
x = 56.25
Solution
$90 is 56.25% of 160.
An alternative to the percent proportion is to use algebra. For example, if you are trying to figure out what numberis 30% of 90, you could use "x" to stand for "what number" and "0.30" in place of 35% and write the followingequation
x = 0.30 ·90
x = 27
Video Overview of Percents, Decimals and Fractions
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Video Overview of the Percent Equation
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Lesson Summary
• A percent is simply a ratio with a base unit of 100, i.e. 13% = 13100 .
• The percent equation is: percent100 = is
o f
Homework
1. Express the following decimals as a percent.
a. 0.011b. 0.001c. 0.91d. 1.75e. 20
2. Express the following fractions as a percent (round to two decimal places when necessary).
a. 16
b. 524
c. 67
d. 117
e. 1397
3. Express the following percentages as a reduced fraction.
a. 11%b. 65%c. 16%d. 12.5%e. 87.5%
4. Find the following.
a. What number is 30% of 90b. 14 is 40% of what number?c. 14 is what % of 52?d. What number is 5% of 80
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1.13. Multi-Step Inequalities www.ck12.org
1.13 Multi-Step Inequalities
What You Will Learn
• Solve an inequality.• Graph an inequality• Write the solution to an inequality in interval notation
Procedure to Solve an Inequality:
a. Remove any parentheses by using the Distributive Property.b. Simplify each side of the inequality by combining like terms.c. Get the variable on one side of the inequality sign and the numerical values on the other.d. Isolate the variablee. Graph the solutionf. Write the answer in interval notation
Remember to reverse the inequality sign if you are multiplying or dividing by a negative number.
Example A
Solve for w : 6x−5 < 7.
Solution: Begin by using the checklist above.
1. Parentheses? No
2. Like terms on the same side of inequality? No
3. Isolate the ax term using the Addition Property.
6x−5+5 < 7+5
Simplify.
6x < 12
4. Isolate the variable using the Multiplication or Division Property.
6x6
<126
x < 2
5. Check your solution. Choose a number less than 2, say 0, and check using the original inequality.
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www.ck12.org Chapter 1. Unit 1 - Review
6(0)−5 < 7?
−5 < 7
Yes, the answer checks. x < 2
6. Graph your solution
FIGURE 1.1
7. Write your solution in interval notation
(−∞,2)
Example B
Solve for x : −9x <−5x−16.
Solution: Begin by using the checklist above.
1. Parentheses? No
2. Like terms on the same side of inequality? No
3. Isolate the ax term using the Addition Property.
−9x+5x <−5x+5x−16
Simplify.
−4x <−15
4. Isolate the variable using the Multiplication or Division Property.
−4x−4
<−16−4
Because the number you are dividing by is negative, you must reverse the inequality sign.
x > 4
5. Check your solution by choosing a number larger than 4, such as 10.
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1.13. Multi-Step Inequalities www.ck12.org
FIGURE 1.2
−9(10)<−5(10)−16?
X −90 <−66
6. Graph your solution
7. Write your solution in interval notation
(4,∞)
Example C
Solve for x : 4x−2(3x−9)≤−4(2x−9).
Solution: Begin by using the previous checklist.
1. Parentheses? Yes. Use the Distributive Property to clear the parentheses.
4x+(−2)(3x)+(−2)(−9)≤−4(2x)+(−4)(−9)
Simplify.
4x−6x+18≤−8x+36
2. Like terms on the same side of inequality? Yes. Combine these.
−2x+18≤−8x+36
3. Isolate the ax term using the Addition Property.
−2x+8x+18≤−8x+8x+36
Simplify.
6x+18≤ 36
6x+18−18≤ 36−18
6x≤ 18
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www.ck12.org Chapter 1. Unit 1 - Review
4. Isolate the variable using the Multiplication or Division Property.
6x6≤ 18
6→ x≤ 3
5. Check your solution using x=0, you will see that it does work
6. Graph your solution
FIGURE 1.3
7. Write your solution in interval notation
(−∞,3]
Video Overview
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Homework
In 1 – 15, solve each of the inequalities, graph the solution set and write it in interval notation.
1. 6x−5 < 102. −9x <−5x−153. −9x
5 ≤ 244. 9x
5 −7≥−3x+125. 5x−1
4 >−2(x+5)6. 4x+3 <−17. 2x < 7x−368. 5x > 8x+279. 5− x < 9+ x
10. 4−6x≤ 2(2x+3)11. 5(4x+3)≥ 9(x−2)− x12. 2(2x−1)+3 < 5(x+3)−2x13. 8x−5(4x+1)≥−1+2(4x−3)14. 2(7x−2)−3(x+2)< 4x− (3x+4)15. 2
3 x− 12(4x−1)≥ x+2(x−3)
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1.14. Absolute Value Equations www.ck12.org
1.14 Absolute Value Equations
What You Will Learn
• How to solve an equation that has an absolute value sign in it.
Absolute Value Equations
Equations can be used to represent absolute value situtations. For example, suppose the distance from zero is 16.What two points can this represent?
Begin by writing an absolute value sentence to represent this situation.
16 = |n|, where n = the missing value
Which two numbers are 16 units from zero?
n = 16 or n =−16
Absolute value situations can also involve distances from points other than zero. We treat such cases as compoundinequalities, separating the two independent equations and solving separately.
Example A
Solve: |x−4|= 5.
Solution: This equation looks like the distance definition:
distance = |x− y| or |y− x|
The distance is 5, and the value of y is 4. We are looking for two values that are five units away from four on anumber line.
Visually, we can see the answers are –1 and 9.
Algebraically, we separate the two absolute value equations and solve.
x−4 = 5 and x−4 =−(5)
By solving each, the solutions become:
x = 9 and x =−1
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www.ck12.org Chapter 1. Unit 1 - Review
Example B
Solve: |2x−7|=−6.
Solution:
Begin by separating this into its separate equations.
2x−7 =−6 and 2x−7 =−(−6) = 6
Solve each equation independently.
2x−7 = 6 2x−7 =−6
2x−7+7 = 6+7 2x−7+7 =−6+7
2x = 13 2x = 1
x =132
x =12
Example C
Solve: 3|x−4|+2 = 11
Solution: This equation is more complicated. Before we can break it into two separate equations, we need to isolatethe absolute value. First, we will subtract 2 from each side
3|x−4|+2−2 = 11−2
3|x−4|= 9
Next, we will divide each side by 3
3|x−4|3
=93
|x−4|= 3
Now that the absolute value is by itselft, we can write two separate equations and solve
x−4 = 3 x−4 =−3
x−4+4 = 3+4 x−4+4 =−3+4
x = 7 x = 1
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1.14. Absolute Value Equations www.ck12.org
Example D
A company packs coffee beans in airtight bags. Each bag should weigh 16 ounces but it is hard to fill each bag to theexact weight. After being filled, each bag is weighed and if it is more than 0.25 ounces overweight or underweight,it is emptied and repacked. What are the lightest and heaviest acceptable bags?
FIGURE 1.4
Solution:
The varying quantity is the weight of the bag of coffee beans. Choosing a letter to represent this quantity and writingan absolute value equation yields:
|w−16|= 0.25
Separate and solve.
w−16 = 0.25 w−16 =−0.25
w = 16.25 w = 15.75
The lightest bag acceptable is 15.75 ounces and the heaviest bag accepted is 16.25 ounces.
Homework
In 1 – 10, solve the absolute value equations.
1. |7u|= 772. |x−5|= 103. |5r−6|= 94. 1 = |6+5z|
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www.ck12.org Chapter 1. Unit 1 - Review
5. |8x|= 326. |m8 |= 17. |x+2|−3 = 68. 2|5x−2|−1 = 39. 51 = |1−5b|
10. 8 = 3+ 12 |10y+5|
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1.15. References www.ck12.org
1.15 References
1. . . CC BY-NC-SA2. . . CC BY-NC-SA3. . . CC BY-NC-SA
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www.ck12.org Chapter 2. Unit 2 - Linear Functions
CHAPTER 2 Unit 2 - Linear FunctionsChapter Outline
2.1 INTRODUCTION TO FUNCTIONS
2.2 FINDING THE DOMAIN AND RANGE OF FUNCTIONS
2.3 GRAPHING USING TABLE OF VALUES
2.4 GRAPHING USING INTERCEPTS
2.5 SLOPE AND RATE OF CHANGE
2.6 GRAPHS USING SLOPE-INTERCEPT FORM
2.7 WRITING EQUATIONS OF LINES
2.8 PARALLEL AND PERPENDICULAR LINES
2.9 GRAPHING LINEAR INEQUALITIES
2.10 GRAPHS OF ABSOLUTE VALUE EQUATIONS
2.11 FITTING A LINE TO DATA
2.12 PROBLEM SOLVING WITH LINEAR MODELS
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2.1. Introduction to Functions www.ck12.org
2.1 Introduction to Functions
What You Will Learn
• Definition of a function• How to find the value of a function
Definition of a Function
Definition: A function is a relationship between two variables such that the input value has ONLY one uniqueoutput value.
Let’s look at a relationship between variables and see what that means
y = 2x
This relationship means y will always be two times x. If x is 1, y is 2 and so on for any value of x. x is called theindependant variable because that is the variable that is chosen first. y is the dependant variable, it’s value dependson x, it will always be twice the value of x.
Function notation replaces the dependant variable y with its function name, usually f (x). These parentheses do notmean multiplication. They separate the function name from the independent variable, x.
input
↓f (x)︸︷︷︸= y← out put
f (x) is read “the function f of x” or simply “ f of x.”
If the function looks like this: h(x) = 3x−1, it would be read h of x equals 3 times x minus 1.
How to Find the Value of a Function
Function notation allows you to easily see the input value for the independent variable inside the parentheses.
Example A
Consider the function f (x) =−12 x2.
Evaluate f (4).
Solution: The value inside the parentheses is the value of the variable x. Use the Substitution Property to evaluatethe function for x = 4.
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www.ck12.org Chapter 2. Unit 2 - Linear Functions
f (4) =−12(42)
f (4) =−12·16
f (4) =−8
To use function notation, the equation must be written in terms of x. This means that the y−variable must be isolatedon one side of the equal sign.
Example B
Rewrite 9x+3y = 6 using function notation.
Solution: The goal is to rearrange this equation so the equation looks like y =. Then replace y = with f (x) =.
9x+3y = 6 Subtract 9x from both sides.
3y = 6−9x Divide by 3.
y =6−9x
3= 2−3x
f (x) = 2−3x
Functions as Machines
You can think of a function as a machine. You start with an input (some value), the machine performs the operations(it does the work), and your output is the answer. For example, f (x) = 3x+2 takes some number, x, multiplies it by3 and adds 2. As a machine, it would look like this:
When you use the function machine to evaluate f (2), the solution is f (2) = 8.
Example C
A function is defined as f (x) = 6x−36. Determine the following:
a) f (2)
b) f (p)
Solution:
a) Substitute x = 2 into the function f (x) : f (2) = 6 ·2−36 = 12−36 =−24.
b) Substitute x = p into the function f (x) : f (p) = 6p−36.
Practice Exercise
Rewrite the equation 2y− 4x = 10 in function notation where f (x) = y, and then evaluate f (−1), f (2), f (0), andf (z).
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Solution:
First we need to solve for y.
Adding 4x to both sides gives 2y = 4x+10, and dividing by 2 gives y = 2x+5.
Now we just replace the y with f (x) to get f (x) = 2x+5..
Now we can evaluate f (x) = y = 2x+5 for f (−1), f (2), f (0), and f (z):
f (−1) = 2(−1)+5 =−2+5 = 3
f (2) = 2(2)+5 = 4+5 = 9
f (0) = 2(0)+5 = 5
f (z) = 2z+5
Video Overview
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Homework
Part 1
Solve each equation for y, then rewrite using function notation
1. .y = 7x−212. 6x+8y = 363. x = 9y+34. y = 65. 5x−3y = 15
Part 2
Evaluate f (−3), f (0), and f (z)
1. f (x) =−2x+32. f (x) = 0.7x+3.23. f (x) = 5(2−x)
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Part 3
Evalute g(1)
1. g(t) = 12 t2 +4
2. g(x) = 3− 12 x
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2.2 Finding the Domain and Range of Func-tions
What You Will Learn
• How to determine if a relation is a function• How to find the domain and range of a function
How to Determine if a Relation is a Function
We saw in the last section that a relationship between numbers was a function if each x value in the relation resultedin a unique value for y. If a set of points is given, look at each x and see if it has a different y with it. If each x doeshave a unique value for y, than the relationship is a function. If not, it is a relation.
Example A
Determine if {(4, 2), (6, -3), (4, -6), (5, 4), (6, -7)} is a function.
It is not because an x of 6 results in two different y values.
If you are looking at a graphed equation instead of a set of points, you can use the Vertical Line Test to determineif it is a function
Example B
Is this graphed relation a function?
By drawing a vertical line (the red line) through the graph, we can see that the vertical line intersects the circle morethan once. Therefore, each value of x does not have a unique value of y and this graph is NOT a function.
Here is a second example:
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No matter where a vertical line is drawn through the graph, there will be only one intersection. Therefore, this graphis a function.
Practice Exercise
Determine if the graphed relation is a function.
Solution:
Imagine moving a vertical line across the plane. Do you see anywhere that this vertical line would intersect the graphat more than one place?
There is no place on this graph where a vertical line would intersect the graph at more than one place. Using thevertical line test, we can conclude the relation is a function.
Domain and Range
The input and output of a function is also called the domain and range. The domain of a function is the set of allinput values. The range of a function is the set of all output values. Sometimes, a function is a set of points. In thiscase, the domain is all the x−values and the range is all the y−values. Functions can also be linear and polynomialequations. In these instances, you need to graph the function to see where it is defined.
Example A
Determine if {(9, 2), (7, -3), (4, -6), (-10, 4), (-2, -7)} is a function. If so, find the domain and range.
Solution: First, this is a function because the x−values do not repeat. To find the domain, we need to list all thex−values. The range is all the y−values. Typically, you would list the values in the order they appear.
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Domain 9,7,4,−10,−2
Range 2,−3,−6,4,−7
Example B
Find the domain and range of y = x−3.
Solution: This graph is a function because it passes the vertical line test. All lines continue forever in both directions,as indicated by the arrows.
Notice the line is solid, there are no dashes or breaks, this means that it is continuous. A continuous function has avalue for every x, or the domain is all real numbers. Can you plug in ANY value for x and get a y−value? Yes.
Domain: (−∞,∞), x is all reals
In words, x is an element in the set of real numbers.
The range of this function is also continuous. Therefore, the range is also the set of all real numbers. We can writethe range in the same ways we wrote the domain, but with y instead of x.
Range: (−∞,∞)
Practice Exercises
Find the domain and range of the following functions.
1. {(8, 3), (-4, 2), (-6, 1), (5, 7)}
2. y =−12 x+4
3.
4.
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Video Overview of Vertical Line Test
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Video Overview of Domain and Range
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Homework
Determine if the following are functions. If so, state the domain and range.
1. {(5, 6), (-1, 5), (7, -3), (0, 9)}2. {(9, 8), (-7, 8), (-7, 9), (8, 8)}3. {(6, 2), (-5, 6), (-5, 2)}4. {(-1, 2), (-6, 3), (10, 7), (8, 11)}5. {(-3, -4), (-5, -6), (1, 2), (2, -6)}6. y = 3x−2
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Are the following graphs functions? If so, state the domain and range.
7.
8.
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2.3 Graphing using Table of Values
What You Will Learn
• Graph a linear function using an equation.• Test if a point is on the line• Solve for y before graphing• Write equations and graph horizontal and vertical lines.
Graph a Linear Equation
A linear equation tells you a relationship between x and y. If you graph that equation, you will get a picture of howx and y are related.
Example 1
A taxi fare costs more the further you travel. Taxis usually charge a fee on top of the per-mile charge to cover hireof the vehicle. In this case, the taxi charges $3 as a set fee and $0.80 per mile traveled. Here is the equation linkingthe cost in dollars (y) to hire a taxi and the distance traveled in miles (x).
y = 0.8x+3
Graph the equation and use your graph to estimate the cost of a seven mile taxi ride.
We will start by making a table of values. We will take a few values for x {0, 1, 2, 3, 4}, find the corresponding yvalues and then plot them. Since the question asks us to find the cost for a seven mile journey, we will choose a scalethat will accommodate this.
TABLE 2.1:
x y0 3
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TABLE 2.1: (continued)
x y1 3.82 4.63 5.44 6.2
The graph is shown to the right. To find the cost of a seven mile journey we first locate x = 7 on the horizontal axisand draw a line up to our graph. Next we draw a horizontal line across to the y axis and read where it hits. It appearsto hit around half way between y = 8 and y = 9. Let’s say it is 8.5.
Solution
A seven mile taxi ride would cost approximately $8.50 ($8.60 exactly).
There are a few interesting points that you should notice about this graph and the formula that generated it.
• The graph is a straight line (this means that the equation is linear), although the function is discrete and willgraph as a series of points.
• The graph crosses the y−axis at y = 3 (look at the equation - you will see a +3 in there!). This is the base costof the taxi.
• Every time we move over by one square we move up by 0.8 squares (look at the coefficient of x in theequation). This is the rate of charge of the taxi (cost per mile).
• If we move over by three squares, we move up by 3×0.8 squares.• This graph is a function, it passes the vertical line test. This means it could also be written f (x) = 0.8x+3
Example 2
A student borrows $500 to buy a computer. She predicts that it can pay off the debt at a rate of $85 per monthaccording to the following equation..
y =−85x+500
Graph the above equation and use your graph to predict when the debt will be fully paid.
First, we start with our table of values. We substitute in x−values and calculate our corresponding y−values.
TABLE 2.2:
x y0 5001 4152 3303 2454 160
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Then we plot our points and draw the line that goes through them.
Take note of the scale that has been chosen. There is no need to have any points above y = 500, but it is still wise toallow a little extra.
We need to determine how many months (the x value) that it takes the debt (y value) to reach zero. We know that itis greater than four (since at x = 4 the y value is still positive), so we need an x scale that goes well past x = 4. Inthis case the x value runs from 0 to 12, though there are plenty of other choices that would work well.
To read the time that the debt is paid off, we simply read the point where the line hits y = 0 (the x axis). It looks asif the line hits pretty close to x = 6. So the debt will definitely be paid off in six months.
Solution
The debt will be paid off in six months
Testing if a point is on the line
It is common to ask if a random point will be on a certain line. To do this, you put the given x and y values into theequation and see if it comes out to be true. If it does, then the point is on the line.
Example 3
2x+ y = 5 Is the point (1,2) on the line?2(1)+(2) = 5 Replace x and y with 1 and 24 = 5 This is not true, so (1,2) is not on the line
Solving for y before graphing
All of the equations we have graphed so far have had y on the left side. Sometimes the equation is messier and weneed to use algebra to get y by itself.
Example 4
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3x+ y = 5
−3x =−3x Subtract 3x from both sides.
y = 5−3x
Now you could graph this using the same procedure as the previous section
Example 5
2y−5x = 8
+5x =+5x Add 5x to both sides.2y = 8+5x2y2
=8+5x
2Divide both sides by 2.
y = 4+52
x
Graphs and Equations of Horizontal and Vertical Lines
Example 6
“Mad-cabs” have an unusual offer going on. They are charging $7.50 for a taxi ride of any length within the citylimits. Graph the function that relates the cost of hiring the taxi (y) to the length of the journey in miles (x).
To proceed, the first thing we need is an equation. You can see from the problem that the cost of a journey does notdepend on the length of the journey. It should come as no surprise that the equation then, does not have x in it. Infact, any value of x results in the same value of y (7.5). Here is the equation.
y = 7.5
The graph of this function is shown to the right. You can see that the graph y = 7.5 is simply a horizontal line.
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Any time you see an equation of the form y = constant then the graph is a horizontal line that intercepts the y−axisat the value of the constant.
Similarly, when you see an equation of the form x = constant then the graph is a vertical line that intercepts thex−axis at the value of the constant. Notice that this a relation, and not a function because each x value (there’s onlyone in this case) corresponds to many (actually an infinite number) y values.
Example 6
Plot the following graphs.
(a) y = 4
(b) y =−4
(c) x = 4
(d) x =−4
(a) y = 4 is a horizontal line that crosses the y−axis at 4
(b) y =−4 is a horizontal line that crosses the y−axis at -4
(c) x = 4 is a vertical line that crosses the x−axis at 4
(d) x =−4 is a vertical line that crosses the x−axis at -4
Example 7
Find an equation for the x−axis and the y−axis.
Look at the axes on any of the graphs from previous examples. We have already said that they intersect at the origin(the point where x = 0 and y = 0). The following definition could easily work for each axis.
x− axis: A horizontal line crossing the y−axis at zero.
y− axis: A vertical line crossing the x−axis at zero.
So using example 3 as our guide, we could define the x−axis as the line y = 0 and the y−axis as the line x = 0.
Video Overview of Graphing with Table of Values
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Video Overview of Horizontal and Vertical Lines
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Homework
1. Make a table of values for the following equations and then graph them.
a. y = 2x+7b. y = 1
2 x−4c. f (x) = 6−3xd. y = 2
5 x−1
2. Solve the following equations for y, make a table and then graph them.
(a) x+ y = 4(b) 6 = 2y+ x(c) 5− y = 4x
3. Check if the following points are on the suggested line
(a) Is the point (3,-2) on the line x+y=1?(b) Is the point (1,5) on the line y=2x-3?(c) Is the point (-6,7) on the line 3x=2y-4?
4. Graph the following lines
(a) y=2(b) x=3(c) x=-4(d) y=0
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2.4 Graphing Using Intercepts
What You Will Learn
• Find intercepts of the graph of an equation.• Use intercepts to graph an equation.
Finding x and y Intercepts and Graphing
Example 1
Find the intercepts of the line y = 13− x and use them to graph the function.
Find the y intercept by replacing x with 0. Substituting gives:
y = 13−0 = 13 (0,13) is the x− intercept.
Find the x intercept by replacing y with zero.
0 = 13− x To isolate x subtract 13 from both sides.
−13 =−x Divide by−1.
Solution
(13, 0) is the x−intercept.
To draw the graph simply plot these points and join them with a line.
Example 2
Graph the following functions by finding intercepts.
a. y = 2x+3
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b. y = 7−2x
c. 4x−2y = 8
d. 2x+3y =−6
a. Find the y−intercept by plugging in x = 0.
y = 2 ·0+3 = 3 The y− intercept is (0,3)
Find the x−intercept by plugging in y = 0.
0 = 2x+3 Subtract 3 from both sides.
−3 = 2x Divide by 2.
−32= x The x− intercept is (−1.5,0).
b. Find the y−intercept by substituting in x = 0.
y = 7−2 ·0 = 7 The y− intercept is (0,7).
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Find the x−intercept by substituting in y = 0.
0 = 7−2x Subtract 7 from both sides.
−7 =−2x Divide by −2.72= x The x− intercept is (3.5,0).
c. Find the y−intercept by substituting in x = 0.
4 ·0−2y = 8
−2y = 8 Divide by −2.
y =−4 The y− intercept is (0,−4).
Find the x−intercept by substituting in y = 0.
4x−2 ·0 = 8
4x = 8 Divide by 4.
x = 2 The x− intercept is (2,0).
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d. Find the y−intercept by substituting in x = 0.
2 ·0+3y =−6
3y =−6 Divide by 3.
y =−2 The y− intercept is (0,−2).
Find the x−intercept by substituting in y = 0.
2x+3 ·0 =−6
2x =−6 Divide by 2.
x =−3 The x− intercept is (−3,0)
Video Overview
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Homework
Find the x and y intercepts for the following equations and graph them
1. y = 3x−62. y =−2x+43. y = 7x−214. 3x−4y =−125. 2x+7y = 146. 5x+10y = 25
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2.5 Slope and Rate of Change
What You Will Learn
• Find positive and negative slopes.• Recognize and find slopes for horizontal and vertical lines.• Understand rates of change.• Interpret graphs and compare rates of change.
Introduction
We come across many examples of slope in everyday life. For example, a slope is in the pitch of a roof, the grade orincline of a road, and the slant of a ladder leaning on a wall. In math, we use the word slope to define steepness in aparticular way.
Slope =distance moved vertically
distance moved horizontally
This is often reworded to be easier to remember:
Slope =riserun
Essentially, slope is the change in y if x increases by 1.
In the picture to the right, the slope would be the ratio of the height of the hill (the rise) to the horizontal length ofthe hill (the run).
Slope =34= 0.75
If the car were driving to the right it would climb the hill. We say this is a positive slope. Anytime you see the graphof a line that goes up as you move to the right, the slope is positive.
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If the car were to keep driving after it reached the top of the hill, it may come down again. If the car is driving tothe right and descending, then we would say that the slope is negative. The picture at right has a negative slope of-0.75.
Do not get confused! If the car turns around and drives back down the hill shown, we would still classify the slopeas positive. This is because the rise would be -3, but the run would be -4 (think of the x−axis - if you move fromright to left you are moving in the negative x−direction). Our ratio for moving left is:
Slope =−3−4
= 0.75 A negative divided by a negative is a positive.
So as we move from left to right, positive slopes increase while negative slopes decrease.
Find a Positive Slope
We have seen that a function with a positive slope increases in y as we increase x. A simple way to find a valuefor the slope is to draw a right angled triangle whose hypotenuse runs along the line. It is then a simple matter ofmeasuring the distances on the triangle that correspond to the rise (the vertical dimension) and the run (the horizontaldimension).
Example 1
Find the slope of the line that passes through the points (1, 2) and (4, 7).
We already know how to graph a line if we are given two points. We simply plot the points and connect them with aline. Look at the graph shown at right.
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Since we already have coordinates for our right triangle, we can quickly work out that the rise would be 5 and therun would be 3 (see diagram). Here is our slope.
Slope =7−24−1
=53
If you look closely at the calculations for the slope you will notice that the 7 and 2 are the y−coordinates of the twopoints and the 4 and 1 are the x−coordinates. This suggests a pattern we can follow to get a general formula for theslope between two points (x1,y1) and (x2,y2).
Slope between (x1,y1) and (x2,y2) =y2−y1x2−x1
or m = 4y4x
In the second equation, the letter m denotes the slope (you will see this a lot in this chapter) and the Greek letterdelta (∆) means change. So another way to express slope is change in y divided by change in x. In the next section,you will see that it does not matter which point you choose as point 1 and which you choose as point 2.
Find a Negative Slope
Any function with a negative slope is simply a function that decreases as we increase x. If you think of the functionas the incline of a road a negative slope is a road that goes downhill as you drive to the right.
Example 2
Find the slopes of the following two lines.
Look at the lines. Both functions fall (or decrease) as we move from left to right. Both of these lines have a negativeslope.
Neither line passes through a great number of lattice points, but by looking carefully you can see a few points thatlook to have integer coordinates. These points have been identified (with rings) and we will use these to determinethe slope. We will also do our calculations twice, to show that we get the same slope whichever way we choose point1 and point 2.
For line A:
(x1,y1) = (−6,3) (x2,y2) = (5,−1) (x1,y1) = (5,−1) (x2,y2) = (−6,3)
m =y2− y1
x2− x1=
(−1)− (3)(5)− (−6)
=−411≈−0.364 m =
y2− y1
x2− x1=
(3)− (−1)(−6)− (−5)
=−411≈−0.364
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For line B:
(x1,y1) = (−4,6) (x2,y2) = (4,−5) (x1,y1) = (4,−5) (x2,y2) = (−4,6)
m =y2− y1
x2− x1=
(−5)− (6)(4)− (−4)
=−11
8=−1.375 m =
y2− y1
x2− x1=
(6)− (−5)(−4)− (4)
=11−8
=−1.375
You can see that whichever way you select the points, the answers are the same!
Solution
Line A has slope -0.364. Line B has slope -1.375.
Find the Slopes of Horizontal and Vertical lines
Example 3
Determine the slopes of the two lines on the graph at the right.
There are two lines on the graph. A(y = 3) and B(x = 5).
Let’s pick two points on line A. say, (x1,y1) = (−4,3) and (x2,y2) = (5,3) and use our equation for slope.
m =y2− y1
x2− x1=
(3)− (3)(5)− (−4)
=09= 0
If you think about it, this makes sense. If there is no change in y as we increase x then there is no slope, or to becorrect, a slope of zero. You can see that this must be true for all horizontal lines.
Horizontal lines (y = constant) all have a slope of 0.
Now consider line B. Pick two distinct points on this line and plug them in to the slope equation.
(x1,y1) = (5,−3) and (x2,y2) = (5,4).
m =y2− y1
x2− x1=
(4)− (−3)(5)− (5)
=70
A division by zero!
Divisions by zero lead to infinities. In math we often use the term undefined for any division by zero.
Vertical lines (x = constant) all have an infinite (or undefined) slope.
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Find a Rate of Change
The slope of a function that describes real, measurable quantities is often called a rate of change. In that case, theslope refers to a change in one quantity (y) per unit change in another quantity (x).
Example 4
Andrea has a part time job at the local grocery store. She saves for her vacation at a rate of $15 every week. Expressthis rate as money saved per day and money saved per year.
Converting rates of change is fairly straight forward so long as you remember the equations for rate (i.e. the equationsfor slope) and know the conversions. In this case 1 week = 7 days and 52 weeks = 1 year.
rate =$15
1 week· 1 week
7 days=
$157 days
=157
dollars per day≈ $2.14 per day
rate =$15
1 week· 52 week
1 year= $15 · 52
year= $780 per year
Example 5
A candle has a starting length of 10 inches. Thirty minutes after lighting it, the length is 7 inches. Determine the rateof change in length of the candle as it burns. Determine how long the candle takes to completely burn to nothing.
In this case, we will graph the function to visualize what is happening.
We have two points to start with. We know that at the moment the candle is lit (time = 0) the length of the candle is10 inches. After thirty minutes (time = 30) the length is 7 inches. Since the candle length is a function of time wewill plot time on the horizontal axis, and candle length on the vertical axis. Here is a graph showing this information.
The rate of change of the candle is simply the slope. Since we have our two points (x1,y1) = (0,10) and (x2,y2) =(30,7) we can move straight to the formula.
Rate of change =y2− y1
x2− x1=
(7 inches)− (10 inches)(30 minutes)− (0 minutes)
=−3 inches30 minutes
=−0.1 inches per minute
The slope is negative. A negative rate of change means that the quantity is decreasing with time.
We can also convert our rate to inches per hour.
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rate =−0.1 inches
1 minute· 60 minutes
1 hour=−6 inches
1 hour=−6 inches per hour
To find the point when the candle reaches zero length we can simply read off the graph (100 minutes). We can usethe rate equation to verify this algebraically.
Length burned = rate× time
0.1×100 = 10
Since the candle length was originally 10 inches this confirms that 100 minutes is the correct amount of time.
Video Overview
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Homework
1. Use the slope formula to find the slope of the line that passes through each pair of points.
a. (-5, 7) and (0, 0)b. (-3, -5) and (3, 11)c. (3, -5) and (-2, 9)d. (-5, 7) and (-5, 11)e. (9, 9) and (-9, -9)f. (3, 5) and (-2, 7)
2. Determine the slopes of the following lines.
FIGURE 2.1
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FIGURE 2.2
FIGURE 2.3
4.
FIGURE 2.4
3. Using the following graph about John’s savings account, how much is he saving per month?In 1999, Lydia purchased a house for $160,000. In 2013, the house was worth $210,000. Find the average
annual rate of change in dollars per year in the value of the house.
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2.6 Graphs Using Slope-Intercept Form
What Will You Learn
• Identify the slope and y−intercept of equations and graphs.• Graph an equation in slope-intercept form.• Understand what happens when you change the slope or intercept of a line.
Identify Slope and y-intercept
One of the most common ways of writing linear equations prior to graphing them is called slope-intercept form.We have actually seen several slope-intercept equations so far. They take the following form:
y = mx+b where m is the slope and the point (0,b) is the y−intercept.
We know that the y−intercept is the point at which the line passes through the y−axis. The slope is a measure of thesteepness of the line. Hopefully, you can see that if we know one point on a line and the slope of that line, we knowwhat the line is. Being able to quickly identify the y−intercept and slope will aid us in graphing linear functions.
Example 1
Identify the slope and y−intercept of the following equations.
a) y = 3x+2
b) y = 0.5x−3
c) y =−7x
d) y =−4
Solution
a)
Comparing, we see that m = 3 and b = 2.
y = 3x+2 has a slope of 3 and a y− intercept of (0, 2)
b)
has a slope of 0.5 and a y− intercept of (0, -3).
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Note that the y−intercept is negative. The b term includes the sign of the operator in front of the number. Justremember that y = 0.5x−3 is identical to y = 0.5x+(−3) and is in the form y = mx+b.
c) At first glance, this does not appear to fit the slope-intercept form. To illustrate how we deal with this, letus rewrite the equation.
We now see that we get a slope of -7 and a y− intercept of (0, 0).
Note that the slope is negative. The (0, 0) intercept means that the line passes through origin.
d) Rewrite as y = 0x−4, giving us a slope of 0 and an intercept of (0, -4).
Remember:
• When m < 0 the slope is negative.
For example, y =−3x+2 has a slope of -3.
• When b < 0 the intercept is below the x axis.
For example, y = 4x−2 has a y−intercept of (0, -2).
• When m = 0 the slope is zero and we have a horizontal line.
For example, y = 3 can be written as y = 0x+3.
• When b = 0 the graph passes through the origin.
For example, y = 4x can be written as = 4x+0.
Example 2
Identify the slope and y−intercept of the lines on the graph shown to the right.
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The intercepts have been marked, as have a number of lattice points that lines pass through.
a. The y−intercept is (0, 5). The line also passes through (2, 3).
slope m =∆y∆x
=−22
=−1
b. The y−intercept is (0, 2). The line also passes through (1, 5).
slope m =∆y∆x
=31= 3
c. The y−intercept is (0, -1). The line also passes through (2, 3).
slope m =∆y∆x
=42= 2
d. The y−intercept is (0, -3). The line also passes through (4, -4).
slope m =∆y∆x
=−14
=−14
or −0.25
Graph an Equation in Slope-Intercept Form
Once we know the slope and intercept of a line it is easy to graph it. Just remember what slope means. Let’s lookback at this example from Lesson 4.1.
Example 3
Ahiga is trying to work out a trick that his friend showed him. His friend started by asking him to think of a number.Then double it. Then add five to what he got. Ahiga has written down a rule to describe the first part of the trick. Heis using the letter x to stand for the number he thought of and the letter y to represent the result of applying the rule.His rule is:
y = 2x+5
Help him visualize what is going on by graphing the function that this rule describes.
In that example, we constructed the following table of values.
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TABLE 2.3:
x y0 2.0+5 = 0+5 = 51 2.1+5 = 2+5 = 72 2.2+5 = 4+5 = 93 2.3+5 = 6+5 = 11
The first entry gave us our y intercept (0, 5). The other points helped us graph the line.
We can now use our equation for slope, and two of the given points.
Slope between (x1,y1) = (0,5) and (x2,y2) = (3,11).
m =y2− y1
x2− x1=
11−53−0
=63= 2
Thus confirming that the slope, m = 2.
An easier way to graph this function is the slope-intercept method. We can now do this quickly, by identifying theintercept and the slope.
Look at the graph we drew, the line intersects the y−axis at 5, and every time we move to the right by one unit, wemove up by two units.
So what about plotting a function with a negative slope? Just remember that a negative slope means the functiondecreases as we increase x.
Example 4
Graph the following function. f (x) =−3x+5
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• Identify y−intercept b = 5• Plot intercept (0, 5)• Identify slope m =−3• Draw a line through the intercept that has a slope of -3.
To do this last part remember that slope = riserun so for every unit we move to the right the function increases by -3
(in other words, for every square we move right, the function comes down by 3).
Changing the Slope of a Line
Look at the graph on the right. It shows a number of lines with different slopes, but all with the same y−intercept(0, 3).
You can see all the positive slopes increase as we move from left to right while all functions with negative slopes fallas we move from left to right.
Notice that the higher the value of the slope, the steeper the graph.
The graph of y = 2x+3 appears as the mirror image of y =−2x+3. The two slopes are equal but opposite.
Fractional Slopes and Rise Over Run
Look at the graph of y = 0.5x+ 3. As we increase the x value by 1, the y value increases by 0.5. If we increasethe x value by 2, then the y value increases by 1. In fact, if you express any slope as a fraction, you can determinehow to plot the graph by looking at the numerator for the rise (keep any negative sign included in this term) and thedenominator for the run.
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Example 5
Find integer values for the rise and run of following slopes then graph lines with corresponding slopes.
a. m = 3
b. m =−2
c. m = 0.75
d. m =−0.375
Solution:
a.
b.
c.
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d.
Changing the Intercept of a Line
When we take an equation (such as y = 2x) and change the y intercept (leaving the slope intact) we see the followingpattern in the graph on the right.
Notice that changing the intercept simply translates the graph up or down. Take a point on the graph of y = 2x, suchas (1, 2). The corresponding point on y = 2x+3 would be (1, 4). Similarly the corresponding point on the y = 2x−3line would be (1, -1).
Let’s look at some graphs of lines and see if we can find the slope and y intercept.
Example 6
This line crosses the y axis at -4, so b = -4. To find the slope, let’s start at -4 and see the next place it crosses a gridpoint, that coordinate is (1, -2). The slope between the y intercept and the point (1,-2) has a rise of 2 and a run ofone, so the slope is 2. Since the y intercept of this graph is -4 and the slope is 2, the equation is y = 2x−4.
Example 7
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This line crosses the y axis at 6, so b = 6. To find the slope, let’s start at 6 and look for the next place it crosses agrid point, that coordinate is (3, -4). The slope between the y intercept and the point (3, -4) has a rise of -2 and arun of 3, so the slope is −2
3 . This means the equation of the line is y =−23 x+6
Video Overview
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Homework
1. Identify the slope and y−intercept for the following equations.
a. y = 2x+5b. f (x) = 1
2 x+7c. y = xd. 3x−2y = 6
2. Put the following equations in slope-intercept form and graph them
a. y = 2x+5b. x− y = 3c. 4y = x−8d. 4x = 3y−12
3. Find the equations of the following lines
(a)
(b)
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2.7 Writing Equations of Lines
What You WIll Learn
• Write an equation given slope and y-intercept.• Write an equation given the slope and a point.• Write an equation given two points.
Write an Equation Given Slope and y-intercept
If you are given the slope and y-intercept of a line:
a. Start with the slope–intercept form of the line y = mx+b.b. Substitute the given values of m and b into the equation.
Example 1
Write an equation with a slope of 4 and a y intercept of -3
Solution
We are given m = 4 and b =−3. Substitute these values into the slope–intercept form y = mx+b.
y = 4x−3
You can also write an equation in slope-intercept form if you are given the graph of the line.
Example 2
Write the equation of each line in slope–intercept form.
a)
b)
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c)
d)
a) The y−intercept =−4 and the slope =−52 . Substitue these values into the slope–intercept form y = mx+b.
Solution
y =−52 x−4
b) The y−intercept = 2 and the slope = 31 . Substitute these values into the slope–intercept form y = mx+b.
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Solution
y = 3x+2
c) The y−intercept = 4 and the slope =−11 . Substitute these values into the slope–intercept form y = mx+b.
Solution
y =−x+4
d) The y−intercept =−2 and the slope = 12 . Substitute these values into the slope–intercept form y = mx+b.
Solution
y = 12 x−2.
Write an Equation Given the Slope and a Point
Often, we don’t know the value of the y−intercept, but we know the value of y for a non-zero value of x. In this casewe can still use the slope–intercept form to find the equation of the line.
For example, we are told that the slope of a line is two and that the line passes through the point (1, 4). To find theequation of the line, we start with the slope–intercept form of a line.
y = mx+b
Substitute the value of the slope.
We don’t know the value of b but we know that the slope is two, and that point (1, 4) is on this line. Where the xvalue is one, and the y value is four. We put this point in the equation and solve for b.
4 = 2(1)+b
4 = 2+b
−2 =−2
2 = b
Therefore the equation of this line is y = 2x+2.
If you are given the slope and a point on the line:
a. Start with the slope–intercept form of the line y = mx+b.b. Substitute the given value of m into the equation.c. Substitute the x and y values of the given point and solve for b.d. Substitute the value of b into the equation.
Example 3
Write the equation of the line in slope–intercept form.
a) The slope of the line is 4 and the line contains point (-1, 5).
b) The slope of the line is −23 and the line contains point (2, -2).
c) The slope of the line is -3 and the line contains point (3, -5).
Solution
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a)
Start with the slope intercept form of the line y = mx+b
Substitute the slope. y = 4x+b
Put point (−1,5) into the equation. 5 = 4(−1)+b⇒ b = 9
Substitute the value of b into the equation. y = 4x+9
b)
Start with the slope intercept form of the line y = mx+b
Substitute the slope. y =−23
x+b
Put point (2,−2) into the equation. −2 =−23(2)+b⇒ b =−2+
43=−2
3
Substitute the value of b into the equation. y =−23
x− 23
c)
Start with the slope intercept form of the line y = mx+b
Substitute the slope. y =−3x+b
Put in point (3,−5) into the equation. −5 =−3(3)+b⇒ b = 4
Substitute the value of b into the equation. y =−3x+4
Write an Equation Given Two Points
One last case is when we are just given two points on the line and we are asked to write the line of the equation inslope–intercept form.
For example, we are told that the line passes through the points (-2, 3) and (5, 2). To find the equation of the line westart with the slope–intercept form of a line
y = mx+b
Since we don’t know the slope, we find it using the slope formula m = y2−y1x2−x1
.
Now substitute the x1 and x2 and the y1 and y2 values into the slope formula to solve for the slope.
m =2−3
5− (−2)=−1
7
We substitute the value of the slope into the slope–intercept form y =−17 x+b
We don’t know the value of b but we know two points on the line. We can substitute either point into the equationand solve for b. Let’s use point (-2, 3).
Therefore, the equation of this line is y =−17 x+ 19
7 .
If you are given two points on the line:
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a. Start with the slope–intercept form of the line y = mx+bb. Use the two points to find the slope using the slope formula m = y2−y1
x2−x1.
c. Substitute the given value of m into the equation.d. Substitute the x and y values of one of the given points into the equation and solve for b.e. Substitute the value of b into the equation.f. Substitute the other point into the equation to check the values of m and b.
Example 4
Write the equations of each line in slope–intercept form.
a) The line contains the points (3, 2) and (-2, 4).
b) The line contains the points (-4, 1) and (-2, 3).
Solution:
a)
a. Start with the slope–intercept form of the line y = mx+b.b. Find the slope of the line. m = y2−y1
x2−x1= 4−2−2−3 =−2
5c. Substitute the value of the slope. y =−2
5 x+bd. Substitute point (3, 2) into the equation. 2 =−2
5(3)+b⇒ b = 2+ 65 = 16
5e. Substitute the value of b into the equation. y =−2
5 x+ 165
f. Substitute point (-2, 4) into the equation to check. 4 =−25(−2)+ 16
5 = 45 +
165 = 20
5 = 4
b)
a. Start with the slope–intercept form of the line y = mx+b.b. Find the slope of the line. m = y2−y1
x2−x1= 3−1−2−(−4) =
22 = 1
c. Substitute the value of the slope. y = x+bd. Substitute point (-2, 3) into the equation. 3 =−2+b⇒ b = 5e. Substitute the value of b into the equation. y = x+5f. Substitute point (-4, 1) into the equation to check. 1 =−4+5 = 1
Video Overview of Writing an Equation with Slope and Intercept (from graph)
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Video Overview of Writing an Equation Given a Point and the Slope
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Video Overview of Writing an Equation Given Two Points
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Homework
Part 1
Find the equation of the line in slope–intercept form.
1. The line has slope of 7 and y−intercept of -2.2. The line has slope of -5 and y−intercept of 6.3. The line has slope of −1
4 and contains point (4, -1).4. The line has slope of 2
3 and contains point(1
2 ,1).
5. The line has slope of -1 and contains point(4
5 ,0).
6. The line contains points (2, 6) and (5, 0).7. The line contains points (5, -2) and (8, 4).8. The line contains points (3, 5) and (-3, 0).9. The line contains points (10, 15) and (12, 20).
Part 2
Write the equation of each line in slope-intercept form.
1.
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2.
Applications
1. To buy a car, Andrew puts a down payment of $1500 and pays $350 per month in installments. Write anequation describing this problem in slope-intercept form. How much money has Andrew paid at the end ofone year?
2. Anne transplants a rose seedling in her garden. She wants to track the growth of the rose so she measures itsheight every week. On the third week, she finds that the rose is 10 inches tall and on the eleventh week shefinds that the rose is 14 inches tall. Assuming the rose grows linearly with time, write an equation describingthis problem in slope-intercept form. What was the height of the rose when Anne planted it?
3. Ravi hangs from a giant spring whose length is 5 m. When his child Nimi hangs from the spring its length is 2m. Ravi weighs 160 lbs. and Nimi weighs 40 lbs. Write the equation for this problem in slope-intercept form.What should we expect the length of the spring to be when his wife Amardeep, who weighs 140 lbs., hangsfrom it?
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2.8 Parallel and Perpendicular Lines
What You Will Learn
• Properties of parallel and perpendicular lines.• Determining whether lines are parallel or perpendicular• Writing equations of parallel and perpendicular lines.
Properties of Parallel and Perpendicular Lines
Let’s start by looking at a graph of two parallel lines.
The two lines will never meet because they are parallel. We can clearly see that the two lines have differenty−intercepts, more specifically 6 and -4.
How about the slopes of the lines? Are they related in any way? Because the lines never meet, they must rise at thesame rate. This means that the slopes of the two lines are the same.
Indeed, if we calculate the slopes of the lines, we find the following results.
Line A: m = 6−20−(−2) =
42 = 2
Line B: m = 0−(−4)2−0 = 4
2 = 2
For Parallel Lines: the slopes are the same, m1 = m2, and the y−intercepts are different.
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Now let’s look at a graph of two perpendicular lines.
We find that we can’t say anything about the y−intercepts. In this example, they are different, but they would be thesame if the lines intersected at the y−intercept.
Now we want to figure out if there is any relationship between the slopes of the two lines.
First of all we see that the slopes must have opposite signs, one negative and one positive.
To find the slope of line A, we pick two points on the line and draw the blue (upper) right triangle. The legs of thetriangle represent the rise and the run.
Looking at the figure m1 =ba
To find the slope of line B, we pick two points on the line and draw the red (lower) right triangle. If we look at thefigure, we see that the two triangles are identical, only rotated by 90◦.
Looking at the diagram m2 =−ab
For Perpendicular Lines: the slopes are negative reciprocals of each other. m1 =− 1m2
or m1m2 =−1
Determine Whether Lines are Parallel or Perpendicular
You can find whether lines are parallel or perpendicular by comparing the slopes of the lines. If you are given pointson the line you can find the slope using the formula. If you are given the equations of the lines, rewrite each equationin a form so that it is easy to read the slope, such as the slope-intercept form.
Example 1
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Determine whether the lines are parallel or perpendicular or neither.
a) One line passes through points (2, 11) and (-1, 2); another line passes through points (0, -4) and (-2, -10).
b) One line passes through points (-2, -7) and (1, 5); another line passes through points (4, 1) and (-8, 4).
c) One lines passes through points (3, 1) and (-2, -2); another line passes through points (5, 5) and (4, -6).
Solution
Find the slope of each line and compare them.
a) m1 =2−11−1−2 = −9
−3 = 3 and m2 =−10−(−4)−2−0 = −6
−2 = 3
The slopes are equal, so the lines are parallel.
b) m1 =5−(−7)1−(−2) =
123 = 4 and m2 =
4−1−8−4 = 3
−12 =−14
The slopes are negative reciprocals of each other, so the lines are perpendicular.
c) m1 =−2−1−2−3 = −3
−5 = 35 and m2 =
−6−54−5 = −11
−1 = 11
The slopes are not the same or negative reciprocals of each other, so the lines are neither parallel nor perpendicular.
Example 2
Determine whether the lines are parallel or perpendicular or neither.
a) Line 1: 3x+4y = 2 Line 2: 8x−6y = 5
b) Line 1: 2x = y−10 Line 2: y =−2x+5
c) Line 1: 7y+1 = 7x Line 2: x+5 = y
Solution
Write each equation in slope-intercept form.
a) Line 1 3x+4y = 2⇒ 4y =−3x+2⇒ y =−34 x+ 1
2 ⇒ slope =−34
Line 2: 8x−6y = 5⇒ 8x−5 = 6y⇒ y = 86 x− 5
6 ⇒ y = 43 x− 5
6 ⇒ slope = 43
The slopes are negative reciprocals of each other, so the lines are perpendicular to each other.
b) Line 1 2x = y−10⇒ y = 2x+10⇒ slope = 2
Line 2 y =−2x+5⇒ slope =−2
The slopes are not the same or negative reciprocals of each other, so the lines are neither parallel nor perpendicular.
c) Line 1 7y+1 = 7x⇒ 7y = 7x−1⇒ y = x− 17 ⇒ slope = 1
Line 2: x+5 = y⇒ y = x+5⇒ slope = 1
The slopes are the same so the lines are parallel.
Writing Equations of Parallel and Perpendicular lines
You can find the equations of lines that are parallel or perpendicular to each other by using what you know about therelationships of their slopes.
Example 1
Find the equation of a line that goes through the point (-2, 4) and is parallel to the line y = 12 x+5
We learned in the last section how to find the equation of a line that had a given slope and went through a point.Here, we are given a parallel line, so we can use our knowledge of parallel lines to determine that the slope weshould use is 1
2 . We can now find the equation of the line.
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Start with the slope intercept form of the line y = mx+b
Substitute the slope. y =12
x+b
Put point (−2,4) into the equation. 4 =12(−2)+b⇒ b = 5
Substitute the value of b into the equation. y =12
x+5
A similar procedure would be used for perpendicular lines, except this time the slope would be the opposite reciprocalof the given line.
Video Overview of Parallel and Perpendicular Lines
MEDIAClick image to the left for more content.
Homework
Part 1
Determine whether the lines are parallel, perpendicular or neither.
1. One line passes through points (-1, 4) and (2, 6); another line passes through points (2, -3) and (8, 1).2. One line passes through points (4, -3) and (-8, 0); another line passes through points (-1, -1) and (-2, 6).3. One line passes through points (-3, 14) and (1, -2); another line passes through points (0, -3) and (-2, 5).4. One line passes through points (3, 3) and (-6, -3); another line passes through points (2, -8) and (-6, 4).5. Line 1: 4y+ x = 8 Line 2: 12y+3x = 16. Line 1: 5y+3x = 1 Line 2: 6y+10x =−37. Line 1: 2y−3x+5 = 0 Line 2: y+6x =−3
Part 2
Find the equations of the following lines
1. Parallel to the line y = 3x−2 and through the point (5, -3)2. Perpendicular to the line y = 4
5 x+124 and throught the point (8, 1)3. Parallel to the line 5y−15 = 2x and through the point (15,7)4. Perpendicular to the line 3x−8y = 16 and through the point (-6,4)
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2.9 Graphing Linear Inequalities
What You Will Learn
• Graph linear inequalities in one variable on the coordinate plane.• Graph linear inequalities in two variables.• Solve real-world problems using linear inequalities
Graph Linear Inequalities in One Variable in the Coordinate Plane
In previous sections, we graphed inequalities in one variable on the number line. We can also graph inequalities inone variable on the coordinate plane. We just need to remember that when we graph an equation of the type x = awe get a vertical line and when we graph an equation of the type y = b we get a horizontal line.
Example 1
Graph the inequality x > 4 on the coordinate plane.
Solution
First, let’s remember what the solution to x > 4 looks like on the number line.
The solution to this inequality is the set of all real numbers x that are bigger than four but not including four. Thesolution is represented by a line.
In two dimensions we are also concerned with values of y, and the solution to x > 4 consists of all coordinate pointsfor which the value of x is bigger than four. The solution is represented by the half plane to the right of x = 4.
The line x = 4 is dashed because the equal sign is not included in the inequality and therefore points on the line arenot included in the solution.
Example 2
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Graph the inequality y≤ 6 on the coordinate plane.
Solution
The solution is all coordinate points for which the value of y is less than or equal than 6. This solution is representedby the half plane below the line y = 6.
The line y = 6 is solid because the equal sign is included in the inequality sign and the points on the line are includedin the solution.
Graph Linear Inequalities in Two Variables
The general procedure for graphing inequalities in two variables is as follows.
Step 1: Re-write the inequality in slope-intercept form y = mx+b. Writing the inequality in this form lets you knowthe direction of the inequality
Step 2 Graph the line of equation y = mx+b using your favorite method. (For example, plotting two points, usingslope and y−intercept, using y−intercept and another point, etc.). Draw a dashed line if the equal sign in not includedand a solid line if the equal sign is included.
Step 3 Shade the half plane above the line if the inequality is greater than. Shade the half plane under the line if theinequality is less than.
Example 5
Graph the inequality y≥ 2x−3.
Solution
Step 1
The inequality is already written in slope-intercept form y≥ 2x−3.
Step 2
Graph the equation y = 2x−3 using slope intercept form.
Step 3
Graph the inequality. We shade the plane above the line because y is greater than. The value 2x−3 defines the line.The line is solid because the equal sign is included.
Example 6
Graph the inequality 5x−2y > 4.
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Solution
Step 1
Rewrite the inequality in slope-intercept form.
−2y >−5x+4
y >52
x−2
Notice that the inequality sign changed direction due to division of negative sign.
Step 2
Graph the equation y > 52 x−2 using slope intercept form.
Step 3
Graph the inequality. We shade the plane below the line because the inequality in slope-intercept form is less than.The line is dashed because the equal sign in not included.
Example 7
Find the equation of this graph
First we will find the equation of the line at the border of this graph, then we will think about the shading. The linecrosses the y axis at 3, so b = 3. The slope up to the next grid point has a rise of 2 and a run of one, so the slope is2. This means the equation of the line at the border is y = 2x+3. The shading goes below the line and it is dotted,so the inequality in the graph is y≤ 2x+3.
Solve Real-World Problems Using Linear Inequalities
In this section, we see how linear inequalities can be used to solve real-world applications.
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Example 8
A pound of coffee blend is made by mixing two types of coffee beans. One type costs $9 per pound and another typecosts $7 per pound. Find all the possible mixtures of weights of the two different coffee beans for which the blendcosts $8.50 per pound or less.
Solution
Let’s apply our problem solving plan to solve this problem.
Step 1:
Let x = weight of $9 per pound coffee beans in pounds
Let y = weight of $7 per pound coffee beans in pounds
Step 2
The cost of a pound of coffee blend is given by 9x+7y.
We are looking for the mixtures that cost $8.50 or less.
We write the inequality 9x+7y≤ 8.50.
Step 3
To find the solution set, graph the inequality 9x+7y≤ 8.50.
Rewrite in slope-intercept y≤−1.29x+1.21.
Graph y =−1.29x+1.21 by making a table of values.
TABLE 2.4:
x y0 1.211 -0.082 -1.37
Step 4
Graph the inequality. The line will be solid. We shade below the line.
Notice that we show only the first quadrant of the coordinate plane because the weight values should be positive.
The blue-shaded region tells you all the possibilities of the two bean mixtures that will give a total less than or equalto $8.50.
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Example 9
Julian has a job as an appliance salesman. He earns a commission of $60 for each washing machine he sells and$130 for each refrigerator he sells. How many washing machines and refrigerators must Julian sell in order to make$1000 or more in commission?
Solution Let’s apply our problem solving plan to solve this problem.
Step 1
Let x = number of washing machines Julian sells
Let y = number of refrigerators Julian sells
Step 2
The total commission is given by the expression 60x+130y.
We are looking for total commission of $1000 or more. We write the inequality. 60x+130y≥ 1000.
Step 3
To find the solution set, graph the inequality 60x+130y≥ 1000.
Rewrite it in slope-intercept y≥−.46x+7.7.
Graph y =−.46x+7.7 by making a table of values.
TABLE 2.5:
x y0 7.72 6.784 5.86
Step 4
Graph the inequality. The line will be solid. We shade above the line.
Notice that we show only the first quadrant of the coordinate plane because dollar amounts should be positive. Also,only the points with integer coordinates are possible solutions.
Video Overview
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Homework
Graph the following inequalities on the coordinate plane.
1. x < 202. y≥−53. y≤ 4x+34. y >− x
2 −65. 3x−4y≥ 126. x+7y < 57. 6x+5y > 18. y+5≤−4x+109. x− 1
2 y≥ 510. 30x+5y < 10011. What is the equation of the following graph?
FIGURE 2.5
12. An ounce of gold costs $670 and an ounce of silver costs $13. Find all possible weights of silver and gold thatmakes an alloy that costs less than $600 per ounce.
13. A phone company charges 50 cents cents per minute during the daytime and 10 cents per minute at night.How many daytime minutes and night time minutes would you have to use to pay more than $20 over a 24hour period?
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2.10 Graphs of Absolute Value Equations
What You Will Learn
• Graph an equation with absolute value
Graphing Absolute Value Equations
Absolute value equations can be graphed in a way that is similar to graphing linear equations. By making a table ofvalues, you can get a clear picture of what an absolute value equation will look like.
Example A
Graph the solutions to y = |x|.
Solution:
Make a table of values and graph the coordinate pairs.
TABLE 2.6:
x y = |x|–2 |−2|= 2–1 |−1|= 10 |0|= 01 |1|= 12 |2|= 23 |3|= 3
The Shape of an Absolute Value Graph
Every absolute value graph will make a “V”-shaped figure. It consists of two pieces: one with a negative slope
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and one with a positive slope. The point of their intersection is called the vertex. An absolute value graph issymmetrical, meaning it can be folded in half on its line of symmetry. The function y = |x| is the parent function,or most basic function, of absolute value functions.
Example B
Graph the solutions to y = |x−1|.
Solution:
Make a table of values and plot the ordered pairs.
TABLE 2.7:
x y = |x−1|–2 |−2−1|= 3–1 |−1−1|= 20 |0−1|= 11 |1−1|= 02 |2−1|= 13 |3−1|= 2
The graph of this function is seen below in green, graphed with the parent function in red.
Notice that the green function is just the parent function shifted over. The vertex is shifted over 1 unit to the right.This is because when x= 1, |x−1|= 0 since |1−1|= 0. The vertex will always be where the value inside the absolutevalue is zero.
Graphing by Finding the Vertex
Absolute value equations can always be graphed by making a table of values. However, you can use the vertex andsymmetry to help shorten the graphing process.
Step 1: Find the vertex by determining which value of x makes the distance zero.
Step 2: Using this value as the center of the x−values, choose several values greater than this value and severalvalues less than this value.
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Example C
Graph y =−3|x+2|−1.
Solution:
Start by considering where the vertex would be by solving for when the absolute value equals zero:
0 = |x+2|⇒ x =−2
The vertex is at x =−2. Choose some points on either side of that to make a table of values.
TABLE 2.8:
x y = |x+2|+3–4 −3|−4+2|−1 =−7–3 −3|−3+2|−1 =−4–2 −3|−2+2|−1 =−1–1 −3|−1+2|−1 =−40 −3|0+2|−1 =−7
Plot the points to make the graph.
Video Overview
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Homework
In 1 – 11, graph the function.
1. y = |x+3|2. y = |x−6|3. y = |4x+2|4. y = |x−4|−35. |x−4|= y
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6. −|x−2|= y7. y = |x|−28. y = 2|x|+39. y =−4|x+2|+3
10. y =−|x+1|+5
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2.11 Fitting a Line to Data
What You Will Learn
• Make a scatter plot.• Fit a line to data and write an equation for that line.• Perform linear regression with a graphing calculator.• Solve real-world problems using linear models of scattered data.
Make a Scatter Plot
A scatter plot is a plot of all the ordered pairs in the table. This means that a scatter plot is a relation, and notnecessarily a function. Also, the scatter plot is discrete, as it is a set of distinct points. Even when we expect therelationship we are analyzing to be linear, we should not expect that all the points would fit perfectly on a straightline. Rather, the points will be “scattered” about a straight line.
Example 1
Make a scatter plot of the following ordered pairs: (0, 2), (1, 4.5), (2, 9), (3, 11), (4, 13), (5, 18), (6, 19.5)
Solution
We make a scatter plot by graphing all the ordered pairs on the coordinate axis.
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Fit a Line to Data
Notice that the points look like they might be part of a straight line, although they would not fit perfectly on astraight line. If the points were perfectly lined up it would be quite easy to draw a line through all of them and findthe equation of that line. However, if the points are “scattered”, we try to find a line that best fits the data.
You see that we can draw many lines through the points in our data set. These lines have equations that are verydifferent from each other. We want to use the line that is closest to all the points on the graph. The best candidate inour graph is the red line A. We want to minimize the sum of the distances from the point to the line of fit as you cansee in the figure below.
Finding this line mathematically is a complex process and is not usually done by hand. We usually “eye-ball” theline or find it exactly by using a graphing calculator or computer software such as Excel. The line in the graph aboveis “eye-balled,” which means we drew a line that comes closest to all the points in the scatter plot.
When we use the line of best fit we are assuming that there is a continuous linear function that will approximate thediscrete values of the scatter plot. We can use this to interpret unknown values.
Write an Equation for a Line of Best Fit
Once you draw the line of best fit, you can find its equation by using two points on the line. Finding the equation ofthe line of best fit is also called linear regression.
Caution: Make sure you don’t get caught making a common mistake. In many instances the line of best fit will notpass through many or any of the points in the original data set. This means that you can’t just use two random pointsfrom the data set. You need to use two points that are on the line.
We see that two of the data points are very close to the line of best fit, so we can use these points to find the equationof the line (1, 4.5) and (3, 11).
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Start with the slope-intercept form of a line y = mx+b.
Find the slope m = 11−4.53−1 = 6.5
2 = 3.25
Then y = 3.25x+b
Plug (3, 11) into the equation. 11 = 3.25(3)+b⇒ b = 1.25
The equation for the line that fits the data best is y = 3.25x+1.25.
Perform Linear Regression with a Graphing Calculator
Drawing a line of fit can be a good approximation but you can’t be sure that you are getting the best results becauseyou are guessing where to draw the line. Two people working with the same data might get two different equationsbecause they would be drawing different lines. To get the most accurate equation for the line, we can use a graphingcalculator. The calculator uses a mathematical algorithm to find the line that minimizes the sum of the squares.
Example 2
Use a graphing calculator to find the equation of the line of best fit for the following data (3, 12), (8, 20), (1, 7), (10,23), (5, 18), (8, 24), (11, 30), (2, 10).
Solution
Step 1 Input the data in your calculator.
Press [STAT] and choose the [EDIT] option.
Input the data into the table by entering the x values in the first column and the y values in the second column.
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Step 2 Find the equation of the line of best fit.
Press [STAT] again use right arrow to select [CALC] at the top of the screen.
Chose option number 4: LinReg(ax+b) and press [ENTER]
The calculator will display LinReg(ax+b)
Press [ENTER] and you will be given the a and b values.
Here a represents the slope and b represents the y−intercept of the equation. The linear regression line is y =2.01x+5.94.
Step 3 Draw the scatter plot.
To draw the scatter plot press [STATPLOT] [2nd] [Y=].
Choose Plot 1 and press [ENTER].
Press the On option and choose the Type as scatter plot (the one highlighted in black).
Make sure that the X list and Y list names match the names of the columns of the table in Step 1.
Choose the box or plus as the mark since the simple dot may make it difficult to see the points.
Press [GRAPH] and adjust the window size so you can see all the points in the scatter plot.
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Step 4 Draw the line of best fit through the scatter plot.
Press [Y=]
Enter the equation of the line of best fit that you just found Y1 = 2.01X +5.94
Press [GRAPH].
Solve Real-World Problems Using Linear Models of Scattered Data
In a real-world problem, we use a data set to find the equation of the line of best fit. We can then use the equation topredict values of the dependent or independent variables. The usual procedure is as follows.
a. Make a scatter plot of the given data.b. Draw a line of best fit.c. Find an equation of a line either using two points on the line or the TI-83/84 calculator.d. Use the equation to answer the questions asked in the problem.
Example 3
Gal is training for a 5 K race (a total of 5000 meters, or about 3.1 miles). The following table shows her times foreach month of her training program. Assume here that her times will decrease in a straight line with time (does thatseem like a good assumption?) Find an equation of a line of fit. Predict her running time if her race is in August.
TABLE 2.9:
Month Month number Average time (minutes)January 0 40February 1 38March 2 39April 3 38May 4 33June 5 30
Solution
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Let’s make a scatter plot of Gal’s running times. The independent variable, x, is the month number and the dependentvariable, y, is the running time in minutes. We plot all the points in the table on the coordinate plane.
Draw a line of fit.
Choose two points on the line (0, 41) and (4, 34).
Find the equation of the line.
m =34−414−0
=−74=−1
34
y =−74
x+b
41 =−72(0)+b⇒ b = 41
y =−74
x+41
In a real-world problem, the slope and y−intercept have a physical significance.
Slope =number o f minutes
month
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Since the slope is negative, the number of minutes Gal spends running a 5 K race decreased as the months pass. Theslope tells us that Gal’s running time decreases by 7
4 or 1.75 minutes per month.
The y−intercept tells us that when Gal started training, she ran a distance of 5 K in 41 minutes, which is just anestimate, since the actual time was 40 minutes.
The problem asks us to predict Gal’s running time in August. Since June is assigned to month number five, thenAugust will be month number seven. We plug x = 7 into the equation of the line of best fit.
y =−74(7)+41 =−49
4+41 =−49
4+
1644
=1154
= 2834
The equation predicts that Gal will be running the 5 K race in 28.75 minutes.
In this solution, we eye-balled a line of best fit. Using a graphing calculator, we found this equation for a line of fity =−2.2x+43.7.
If we plug x = 7 in this equation, we get y = −2.2(7)+ 43.7 = 28.3. This means that Gal ran her race in 28.3minutes. You see that the graphing calculator gives a different equation and a different answer to the question. Thegraphing calculator result is more accurate but the line we drew by hand still gives a good approximation to theresult.
Example 4
Baris is testing the burning time of “BriteGlo” candles. The following table shows how long it takes to burn candlesof different weights. Assume it’s a linear relation and we can then use a line to fit the data. If a candle burns for 95hours, what must be its weight in ounces?
TABLE 2.10: Candle Burning Time Based on Candle Weight
Candle weight (oz) Time (hours)2 153 204 355 3610 8016 10022 12026 180
Solution
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Let’s make a scatter plot of the data. The independent variable, x, is the candle weight in ounces and the dependentvariable, y, is the time in hours it takes the candle to burn. We plot all the points in the table on the coordinate plane.
Then we draw the line of best fit.
Now pick two points on the line (0,0) and (30, 200).
Find the equation of the line:
m =20030
=203
y =203
x+b
0 =203(0)+b⇒ b = 0
y =203
x
In this problem the slope is burning time divided by candle weight. A slope of 203 = 6 2
3 tells us for each extra ounceof candle weight, the burning time increases by 6 2
3 hours.
A y−intercept of zero tells us that a candle of weight 0 oz will burn for 0 hours.
The problem asks for the weight of a candle that burns 95 hours. We are given the value of y = 95. We need to usethe equation to find the corresponding value of x.
y =203
x⇒=203
x⇒ x =28520
=574
= 1414
A candle that burns 95 hours weighs 14.25 oz.
The graphing calculator gives the linear regression equation as y = 6.1x+5.9 and a result of 14.6 oz.
Notice that we can use the line of best fit to estimate the burning time for a candle of any weight.
Video Overview
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Homework
For each data set, draw the scatter plot and find the equation of the line of best fit for the data set by hand.
1. (57, 45) (65, 61) (34, 30) (87, 78) (42, 41) (35, 36) (59, 35) (61, 57) (25, 23) (35, 34)2. (32, 43) (54, 61) (89, 94) (25, 34) (43, 56) (58, 67) (38, 46) (47, 56) (39, 48)3. (12, 18) (5, 24) (15, 16) (11, 19) (9, 12) (7, 13) (6, 17) (12, 14)4. (3, 12) (8, 20) (1, 7) (10, 23) (5, 18) (8, 24) (2, 10)
For each data set, use a graphing calculator to find the equation of the line of best fit.
1. (57, 45) (65, 61) (34, 30) (87, 78) (42, 41) (35, 36) (59, 35) (61, 57) (25, 23) (35, 34)2. (32, 43) (54, 61) (89, 94) (25, 34) (43, 56) (58, 67) (38, 46) (47, 56) (95, 105) (39, 48)3. (12, 18) (3, 26) (5, 24) (15, 16) (11, 19) (0, 27) (9, 12) (7, 13) (6, 17) (12, 14)4. Shiva is trying to beat the samosa eating record. The current record is 53.5 samosas in 12 minutes. The
following table shows how many samosas he eats during his daily practice for the first week of his training.Will he be ready for the contest if it occurs two weeks from the day he started training? What are the meaningsof the slope and the y−intercept in this problem?
TABLE 2.11:
Day No. of Samosas1 302 343 364 365 406 437 45
1. Nitisha is trying to find the elasticity coefficient of a Superball. She drops the ball from different heights andmeasures the maximum height of the resulting bounce. The table below shows her data. Draw a scatter plotand find the equation. What is the initial height if the bounce height is 65 cm? What are the meanings of theslope and the y−intercept in this problem?
TABLE 2.12:
Initial height (cm) Bounce height (cm)30 2235 2640 2945 3450 38
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TABLE 2.12: (continued)
Initial height (cm) Bounce height (cm)55 4060 4565 5070 52
1. The following table shows the median California family income from 1995 to 2002 as reported by the USCensus Bureau. Draw a scatter plot and find the equation. What would you expect the median annual incomeof a Californian family to be in year 2010? What are the meanings of the slope and the y−intercept in thisproblem?
TABLE 2.13:
Year Income1995 538071996 552171997 552091998 554151999 631002000 632062001 637612002 65766
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2.12 Problem Solving with Linear Models
What You Will Learn
• How to use linear regression to model real world situations
A model is an equation that best describes the data graphed in the scatter plot. Sometimes the data fits a linearmodel, often the situation is more complicated and other types of models are necessary. This is a very importantsubject in mathematics. It is often useful to have a mathematical model to describe a real life situation. If you knowmany different types of math equations, you can model many of the situations that actually occur in the world.
Example A
Dana heard something very interesting at school. Her teacher told her that if you divide the circumference of acircle by its diameter you always get the same number. She tested this statement by measuring the circumferenceand diameter of several circular objects. The following table shows her results.
From this data, estimate the circumference of a circle whose diameter is 12 inches.
Solution:
Begin by creating a scatter plot and drawing the line of best fit.
TABLE 2.14: Diameter and Circumference of Various Objects
Object Diameter (inches) Circumference (inches)Table 53 170Soda can 2.25 7.1Cocoa tin 4.2 12.6Plate 8 25.5Straw 0.25 1.2Propane tank 13.3 39.6Hula hoop 34.25 115
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Find the equation of the line of best fit using points (0.25, 1.2) and (8, 25.5).
Slope m =25.5−128−0.25
=24.37.75
= 3.14
amp;y = 3.14x+b
1.2 = 3.14(0.25)+b⇒ b = 0.42
Equation y = 3.14x+0.42
Diameter = 12 inches⇒ y = 3.14(12)+0.42 = 38.1 inches
In this problem, the slope = 3.14. This number should be very familiar to you—it is the number pi rounded tothe hundredths place. Theoretically, the circumference of a circle divided by its diameter is always the same and itequals 3.14 or π.
Example B
Using Dana’s data from Example A, estimate the circumference of a circle whose diameter is 25 inches.
Solution:
The equation y = 3.14x+ 0.42 of the relationship between diameter and circumference from Example A applieshere.
Diameter = 25 inches⇒ y = 3.14(25)+0.42 = 78.92 inches
A circle with a diameter of 25 inches will have a circumference that is approximately 78.92 inches.
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Example C
Using Dana’s data from Example A, estimate the circumference of a circle whose diameter is 60 inches.
Solution:
The equation y = 3.14x+ 0.42 of the relationship between diameter and circumference from Example A applieshere.
Diameter = 60 inches⇒ y = 3.14(60)+0.42 = 188.82 inches
A circle with a diameter of 60 inches will have a circumference that is approximately 188.82 inches.
Guided Practice
A cylinder is filled with water to a height of 73 centimeters. The water is drained through a hole in the bottom of thecylinder and measurements are taken at two-second intervals. The table below shows the height of the water levelin the cylinder at different times.
TABLE 2.15: Water Level in Cylinder at Various Times
Time (seconds) Water level (cm)0.0 732.0 63.94.0 55.56.0 47.28.0 40.010.0 33.412.0 27.414.0 21.916.0 17.118.0 12.920.0 9.422.0 6.324.0 3.926.0 2.028.0 0.730.0 0.1
Find the water level at 15 seconds.
Solution:
Begin by graphing the scatter plot. As you can see below, a straight line does not fit the majority of this data. Wewill learn other types of models later in the year that will better describe this situation.
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Video Overview
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Homework
1. What is a mathematical model?2. What is linear modeling? What are the options to determine a linear model?3. Can you think of a real world situation that could be descibed by a linear model?
Use the Life Expectancy table below to answer the questions.
4. Make a scatter plot of the data.5. Use a line of best fit to estimate the life expectancy of a person born in 1955.6. Use linear interpolation to estimate the life expectancy of a person born in 1955.7. Use a line of best fit to estimate the life expectancy of a person born in 1976.8. Use linear interpolation to estimate the life expectancy of a person born in 1976.9. Use a line of best fit to estimate the life expectancy of a person born in 2012.
10. Use linear extrapolation to estimate the life expectancy of a person born in 2012.11. Which method gives better estimates for this data set? Why?
TABLE 2.16:
Birth Year Life expectancy in years1930 59.71940 62.91950 68.21960 69.71970 70.81980 73.71990 75.42000 77
The table below lists the high temperature for the first day of each month in 2006 in San Diego, California (WeatherUnderground). Use this table to answer the questions.
12. Draw a scatter plot of the data.13. Use a line of best fit to estimate the temperature in the middle of the 4th month (month 4.5).14. Use linear interpolation to estimate the temperature in the middle of the 4th month (month 4.5).15. Use a line of best fit to estimate the temperature for month 13 (January 2007).16. Use linear extrapolation to estimate the temperature for month 13 (January 2007).17. Which method gives better estimates for this data set? Why?
TABLE 2.17:
Month number Temperature (F)1 632 663 614 645 716 787 888 789 81
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TABLE 2.17: (continued)
Month number Temperature (F)10 7511 6812 69
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CHAPTER 3 Unit 3 - Systems of LinearEquations
Chapter Outline3.1 SOLVING LINEAR SYSTEMS BY GRAPHING
3.2 SOLVING LINEAR SYSTEMS BY SUBSTITUTION
3.3 SOLVING LINEAR SYSTEMS BY COMBINATION (ELIMINATION)
3.4 SPECIAL TYPES OF LINEAR SYSTEMS
3.5 APPLICATIONS OF LINEAR SYSTEMS
3.6 SYSTEMS OF LINEAR INEQUALITIES
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3.1 Solving Linear Systems by Graphing
What You Will Learn
• Determine whether an ordered pair is a solution to a system of equations.• Solve a system of equations graphically.• Solve a system of equations graphically with a graphing calculator.
Determine Whether an Ordered Pair is a Solution to a System of Equations
A linear system of equations consists of a set of equations that must be solved together. We will be looking for avalue for x and y that will make BOTH of the equations true.
Consider the following system of equations.
y = x+2
y =−2x+1
Since the two lines are in a system we deal with them together by graphing them on the same coordinate axes. Thelines can be graphed using your favorite method. Let’s graph by making a table of values for each line.
Line 1 y = x+2
TABLE 3.1:
x y0 21 3
Line 2 y =−2x+1
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TABLE 3.2:
x y0 11 -1
A solution for a single equation is any point that lies on the line for that equation. A solution for a system ofequations is any point that lies on both lines in the system.
For Example
• Point A is not a solution to the system because it does not lie on either of the lines.• Point B is not a solution to the system because it lies only on the blue line but not on the red line.• Point C is a solution to the system because it lies on both lines at the same time.
In particular, this point marks the intersection of the two lines. It solves both equations, so it solves the system.For a system of equations, the geometrical solution is the intersection of the two lines in the system. The algebraicsolution is the ordered paid that solves both equations.
You can confirm the solution by substituting it into the system of equations, and confirming that the solution worksin each equation.
Example 1
Determine which of the points (1, 3), (0, 2) or (2, 7) is a solution to the following system of equations.
y = 4x−1
y = 2x+3
Solution
To check if a coordinate point is a solution to the system of equations, we substitute each of the x and y values intothe equations to see if they work.
Point (1, 3)
y = 4x−1
3? = ?4(1)−1
3 = 3
The solution checks.
y = 2x+3
3? = ?2(1)+3
3 6= 5
The solution does not check.
Point (1, 3) is on line y = 4x−1 but it is not on line y = 2x+3 so it is not a solution to the system.
Point (0, 2)
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y = 4x−1
2? = ?4(0)−1
2 6=−1
The solution does not check.
Point (0, 2) is not on line y = 4x− 1 so it is not a solution to the system. Note that it is not necessary to check thesecond equation because the point needs to be on both lines for it to be a solution to the system.
Point (2, 7)
y = 4x−1
7? = ?4(2)−1
7 = 7
The solution checks.
y = 2x+3
7? = ?2(2)+3
7 = 7
The solution checks.
Point (2, 7) is a solution to the system since it lies on both lines.
Answer The solution to the system is point (2, 7).
Determine the Solution to a Linear System by Graphing
The solution to a linear system of equations is the point which lies on both lines. In other words, the solution is thepoint where the two lines intersect.
We can solve a system of equations by graphing the lines on the same coordinate plane and reading the intersectionpoint from the graph.
This method most often offers only approximate solutions. It is exact only when the x and y values of the solutionare integers. However, this method it is a great at offering a visual representation of the system of equations anddemonstrates that the solution to a system of equations is the intersection of the two lines in the system.
Example 2
(The equations are in slope-intercept form)
Solve the following system of equations by graphing.
y = 3x−5
y =−2x+5
Solution
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Graph both lines on the same coordinate axis using any method you like.
In this case, let’s make a table of values for each line.
Line 1 y = 3x−5
TABLE 3.3:
x y1 -22 1
Line 2 y =−2x+5
TABLE 3.4:
x y1 32 1
Answer The solution to the system is given by the intersection point of the two lines. The graph shows that the linesintersect at point (2, 1). So the solution is x = 2,y = 1 or (2, 1).
Example 3
(The equations are in standard form)
Solve the following system of equations by graphing
2x+3y = 6
4x− y =−2
Solution
Graph both lines on the same coordinate axis using your method of choice.
Here we will graph the lines by finding the x− and y−intercepts of each of the lines.
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Line 1 2x+3y = 6
x−intercept set y = 0⇒ 2x = 6⇒ x = 3 which results in point (3, 0).
y−intercept set x = 0⇒ 3y = 6⇒ y = 2 which results in point (0, 2).
Line 2 −4x+ y = 2
x−intercept: set y = 0⇒−4x = 2⇒ x =−12 which results in point
(−1
2 ,0).
y−intercept: set x = 0⇒ y = 2 which results in point (0, 2)
Answer The graph shows that the lines intersect at point (0, 2). Therefore, the solution to the system of equations isx = 0, y = 2.
Example 4:
Solve the following system by graphing.
y = 3
x+ y = 2
Line 1 y = 3 is a horizontal line passing through point (0, 3).
Line 2 x+ y = 2
x−intercept: (2, 0)
y−intercept: (0, 2)
Answer The graph shows that the solution to this system is (-1, 3) x =−1, y = 3.
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These examples are great at demonstrating that the solution to a system of linear equations means the point at whichthe lines intersect. This is, in fact, the greatest strength of the graphing method because it offers a very visualrepresentation of system of equations and its solution. You can see, however, that determining a solution from agraph would require very careful graphing of the lines, and is really only practical when you are certain that thesolution gives integer values for x and y. In most cases, this method can only offer approximate solutions to systemsof equations. For exact solutions other methods are necessary.
Solving a System of Equations Using the Graphing Calculator
A graphing calculator can be used to find or check solutions to a system of equations. In this section, you learnedthat to solve a system graphically, you must graph the two lines on the same coordinate axes and find the point ofintersection. You can use a graphing calculator to graph the lines as an alternative to graphing the equations by hand.
Example 6
Solve the following system of equations using a graphing calculator.
x−3y = 4
2x+5y = 8
In order to input the equations into the calculator, they must be written in slope-intercept form (i.e., y = mx+ bform), or at least you must isolate y.
x−3y = 4 y =13
x− 43
⇒
2x+5y = 8 y =−25
x− 85
Press the [y=] button on the graphing calculator and enter the two functions as:
Y1 =x3− 4
3
Y2 =−2x3− 8
5
Now press [GRAPH]. The window below is set to −5≤ x≤ 10 and −5≤ x≤ 5.
The first screen below shows the screen of a TI-83 family graphing calculator with these lines graphed.
There are a few different ways to find the intersection point.
Option 1 Use [TRACE] and move the cursor with the arrows until it is on top of the intersection point. The valuesof the coordinate point will be on the bottom of the screen. The second screen above shows the values to beX = 4.0957447 and Y = .03191489.
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Use the [ZOOM] function to zoom into the intersection point and find a more accurate result. The third screenabove shows the system of equations after zooming in several times. A more accurate solution appears to be X = 4and Y = 0.
Option 2 Look at the table of values by pressing [2nd] [GRAPH]. The first screen below shows a table of valuesfor this system of equations. Scroll down until the values for and are the same. In this case this occurs at X = 4 andY = 0.
Use the [TBLSET] function to change the starting value for your table of values so that it is close to the intersectionpoint and you don’t have to scroll too long. You can also improve the accuracy of the solution by taking smallervalues of Table 1.
Option 3 Using the [2nd] [TRACE] function gives the screen in the second screen above.
Scroll down and select intersect.
The calculator will display the graph with the question [FIRSTCURVE]? Move the cursor along the first curve untilit is close to the intersection and press [ENTER].
The calculator now shows [SECONDCURVE]?
Move the cursor to the second line (if necessary) and press [ENTER].
The calculator displays [GUESS]?
Press [ENTER] and the calculator displays the solution at the bottom of the screen (see the third screen above).
The point of intersection is X = 4 and Y = 0.
Notes:
• When you use the “intersect” function, the calculator asks you to select [FIRSTCURVE]? and [SECOND-CURVE]? in case you have more than two graphs on the screen. Likewise, the [GUESS]? is requested in casethe curves have more than one intersection. With lines you only get one point of intersection, but later in yourmathematics studies you will work with curves that have multiple points of intersection.
• Option 3 is the only option on the graphing calculator that gives an exact solution. Using trace and table giveyou approximate solutions.
Video Overview of Solving Systems by Graphing
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Video Overview of Solving Systems using the Graphing Calculator
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Homework
Determine which ordered pair satisfies the system of linear equations.
1. y = 3x−2y =−x
a. (1, 4)b. (2, 9)c.(1
2 ,−12
)2. y = 2x−3
y = x+5
a. (8, 13)b. (-7, 6)c. (0, 4)
3. 2x+ y = 85x+2y = 10
a. (-9, 1)b. (-6, 20)c. (14, 2)
4. 3x+2y = 6y = x
2 −3
a.(3,−3
2
)b. (−4,3)c.(1
2 ,4)
Solve the following systems using the graphing method.
1. y = x+3y =−x+3
2. y = 3x−6y =−x+6
3. 2x = 4y =−3
4. y =−x+5− x+ y = 1
5. x+2y = 85x+2y = 0
6. 3x+2y = 124x− y = 5
7. 5x+2y =−4x− y = 2
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8. 2x+4 = 3yx−2y+4 = 0
9. y = x2 −3
2x−5y = 510. y = 4
x = 8−3y
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3.2 Solving Linear Systems by Substitution
What You Will Learn
• Solve systems of equations with two variables by substituting for either variable.
Solving Linear Systems Using Substitution
Graphing two equations to find the solution can be awkward, especially if the answer is a fraction. The method ofsubstitution is often a better choice.
Example 1
Solve the system
2x+3y = 6
−4x+ y = 2
The first step is to solve one of the equations for x or y. y is already by itself in the second equation, so that is thebest place to start.
−4x+ y = 2 Add 4x to both sides.
y = 2+4x
Now substitute this equation in place of y in the first equation.
2x+3(2+4x) = 6 Distribute the 3.
2x+6+12x = 6 Collect like terms.
14x+6 = 6 Subtract 6 from both sides.
14x = 0
x = 0
Substitute back into our expression for y.
y = 2+4 ·0 = 2
The solution is (x = 0,y = 2)
Next, consider a more complicated example. In the following example the solution gives a fractional answer, so itwould be difficult to solve by graphing.
Example 2
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Solve the system
2x+2y = 3
x−2y =−1
Again, we start by looking to isolate one variable in either equation. The second equation has x by itself, so we willstart there.
Solve the second equation for x
x−4y =−1 Add 4y to both sides.
x =−1+4y
Substitute this expression into the first equation and solve.
2(−1+4y)+2y Distribute the 2
−2+8y+2y = 3 Collect like terms.
−2+10y = 3 Add 2 to both sides
10y = 5 Divide both sides by 10
y =12
Substitute this back into the equation that was solved for x
x =−1+4y
x =−1+4(12)
x =−1+2
x = 1
So our solution is, x = 1,y = 12 .
Video Overview, Example 1
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Video Overview, Example 2
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Homework
Solve the following systems
1)
x+2y = 9
3x+5y = 20
2)
x−3y = 10
2x+ y = 13
3)
5x+2y =−1
2x− y =−13
4)
4x+3y =−5
x = y−3
5)
7x+4y = 13
x+ y = 1
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3.3 Solving Linear Systems by Combination(Elimination)
What You Will Learn
• Solve systems by combining the two equations (or multiples of the two equations) to eliminate a variable.
Solving Systems Without Multiplying
Objective
Solve systems by adding the two equations together to eliminate a variable.
Example A
Solve the system using Linear Combination:
2x−3y =−9
5x+3y = 30
Solution: Notice that the coefficients of the y terms are opposites. When we add the two equations together, theseterms will be eliminated because their sum is zero.
2x−��3y =−9
+5x+��3y = 30
7x = 21
Now we can solve for x:
7x = 21
x = 3
Now that we have found x, we can plug this value into either equation to find y:
2(3)−3y =−9 5(3)+3y = 30
6−3y =−9 15+3y = 30
−3y =−15 or 3y = 15
y = 5 y = 5
The solution is therefore: (3, 5).
Remember to check your answer:
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2(3)−3(5) = 6−15 =−9
5(3)+3(5) = 15+15 = 30 X
Example B
Solve the system using Linear Combination:
2x+ y = 2
−3x+ y =−18
Solution: In this case the coefficients of the y terms are the same, not opposites. One way to solve this systemusing linear combination would be to subtract the second equation from the first instead of adding it. Sometimessubtraction results in more errors, however, particularly when negative numbers are involved. Instead of subtracting,multiply the second equation by -1 and then add them together.
−1(−3x+ y =−18)
3x− y = 18
Essentially, we changed all of the signs of the terms in this equation.
Now we can add the two equations together to eliminate y:
2x+�y = 2
+3x−�y = 18
5x = 20
Now we can solve for x:
5x = 20
x = 4
Now that we have found x, plug this value into either equation to find y:
2(4)+ y = 2 −3(4)+ y =−18
8+ y = 2 or −12+ y =−18
y =−6 y =−6
The solution is therefore: (4, -6).
Remember to check your answer:
2(4)+(−6) = 8−6 = 2
−3(4)+(−6) =−12−6 =−18 X
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Solving Systems by Multiplying One Equation to Cancel a Variable
Objective
Solve systems using linear combinations by multiplying one equation by a constant to eliminate a variable.
Example
Solve the system using linear combinations:
4x+ y = 0
x−3y = 26
Solution: Here we have coefficients of y that are opposite signs (one is positive and one is negative). We canget opposite values if we multiple the first equation by 3. Be careful; make sure you multiply the entire equation,including the constant, by 3:
3(4x+ y = 0)
12x+3y = 0
Now we can use this new equation in our system to eliminate y and solve for x:
12x+��3y = 0
+ x−��3y = 26
13x = 26
x = 2
Now, find y:
4(2)+ y = 0
8+ y = 0
y =−8
Solution: (2, -8)
Check your answer:
4(2)+(−8) = 8−8 = 0
(2)−3(−8) = 2+24 = 26 X
∗ Note that we could have used the other equation to find yin the final step.∗∗ From the beginning, we could have multiplied the second equation by -4 instead to cancel out the xvariable.
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Solving Systems by Multiplying Both Equations to Cancel a Variable
Objective
Solve linear systems using linear combinations in which both equations must be multiplied by a constant to cancel avariable.
Example A
Solve the system using linear combinations:
2x−5y = 15
3x+7y = 8
Solution: In this problem we cannot simply multiply one equation by a constant to get opposite coefficients for oneof the variables. Here we will need to identify the least common multiple of the coefficients of one of the variablesand use this value to determine what to multiple each equation by. If we look at the coefficients of x, 2 and 3, theleast common multiple of these numbers is 6. So, we want to have the coefficients of x be 6 and -6 so that they areopposites and will cancel when we add the two equations together. In order to get coefficients of 6 and -6 we canmultiply the first equation by 3 and the second equation by -2 (it doesnt matter which one we make negative.)
3(2x−5y = 15) ⇒ ��6x−15y = 45
−2(3x+7y = 8) + −��6x−14y =−16
−29y = 29
y =−1
Now find x:
2(x)−5(−1) = 15
2x+5 = 15
2x = 10
x = 5
Solution: (5, -1)
Check your answer:
2(5)−5(−1) = 10+5 = 15
3(5)+7(−1) = 15−7 = 8 X
∗ This problem could also be solved by eliminating the y variables first. To do this, find the least common multipleof the coefficients of y, 5 and 7. The least common multiple is 35, so we would multiply the first equation by 7 andthe second equation by 5. Since one of them is already negative, we dont have to multiply by a negative.
Example B
Solve the system using linear combinations:
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7x+20y =−9
−2x−3y = 8
Solution: The first step is to decide which variable to eliminate. Either one can be eliminated but sometimes it ishelpful to look at what we need to multiply by to eliminate each one and determine which is easier to eliminate.In general, it is easier to work with smaller numbers so in this case it makes sense to eliminate x first. The LeastCommon Multiple (LCM) of 7 and 2 is 14. To get 14 as the coefficient of each term, we need to multiple the firstequation by 2 and the second equation by 7:
2(7x+20y =−9) ⇒ ��14x+40y =−18
7(−2x−3y = 8) −��14x−21y = 56
19y = 38
y = 2
7x+20(2) =−9
7x+40 =−9
7x =−49
x =−7
Video Overview
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Homework
Solve the following systems using linear combinations.
1)
x+ y = 2
2x− y =−5
2)
−2x−7y = 15
2x+ y = 9
3)
x− y = 27
2x+ y = 9
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4)
x−7y = 27
2x+ y = 9
5)
x+3y = 31
3x−5y =−5
6)
10x+ y =−6
−7x−5y =−13
7)
2x+4y = 18
x−5y = 9
8)
2x+6y = 8
3x+2y =−23
9)
12x− y = 2
2x+5y = 21
10)
2x+4y = 24
−3x−2y =−26
11)
3x+2y = 19
5x+4y = 23
12)
3x−9y = 13
x−3y = 7
13)
3x+2y =−3
−6x−5y = 4
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14)
5x+4y = 12
3x+5y = 15
15)
3x+5y = 7
5x+4y =−10
16)
2x+3y = 13
5x+2y =−6
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3.4 Special Types of Linear Systems
What You Will Learn
• Identify linear systems by the number of solutions
Types of Solutions to Linear Systems
Solutions to a system of linear equations can have several forms:
• One solution• No solutions• An infinite number of solutions
All of the systems we have looked at so far have one solution This is because the lines crossed at a single pointwhen they were graphed. There are other possibilities, however, and this section will consider those.
Parallel Lines
A system with parallel lines will have no solutions.
Remember that parallel lines have the same slope. When graphed, the lines will have the same steepness withdifferent y−intercepts. Therefore, parallel lines will never intersect, so they have no solution.
Algebraically, a system with no solutions looks like this when solved.
{4y = 5−3x6x+8y = 7
The first equation in this system is “almost” solved for y. Substitution would be appropriate to solve this system.
{4y = 5−3x6x+8y = 7
→
{y = 5
4 −34 x
6x+8y = 7
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Using the Substitution Property, replace the y−variable in the second equation with its algebraic expression inequation #1.
6x+8(
54− 3
4x)= 7
Apply the Distributive Property. 6x+10−6x = 7
Add like terms. 10 = 7
You have solved the equation correctly, yet the answer does not make sense.
When solving a system of parallel lines, the final equation will be untrue.
Because 10 6= 7 and you have done your math correctly, you can say this system has “no solutions.”
Coincident Lines
Coincident lines are lines with the same slope and y−intercept. The lines completely overlap, the graph will looklike the graph of one line.
When solving a system involving coincident lines, the solution has the following result.
{x+ y = 33x+3y = 9
Multiply the first equation by –3:
{−3(x+ y = 3)3x+3y = 9
→
{−3x−3y =−93x+3y = 9
Add the equations together.
0 = 0
There are no variables left and you KNOW you did the math correctly. However, this is a true statement.
When solving a system of coincident lines, the resulting equation will be without variables and the statement will betrue. You can conclude the system has an infinite number of solutions.
Practice Exercises
Determine whether the following system of linear equations has zero, one, or infinitely many solutions:
{2y+6x = 20y =−3x+7
What kind of system is this?
Solution:
It is easier to compare equations when they are in the same form. We will rewrite the first equation in slope-interceptform.
2y+6x = 20⇒ y+3x = 10⇒ y =−3x+10
Since the two equations have the same slope, but different y-intercepts, they are parallel lines. Parallel lines neverintersect, so they have no solutions.
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Video Overview
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Homework
Find the solution of each system of equations using the method of your choice. Please state whether the system hasone solution, no solutions or an infinite number of solutions.
1.
2x−5y = 2
4x+ y = 5
2.
3x5+ y = 3
1.2x+2y = 6
3.
3x−4y = 13
y =−3x−7
4.
3x−3y = 3
x− y = 1
5.
0.5x− y = 30
0.5x− y =−30
6.
4x−2y =−2
3x+2y =−12
7.
3x+2y = 4
−2x+2y = 24
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8.
5x−2y = 3
2x−3y = 10
9.
3x−4y = 13
y =−3x− y
10.
5x−4y = 1
−10x+8y =−30
11.
4x+5y = 0
3x = 6y+4.5
12.
−2y+4x = 8
y−2x =−4
13.
x− y2=
32
3x+ y = 6
14.
0.05x+0.25y = 6
x+ y = 24
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3.5 Applications of Linear Systems
What You Will Learn
• How to solve real-world applications that involve linear systems.
Introduction
We will start this section by writing a system of equations from algebraic sentences.
Example A
The sum of two numbers is 9 and the difference of the same two numbers is -1. What are the two numbers?
We will use the variables x and y to stand for the two numbers. This means we can write the following equations.
x+ y = 9
x− y =−1
If we add these two equations together, the y’s cancel out and we can solve for x
2x = 8
x = 4
Put x back into the first equation and solve for y.
5+ y = 9
y = 4
Solving Real World Applications with Linear Systems
These problems can be challenging to solve. The hard part is usually setting up the equations after you read theproblem. It is important to think about what you trying to find, to carefully define variables for those things and toset up equations using those variables. You can solve these equations using graphing, substitution or combination.Here’s an example.
Example B
The movie rental store CineStar offers customers two choices. Customers can pay a yearly membership of $45 andthen rent each movie for $2 or they can choose not to pay the membership fee and rent each movie for $3.50. Howmany movies would you have to rent before the membership becomes the cheaper option?
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Solution
Let’s translate this problem into algebra. Since there are two different options to consider, we can write two differentequations and form a system.
The choices are “membership” and “no membership.” We’ll call the number of movies you rent x and the total costof renting movies for a year y.
TABLE 3.5:
flat fee rental fee totalmembership $45 2x y = 45+2xno membership $0 3.50x y = 3.5x
The flat fee is the dollar amount you pay per year and the rental fee is the dollar amount you pay when you rent amovie. For the membership option the rental fee is 2x, since you would pay $2 for each movie you rented; for theno membership option the rental fee is 3.50x, since you would pay $3.50 for each movie you rented.
Our system of equations is:
y = 45+2x
y = 3.50x
Here’s a graph of the system:
Now we need to find the exact intersection point. Since each equation is already solved for y, we can easily solvethe system with substitution. Substitute the second equation into the first one:
y = 45+2x
⇒ 3.50x = 45+2x⇒ 1.50x = 45⇒ x = 30 movies
y = 3.50x
You would have to rent 30 movies per year before the membership becomes the better option.
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This example shows a real situation where a consistent system of equations is useful in finding a solution. Rememberthat for a consistent system, the lines that make up the system intersect at single point. In other words, the lines arenot parallel or the slopes are different.
In this case, the slopes of the lines represent the price of a rental per movie. The lines cross because the price ofrental per movie is different for the two options in the problem.
Example C
A baker sells plain cakes for $7 and decorated cakes for $11. On a busy Saturday the baker started with 120 cakes,and sold all but three. His takings for the day were $991. How many plain cakes did he sell that day, and how manywere decorated before they were sold?
Solution:
TABLE 3.6:
plain cakes decorated cakes totalCakes sold p d 120−3 = 117Cost of cakes 7p 11d $991
The system of equations that describes this problem is:
p+d = 117
7p+11d = 991
Let’s solve this system by substituting the second equation into the first equation:
p+d = 117⇒ p = 117−d
7p+11d = 991⇒ 7(117−d)+11d = 991
⇒ 819−7d +11d = 991⇒ 819+4d = 991
4d = 172⇒ d = 43
We can substitute d into the the first equation to solve for p.
p = 117−d = 117− (43) = 74
The baker sold 74 plain cakes and 43 decorated cakes.
Video Overview
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Homework
1. The sum of two number is 3 and the difference of the same two numbers is 7. What are the numbers?2. Juan is considering two cell phone plans. The first company charges $120 for the phone and $30 per month
for the calling plan that Juan wants. The second company charges $40 for the same phone but charges $45 permonth for the calling plan that Juan wants. After how many months would the total cost of the two plans bethe same?
3. Jamal placed two orders with an internet clothing store. The first order was for 13 ties and 4 pairs ofsuspenders, and totaled $487. The second order was for 6 ties and 2 pairs of suspenders, and totaled $232.The bill does not list the per-item price, but all ties have the same price and all suspenders have the same price.What is the cost of one tie and of one pair of suspenders?
4. Teri bought games and cakes. Games cost $12 each while cakes are only $3 each. If Teri spent $93 and onlyended up with 16 items, how many games and cakes did Teri buy?
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3.6 Systems of Linear Inequalities
What You Will Learn
• How to solve a system of linear inequalities by graphing
Solving Systems of Inequalities by Graphing
In the last chapter you learned how to graph a linear inequality in two variables. To do that, you graphed the equationof the straight line on the coordinate plane. The line was solid for ≤ or ≥ signs (where the equals sign is included),and the line was dashed for <or >signs (where the equals sign is not included). Then you shaded above the line (ifthe inequality began with y > or y≥) or below the line (if it began with y < or y≤).
In this section, we’ll see how to graph two or more linear inequalities on the same coordinate plane. The inequalitiesare graphed separately on the same graph, and the solution for the system is the common shaded region betweenall the inequalities in the system. One linear inequality in two variables divides the plane into two half-planes. Asystem of two or more linear inequalities can divide the plane into more complex shapes.
Let’s start by solving a system of two inequalities.
Example A
Solve the following system:
2x+3y≤ 18
x−4y≤ 12
Solution
Solving systems of linear inequalities means graphing and finding the intersections. So we graph each inequality,and then find the intersection regions of the solution.
First, let’s rewrite each equation in slope-intercept form. (Remember that this form makes it easier to tell whichregion of the coordinate plane to shade.) Our system becomes
3y≤−2x+18 y≤−23
x+6
⇒
−4y≤−x+12 y≥ x4−3
Notice that the inequality sign in the second equation changed because we divided by a negative number!
For this first example, we’ll graph each inequality separately and then combine the results.
Here’s the graph of the first inequality:
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The line is solid because the equals sign is included in the inequality. Since the inequality is less than or equal to,we shade below the line.
And here’s the graph of the second inequality:
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The line is solid again because the equals sign is included in the inequality. We now shade above the line because yis greater than or equal to.
When we combine the graphs, we see that the blue and red shaded regions overlap. The area where they overlap isthe area where both inequalities are true. Thus that area (shown below in purple) is the solution of the system.
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The kind of solution displayed in this example is called unbounded, because it continues forever in at least onedirection (in this case, forever upward and to the left).
Example B
There are also situations where a system of inequalities has no solution. For example, let’s solve this system.
y≤ 2x−4
y > 2x+6
Solution
We start by graphing the first line. The line will be solid because the equals sign is included in the inequality. Wemust shade downwards because y is less than.
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Next we graph the second line on the same coordinate axis. This line will be dashed because the equals sign is notincluded in the inequality. We must shade upward because y is greater than.
It doesn’t look like the two shaded regions overlap at all. The two lines have the same slope, so we know they areparallel; that means that the regions indeed won’t ever overlap since the lines won’t ever cross. So this system ofinequalities has no solution.
But a system of inequalities can sometimes have a solution even if the lines are parallel. For example, what happensif we swap the directions of the inequality signs in the system we just graphed?
To graph the system
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y≥ 2x−4
y < 2x+6
,
we draw the same lines we drew for the previous system, but we shade upward for the first inequality and downwardfor the second inequality. Here is the result:
You can see that this time the shaded regions overlap. The area between the two lines is the solution to the system.
Graph a System of More Than Two Linear Inequalities
When we solve a system of just two linear inequalities, the solution is always an unbounded region—one thatcontinues infinitely in at least one direction. But if we put together a system of more than two inequalities, sometimeswe can get a solution that is bounded—a finite region with three or more sides.
Let’s look at a simple example.
Example C
Find the solution to the following system of inequalities.
3x− y < 4
4y+9x < 8
x≥ 0
y≥ 0
Solution
Let’s start by writing our inequalities in slope-intercept form.
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y > 3x−4
y <−94
x+2
x≥ 0
y≥ 0
Now we can graph each line and shade appropriately. First we graph y > 3x−4 :
Next we graph y <−94 x+2 :
Finally we graph x≥ 0 and y≥ 0, and we’re left with the region below; this is where all four inequalities overlap.
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The solution is bounded because there are lines on all sides of the solution region. In other words, the solutionregion is a bounded geometric figure, in this case a triangle.
Notice, too, that only three of the lines we graphed actually form the boundaries of the region. Sometimes when wegraph multiple inequalities, it turns out that some of them don’t affect the overall solution; in this case, the solutionwould be the same even if we’d left out the inequality y > 3x−4. That’s because the solution region of the systemformed by the other three inequalities is completely contained within the solution region of that fourth inequality;in other words, any solution to the other three inequalities is automatically a solution to that one too, so adding thatinequality doesn’t narrow down the solution set at all.
But that wasn’t obvious until we actually drew the graph!
Video Overview
MEDIAClick image to the left for more content.
Homework
Find the solution region of the following systems of inequalities.
1.
x− y <−6
2y≥ 3x+17
2.
4y−5x < 8
−5x≥ 16−8y
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3.
5x− y≥ 5
2y− x≥−10
4.
5x+2y≥−25
3x−2y≤ 17
x−6y≥ 27
5.
2x−3y≤ 21
x+4y≤ 6
3x+ y≥−4
6.
12x−7y < 120
7x−8y≥ 36
5x+ y≥ 12
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CHAPTER 4 Unit 4 - One VariableStatistics
Chapter Outline4.1 DOT PLOTS, FREQUENCY TABLES, HISTOGRAMS AND BAR GRAPHS
4.2 MEASURES OF CENTRAL TENDENCY (MEAN, MEDIAN AND MODE)
4.3 BOX AND WHISKER PLOTS
4.4 SHAPE, CENTER AND SPREAD
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4.1 Dot Plots, Frequency Tables, Histogramsand Bar Graphs
What You Will Learn
• Dot Plots• Frequency Tables• Histograms• Bar Graphs
Dot Plots
A dot plot consists of a horizontal scale (a number line) on which dots are placed to show the numerical values ofthe data. If a data value repeats the dots are piled up at that location. One dot for each repitition. We say that thedot plot displays the distribution of the data.
A dot plot is easier to look at compared to a long list of numbers. By examining the dot plot in the gray box, you cansee that the numbers 25 and 40 are repeated the most. You can see that the number 50 is repeated the least. Not onlythat but you can count very easily how many times each number is repeated. A dot plot is a good way of organizingnumbers and values.
Here are two dot plots. Students were surveyed and asked what they thought their grade would be on a mathexam. The results of the survey are in green. Then the students took the math test and the scores were plotted onthe top graph in purple. The blue line is the halfway point between the scores. What conclusions can we make basedon looking at the two dot plots?
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1. The students tended to think that they would score lower than they actually did.2. Counting the dots, more people scored above 15, where in the survey, more students thought they would score
below 15.3. In the survey, only 14 students thought they would score above 25. On the actual test, 24 people scored above
a 25.4. On the actual math test, scores were spread out a lot more than in the survey scores.
These are only a few of the comparisons you could make by looking at the dot plots.
Example 1
The following tables show the number of cars sold each month in one year for two separate dealers. Create a dotplot for each of the following tables of data using the same grid. Give three inferences based on the graphs and howthey overlap.
First we look at the lowest and highest number of cars sold for both because the graph must be able to handleall data. The lowest number is 23 and the highest is 38. The horizontal axis is made to include 23 - 38. Thevertical axis is left to increase by 1’s.
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Plot the individual data points for each of the tables. Use two colors to represent the two dealers. Make sureyou use a key.
Three inferences based on the dot plots of both sets of data are:
1. Mac’s sold more cars than Sid’s.2. Mac and Sid had three months where they sold the same amount of cars.3. The maximum number of cars sold in one month by Sid was 35 and the maximum number of cars sold by Mac
was 38.
Create your own dot plots with the applet from http://www.shodor.org/interactivate/activities/PlopIt/ . Just enteryour data and see what the dot plot looks like.
Frequency Tables and Histograms
Frequency tables are a great way to record and organize data. Once you have created a frequency table, we can makea histogram to present a visual display of the information in the frequency table.
A frequency table is created by making a table with three separate columns. One column is designated for intervals.The amount of intervals is determined by the range in data values. If the range in data values is not that great, theintervals will be small. If the range in data values is great, the intervals will be larger. It is important that the intervalsare of equal size and do not overlap.
Another column is created for tallied results. This is where you tally the number of times you see a data value fromeach interval.
In the last column, add the tally marks to determine the frequency results.
Let’s look at how we can apply this information.
The data below depicts the amount of time (in minutes) 20 middle school students spent on the computer each day.Arrange the data on a frequency table.
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10, 32, 8, 55, 5, 0, 30, 20, 25, 45, 40, 60, 45, 15, 5, 56, 47, 12, 15, 20
Step 1: Make a table with three separate columns.
• Intervals• Tallied results• Frequency results
Step 2: Looking at the data, tally the number of times a data value occurs.
Step 3: Add the tally marks to record the frequency.
TABLE 4.1:
Number of Minutes on the Com-puter
Tally Frequency
0 – 5 I I I 36 – 10 I I 211 – 15 I I I 316 – 20 I I 221 – 25 I 126 – 30 I 131 – 35 I 136 – 40 I 141 – 45 I I 246 – 50 I 151 – 55 I 156 – 60 I I 2
Now we can make a histogram from this data.
1. Draw the horizontal (x) and vertical (y) axis.
2. Give the graph the title “Minutes Spent on the Computer.”
3. Label the horizontal axis “Minutes.” List the intervals across the horizontal axis.
4. Title the vertical axis “Frequency.” Label the axis by halves (0.5).
5. For each interval on the horizontal access, draw a vertical column to the appropriate frequency value. Recall thaton a histogram, there are no spaces in between vertical columns.
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Bar Graphs
Bar graphs are similar to histograms, except the entries on the x axis aren’t numbers, they are things. In this example,the entries on the x axis will be the names of US presidents.
The data table below depicts the ages of twenty of our nation’s presidents at the time of Inauguration. Create a bargraph to display the data.
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To create a bar graph, first draw a horizontal (x) and vertical (y) axis.
Label the horizontal axis with each president’s last name.
Label the vertical axis with intervals of two, beginning with the number thirty.
Next, draw a vertical column to the appropriate value for each president.
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Examining data using bar graphs, histograms and frequency tables can help us to understand information ina visual way. For people who learn in a visual way, using data displays is a great way to tackle such a task!
Now use this bar graph to answer the following questions about our presidents.
Example A
How many of our presidents were sixty-one years of age or older at the time of the Inauguration?
Solution: Seven presidents
Example B
What was the most popular age for a president at the time of the Inauguration?
Solution: Fifty-seven years old
Example C
What was the age of the youngest president? Who was it?
Solution: Forty-three years old, Kennedy
Video Overview of Dot Plots
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MEDIAClick image to the left for more content.
Video Overview of Bar Graphs
MEDIAClick image to the left for more content.
Video Overview of Histograms
MEDIAClick image to the left for more content.
Homework
1. The heights of 20 basketball players, in inches are given below. Make a dot plot from this data.
68, 70, 70, 71, 75, 80, 81, 82, 84, 75, 75, 80, 75, 77, 75, 80, 83, 80, 71, 70
2. Create a histogram to display the data from the frequency table below.
TABLE 4.2: Monthly Internet Purchases
Data Values Tally Frequency0 – 3 I I I 34 – 7 I I I I 48 – 11 I 112 – 15 I I 2
3. The data collected depicts the number of letters in the last names of twenty people. Create a frequency table todisplay the data.
12 3 5 9 11 2 7 5 6 8 14 4 8 7 5 10 5 9 7 15
4. Create a histogram to display the data.
5. The data collected depicts the number of hours twelve families traveled this summer to their vacation destination.
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Create a frequency table to display the data.
7 3 10 5 12 9 8 4 3 11 3 9
6. Create a histogram to display the data.
7. Write a few sentences to explain any conclusions that you can draw from the data.
Directions: Use the histogram from the Practice Exercise to answer the following questions.
8. How many students spend less than 30 minutes studying?
9. How many students spend greater than 60 minutes studying?
10. How many students spend greater than 90 minutes studying?
11. How many students spend between 2 and 2 1/2 hours studying?
12. If this histogram is the result of a survey,was the number of students surveyed greater than 100?
Directions: Use this bar graph to answer the following questions.
13. If each girl could only vote once, what was the total number of girls surveyed?
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14. If each boy could only vote once, how many boys were surveyed?
15. What fraction of the girls surveyed chose track as their favorite sport?
16. What fraction of the girls surveyed chose soccer as their favorite sport?
17. What percentage of the girls surveyed chose track and soccer as their favorite sport? You may round to thenearest whole percent.
18. What fraction of the boys chose football as their favorite sport?
19. How many boys and girls were surveyed in all?
20. True or false. Basketball is the least popular sport among girls.
21. True or false. It is also the least popular among boys.
22. What is the most popular sport among girls?
23. What is the least popular sport among boys?
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4.2 Measures of Central Tendency (Mean, Me-dian and Mode)
What You Will Learn
• Calculate the mode, median, and mean for a set of data• Understand the differences between each measure of center.
Mode, Mean, and Median
Mode
The mode is defined as the most frequently occurring number in a data set.
Example:
The students in a statistics class were asked to report the number of children that live in their house (includingbrothers and sisters temporarily away at college). The data is recorded below:
1,3,4,3,1,2,2,2,1,2,2,3,4,5,1,2,3,2,1,2,3,6
In this example, the mode could be a useful statistic that would tell us something about the families of statisticsstudents in our school. In this case, 2 is the mode as it is the most frequently occurring number of children in thesample, telling us that a large number of students in our class have 2 children in their home.
Two Issues with the Mode
a. If there is more than one number that is the most frequent than the mode is usually both of those numbers. Forexample, if there were seven 3−child households and seven with 2 children, we would say that the mode is,“2 and 3.” When data is described as being bimodal, it is clustered about two different modes. Technically, ifthere were more than two, they would all be the mode. However, the more of them there are, the more trivialthe mode becomes. In those cases, we would most likely search for a different statistic to describe the centerof such data.
b. If each data value occurs an equal number of times, we usually say, “There is no mode.” Again, this is a casewhere the mode is not at all useful in helping us to understand the behavior of the data.
Mean
You are probably comfortable calculating averages. The average is a measure of center that statisticians call themean. Most students learn early on in their studies that you calculate the mean by adding all of the numbers anddividing by the number of numbers. While you are expected to be able to perform this calculation, most real data setsthat statisticians deal with are so large that they very rarely calculate a mean by hand. It is much more critical thatyou understand why the mean is such an important measure of center. The mean is actually the numerical “balancingpoint” of the data set. A certain math teacher might refer to the mean as being "the center of mass".
We can illustrate this physical interpretation of the mean. Below is a graph of the class data from the last example.
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If you have snap cubes like you used to use in elementary school, you can make a physical model of the graph, usingone cube to represent each student’s family and a row of six cubes at the bottom to hold them together like this:
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There are 22 students in this class and the total number of children in all of their houses is 55, so the mean ofthis data is 55÷ 22 = 2.5. Statisticians use the symbol X to represent the mean when X is the symbol for a singlemeasurement. It is pronounced “x bar.”
It turns out that the model that you created balances at 2.5. In the pictures below, you can see that a block placed at3 causes the graph to tip left, and while one placed at 2 causes the graph to tip right. However, if you place the blockat about 2.5, it balances perfectly!
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Median
The median is simply the middle number in a set of data. Think of 5 students seated in a row in statistics class:
Aliyah Bob Catalina David Elaine
Which student is sitting in the middle? If there were only four students, what would be the middle of the row? Theseare the same issues you face when calculating the numeric middle of a data set using the median.
Let’s say that Ron has taken five quizzes in his statistics class and received the following grades:
80,94,75,90,96
Before finding the median, you must put the data in order. The median is the numeric middle. Placing the data inorder from least to greatest yields:
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75,80,90,94,96
The middle number in this case is the third grade, or 90, so the median of this data is 90. Notice that just bycoincidence, this was also the third quiz that he took, but this will usually not be the case.
Of course, when there is an even number of numbers, there is no true value in the middle. In this case we take thetwo middle numbers and find their mean. If there are four students sitting in a row, the middle of the row is halfwaybetween the second and third students.
Example
Take Rhonda’s quiz grades:
91,83,97,89
Place them in numeric order:
83,89,91,97
The second and third numbers “straddle” the middle of this set. The mean of these two numbers is 90, so the medianof the data is 90.
Mean vs. Median
Both the mean and the median are important and widely used measures of center. So you might wonder why weneed them both. There is an important difference between them that can be explained by the following example.
Suppose there are 5 houses in a community with the following prices: $55,000 $58,000 $60,000 $61,000 $3,200,000. The"median housing price" is far more useful in real estate because it tells us that half of the houses on the market areless and half are more. The mean of this data set $686,000, and it is not at all descriptive of the market. Theimportant thing to understand is that extremely large or small values in a data set can have a large influence on themean.
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Video Overview of Mean, Median and Mode
MEDIAClick image to the left for more content.
Homework
1. In Lois’ 2nd grade class, all of the students are between 45 and 52 inches tall, except one boy, Lucas, who is62 inches tall. Which of the following statements is true about the heights of all of the students?
a. The mean height and the median height are about the sameb. The mean height is greater than the median height.c. The mean height is less than the median height.d. More information is needed to answer this question.e. None of the above is true.
2. Enrique has a 91, 87 and 95 for his statistics grades for the first three quarters. His mean grade for the yearmust be a 93 in order for him to be exempt from taking the final exam. Assuming grades are rounded followingvalid mathematical procedures, what is the lowest whole number grade he can get for the 4th quarter and stillbe exempt from taking the exam?
3. The chart below shows the data from the Galapagos tortoise preservation program with just the number ofindividual tortoises that were bred in captivity and reintroduced into their native habitat.
TABLE 4.3:
Island or Volcano Number of Individuals RepatriatedWolf 40Darwin 0Alcedo 0Sierra Negra 286Cerro Azul 357Santa Cruz 210Española 1293San Cristóbal 55Santiago 498Pinzón 552Pinta 0
Figure: Approximate Distribution of Giant Galapagos Tortoises in 2004 ("Estado Actual De Las Poblaciones deTortugas Terrestres Gigantes en las Islas Galápagos," Marquez, Wiedenfeld, Snell, Fritts, MacFarland, Tapia, yNanjoa, Scologia Aplicada, Vol. 3, Num. 1,2, pp. 98-11).
For this data, calculate each of the following:
(a) mode
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(b) median
(c) mean
(d) upper and lower quartiles
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4.3 Box and Whisker Plots
What You Will Learn
• How to determine the quartiles of a data set• How to make a 5-number summary• How to make a box and whisker plot
In the last section, we saw how data could be organized by using a dot plot or a histogram. Another way to plot datais by using a box and whisker plot (often called a box plot). Histograms and box and whisker plots can be used todraw conclusions about a data set, we will learn more about drawing conclusions in the next section of the book.
A box and whisker plot (often called a box plot) involves plotting 5 specific values. The 5 specific values are oftenreferred to as a five-number summary of the organized data set. The five-number summary consists of the following:
1. The lowest number in the data set (minimum value)2. The lower quartile: Q1 (median of the first half of the data set)3. The median of the entire data set (median)4. The upper quartile: Q3 (median of the second half of the data set)5. The highest number in the data set (maximum value)
The display of the five-number summary produces a box and whisker plot as shown below:
The above model of a box-and-whisker plot shows 2 horizontal lines (the whiskers) that each contain 25% of thedata and are of the same length. In addition, it shows that the median of the data set is in the middle of the box,which contains 50% of the data. The lengths of the whiskers and the location of the median with respect to the centerof the box are used to describe the distribution of the data. It’s important to note that this is just an example. Not allbox-and-whisker plots have the median in the middle of the box and whiskers of the same size.
Making a Box and Whisker Plot from data
Here is a data set showing the number of hours that teenagers work in a part time job.
16, 10, 8, 8, 11, 11, 12, 15, 10, 20, 6, 16, 8
To work with this set of data, the first thing that we need to do is to order it from least to greatest including anyrepeated numbers.
6, 8, 8, 8, 10, 10, 11, 11, 12, 15, 16, 16, 20.
Now we will find the numbers for the five-number summary. The first two numbers to find are the minimum andthe maximum, they are 6 and 20.
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Next, we find the median. Remember that the median is the middle number in a set of data. Here there are 13values. The median is 11.
Lastly, we need to find the lower and upper quartiles. The quartiles divides the data set into four parts. With themedian, our data set is divided into two parts. The first part is the first half up to 10 and the second half starts at 11and goes to 20.
Take a look.
6, 8, 8, 8, 10, 10, 11, 11, 12, 15, 16, 16, 20
The median of the first half of the data is called the lower quartile (Q1). The median of the second half of the datais called the upper quartile (Q3).
6, 8, 8, 8, 10, 10, 11, 11, 12, 15, 16, 16, 20
Q1 is the average between 8 and 8. Q1 is 8.
Q3 is the average between 15 and 16. Q3 is 15.5.
Here are the steps to drawing a box-and-whisker plot.
1. Draw a number line labeled to show the range of data from least to greatest.2. Mark the median, the upper quartile, the lower quartile, the minimum and the maximum on the number line.3. Draw in a box around the quartiles. The median is the middle line of the two boxes.4. Then draw in the whiskers. These are lines that extend from each quartile to the upper and lower extremes.
Here is a picture of a number line with a completed box-and-whisker plot on it.
Now let’s examine this plot. The first box goes from the lower quartile 8 to the median 11. The second box goesfrom the median 11 to the upper quartile 15.5. The whiskers extend out from the lower quartile to the minimum of6, and from the upper quartile to the maximum of 20.
Now that you know how to draw a box-and-whisker plot and find the median, quartiles and extremes of a set of data,we can work the other way around. We can look at a box-and-whisker plot to identify the median, quartiles andextremes.
We can use this chart to examine the data. The median divides the two boxes. The median here is 200. The lowerquartile is 100 and the upper quartile is 300. The minimum is 50 and the maximum is 400.
Guided Practice
a. For the following data sets, determine the five-number summaries:
i. 12, 16, 36, 10, 31, 23, 58
ii. 144, 240, 153, 629, 540, 300
b. Use the data set for part i of the previous question and the five-number summary to construct a box-and-whiskerplot to model the data set.
Answer:
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a. i. The first step is to organize the values in the data set as shown below:
12, 16, 36, 10, 31, 23, 58
10, 12, 16, 23, 31, 36, 58
Now complete the following list:
Minimum value→ 10
Q1→ 12
Median→ 23
Q3→ 36
Maximum value→ 58
ii. The first step is to organize the values in the data set as shown below:
144, 240, 153, 629, 540, 300
144, 153, 240, 300, 540, 629
Now complete the following list:
Minimum value→ 144
Q1→ 153
Median→ 270
Q3→ 540
Maximum value→ 629
b. The five-number summary can now be used to construct a box-and-whisker plot for part i. Be sure to provide ascale on the number line that includes the range from the minimum value to the maximum value.
Minimum value→ 10
Q1→ 12
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Median→ 23
Q3→ 36
Maximum value→ 58
Video Overview
MEDIAClick image to the left for more content.
Homework
Directions: For each set of data, find the five-number summary and draw a box and whisker plot.
(1) 12, 13, 15, 17, 21, 22, 24, 26, 28, 30, 31
(2) 26, 27, 29, 30, 32, 35, 41, 42, 44
(3) 100, 105, 107, 109.110, 120
(4) 87, 85, 89, 92, 94, 97, 102, 105, 105, 113, 108.
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4.4 Shape, Center and Spread
What You Will Learn
• How to determine the shape of a data set• Which measure of central tendency is best to use for a given data set?• Three ways to calculate the spread of a data set
– Range– Interquartile Range– Standard Deviation
• Which measure of spread is best to use for a given data set?• How to determine outliers
Shape
The shape of a data set refers to where the largest amount of the data occurs. If most of the data is in the middle ofthe histogram, the data is symmetric
If the histogram has a "tail" going out to the right, the data is skewed to the right
If the histogram has a "tail" going out to the left, the data is skewed to the left
A box and whisker plot can also provide information about the shape of the data.
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a. If the median is located in the center or near the center of the box, the distribution is approximately symmetric.
b. If the median is located to the left of the center of the box, the distribution is skewed to the right.
c. If the median is located to the right of the center of the box, the distribution is skewed to the left.
Information about the data set that can be determined from the box-and-whisker plot with respect to the length ofthe whiskers includes the following:
a. If the whiskers are the same or almost the same length, the distribution is approximately symmetric.
b. If the right whisker is longer than the left whisker, the distribution is skewed to the right.
c. If the left whisker is longer than the right whisker, the distribution is skewed to the left.
Example A
For each box-and-whisker plot, list the five-number summary and describe the distribution based on the location ofthe median.
a. Minimum value→ 4
Q1→ 6
Median→ 9
Q3→ 10
Maximum value→ 12
The median of the data set is located to the right of the center of the box, which indicates that the distribution isskewed to the left.
b. Minimum value→ 225
Q1→ 250
Median→ 300
Q3→ 325
Maximum value→ 350
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The median of the data set is located to the right of the center of the box, which indicates that the distribution isskewed to the left.
c. Minimum value→ 60
Q1→ 70
Median→ 75
Q3→ 95
Maximum value→ 100
The median of the data set is located to the left of the center of the box, which indicates that the distribution isskewed to the right.
Example B
The numbers of square feet (in 100s) of 10 of the largest museums in the world are shown below:
650, 547, 204, 213, 343, 288, 222, 250, 287, 269
Construct a box-and-whisker plot for the above data set and describe the distribution.
The first step is to organize the data values as follows:
20,400 21,300 22,200 25,000 26,900 28,700 28,800 34,300 54,700 65,000
Now calculate the median, Q1, and Q3.
20,400 21,300 22,200 25,000 26,900 28,700 28,800 34,300 54,700 65,000
Median→ 26,900+28,7002
=55,600
2= 27,800
Q1 = 22,200
Q3 = 34,300
Next, complete the following list:
Minimum value→ 20,400
Q1→ 22,200
Median→ 27,800
Q3→ 34,300
Maximum value→ 65,000
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The right whisker is longer than the left whisker, which indicates that the distribution is skewed to the right.
Which Measure of Central Tendency to Use?
The position of the center of a data set can be described the mean, the median or the mode. The mean is usuallythe best choice if the data is symmetric. If the data is skewed to the right or left, the median is usually best. Afew numbers that are unusually large or small can have a large effect on the mean and then it no longer describes atypical number in the data set. The mode can be the best measure of central tendency if there is significant repetitionin the data.
Example C
Find the mean, median, and mode of the salaries given below. Which measure of central tendency is the best todescribe this set of data? Justify your answer.
TABLE 4.4:
Professional Realm Annual incomeFarming, Fishing, and Forestry $19,630Sales and Related $28,920Architecture and Engineering $56,330Healthcare Practitioners $49,930Legal $69,030Teaching & Education $39,130Construction $35,460Professional Baseball Player* $2,476,590
(Source: Bureau of Labor Statistics, except (*) - The Baseball Players’ Association (playbpa.com)).
Mean (average):
x̄ =19,630+28,920+56,330+49,930+69,030+39,130+35,460+2,476,590
8
x̄ =2,774,750
8x̄ = 346,843.75
mean = 346,843.75
Median (middle):
Arrange the salaries from least to greatest.
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19,630 /28,920 /35,460 /39,130 /49,930 /56,330 /69,030 /2,476,590
There is no middle value. Take the two closest to the middle and average them together.
39,130+49,9302
median = 44,530
Mode (most):
There is no value that occurs the most. Each salary occurs one time in the chart.
T here is no mode.
The mean is $346,843.75. The median is $44,530. There is no mode.
The median is the best measure to describe the data. This is because there is such a wide range in the data.The mean would make people think that any professional would make about $346,843.75. Most professionalsdo not.
Spread
The spread of the data refers to how far apart all of the data is. The simplest measure of spread is the range. Tofind the range, just subtract the minimum value from the maximum value.
Example D
Find the range of the following data:
223, 121, 227, 433, 122, 193, 397, 276, 303, 199, 197, 265, 366, 401, 222
Solution
The first thing to do in this case is to order the data, listing all values in ascending order:
121, 122, 193, 197, 199, 222, 223, 227, 265, 276, 303, 366, 397, 401, 433
Note: It is extremely important to make sure that you don’t skip any values when you reorder the list. Two ways todo this are (i) cross out the numbers in the original list as you write them in the second list, and (ii) count the numberof values in both lists when you are done. In this example, both lists contain 15 values, so we can be sure we didn’tmiss any (as long as we didn’t count any twice!)
The range is found by subtracting the lowest value from the highest: 433−122 = 311.
Interquartile Range (IQR)
If you are told that you are between the 25th percent and 75th percent of your class in height, would you know howto find the other people that are in this range as well? In statistics this is called finding the interquartile range.
Inter is between, quartile is divided into 4 parts, range is the difference between the greatest and least data value. Thatmeans that the interquartile range is the difference between the number at the 75th quartile and the number at the
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25th quartile (like a mini range). Let’s make this clear using the following set of random data. We have alreadyarranged the data from least to greatest.
Example E
To make our four groups we first find the median. Remember that the median is the middle number in the data. Themiddle is also the 50th quartile. Think - 50% is half! Once we have the data split into two halves, we split thosehalves in half. This will make 4 equal parts.
To find the 1st quartile, we would average 22 and 24 to get 23. 23 is the 1st quartile.
To find the 3rd quartile, we would average 41 and 45 to get 43. 43 is the 3rd quartile.
Now we can find the interquartile range, it is the difference between 43 - 23 = 20.
Standard Deviation
The standard deviation is another measure of the spread of the data. It describes more exactly how all the data isrelated to the mean. It is more complicated to calculate, though. Let’s look at the following data values and learnhow to calculate the standard deviation:
11, 13, 14, 15, 19, 22, 24, 26
The mean of these values is 18; of course, the values all differ from 18 by varying amounts. Here’s a list of thevalues’ deviations from the mean:
-7, -5, -4, -3, 1, 4, 6, 8
If we take the mean of these deviations, we find that it is zero:
−7−5−4−3+1+4+6+88
=08= 0
This comes as no surprise. You can see that some of the values are positive and some are negative, as the meanlies somewhere near the middle of the range. You can use algebra to prove (try it!) that the sum of the deviationswill always be zero, no matter what numbers are in the list. So, the sum of the deviations is not a useful tool formeasuring variance.
But if we square the differences, all the negative differences become positive, and then we can tell how great theaverage deviation is. If we do that for this data set, we get the following list:
49, 25, 16, 9, 1, 16, 36, 64
The sum of those squares is 216, so their average is 2168 = 27.
We call this averaging of the square of the differences from the mean (the mean squared deviation) the variance.The variance is a measure of the dispersion, and its value is lower for tightly grouped data than for widely spreaddata. In the example above, the variance is 27. What exactly does that 27 refer to, though? Since it is the meanof the squares of the deviation, it seems logical that taking its square root would be a better way to make sense ofit. This square root of the variance is called the standard deviation, and is given the symbol σ (lower case Greekletter "s").
The standard deviation is often used to to find values within a certain range of the mean. It works out that 68% of adata set will be within one standard deviation of the mean, 95% will be within two standard deviations of the meanand 99.7% will be within three standard deviations of the mean.
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Example F
Suppose that the mean number of calories consumed by an adult in a particular country per day is 3,200, with astandard deviation of 150 calories. Determine each of the following intervals and shade the area under a normaldistribution curve for each interval.
a. The values within 1 standard deviation of the mean
b. The values within 2 standard deviations of the mean
c. The values within 3 standard deviations of the mean
Answer:
a. The values within 1 standard deviation of the mean are from 3,200−150 calories to 3,200+150 calories, or from3,050 calories to 3,350 calories.
b. The values within 2 standard deviations of the mean are from 3,200− (2× 150) calories to 3,200 + (2×150) calories, or from 2,900 calories to 3,500 calories.
c. The values within 3 standard deviations of the mean are from 3,200− (3× 150) calories to 3,200 + (3×150) calories, or from 2,750 calories to 3,650 calories.
Which Measure of Spread to Use?
Standard deviation is designed to be used if the data is symmetric or normally distributed. If the data is skewed, thepoints far out along the tail will affect the standard deviation and it is no longer representative of the "typical" spreadin the data. This is very similar to the way that skewed data can affect the mean. If there is significant skew in the
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data, it is better to use the interquartile range.
Outliers
Outliers are data points that are significantly different from the rest of the data. An example of an outlier wouldbe the salary of a professional baseball player compared to a typical salary, as outlined in a previous example aboutcalculating the mean. Statisticians need a way to define whether or not a number is an outlier. One method forchecking a data set for the presence of an outlier is to follow the procedure below.
1. Organize the given data set and determine the values of Q1 and Q3.2. Calculate the difference between Q1 and Q3. and determine the interquartile range (IQR)3. Multiply the difference by 1.5, subtract this result from Q1, and add it to Q3.4. The results from Step 3 will be the range into which all values of the data set should fit. Any values that are
below or above this range are considered outliers.
Example G
Using the procedure outlined above, check the following data sets for outliers:
a. 18, 20, 24, 21, 5, 23, 19, 22
b. 13, 15, 19, 14, 26, 17, 12, 42, 18
a. Organize the given data set as follows:
18, 20, 24, 21, 5, 23, 19, 22
5, 18, 19, 20, 21, 22, 23, 24
Determine the values for Q1 and Q3.
5, 18, 19 , 20, 21, 22, 23 , 24
Q1 =18+19
2=
372
= 18.5 Q3 =22+23
2=
452
= 22.5
Calculate the difference between Q1 and Q3: Q3−Q1 = 22.5−18.5 = 4.0.
Multiply this difference by 1.5: (4.0)(1.5) = 6.0.
Finally, compute the range.
Q1−6.0 = 18.5−6.0 = 12.5
Q3 +6.0 = 22.5+6.0 = 28.5
.
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Are there any data values below 12.5? Yes, the value of 5 is below 12.5 and is, therefore, an outlier.
Are there any values above 28.5? No, there are no values above 28.5.
b. Organize the given data set as follows:
13, 15, 19, 14, 26, 17, 12, 42, 18
12, 13, 14, 15, 17, 18, 19, 26, 42
Determine the values for Q1 and Q3.
12, 13, 14 , 15, 17 , 18, 19, 26 , 42
Q1 =13+14
2=
272
= 13.5 Q3 =19+26
2=
452
= 22.5
Calculate the difference between Q1 and Q3: Q3−Q1 = 22.5−13.5 = 9.0.
Multiply this difference by 1.5: (9.0)(1.5) = 13.5.
Finally, compute the range.
Q1−13.5 = 13.5−13.5 = 0
Q3 +13.5 = 22.5+13.5 = 36.0
Are there any data values below 0? No, there are no values below 0.
Are there any values above 36.0? Yes, the value of 42 is above 36.0 and is, therefore, an outlier.
Video Overview of Range and Standard Deviation
MEDIAClick image to the left for more content.
Homework
For each of the following box-and-whisker plots, list the five-number summary and comment on the distribution ofthe data:
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1.
2.
3. The following data represents the number of coins that 12 randomly selected people had in their piggy banks:
35 58 29 44 104 39 72 34 50 41 64 54
a. Construct a histogram for this data. Is it skewed? If so, in what direction?
b. Which measure of spread is better to use for this data?
c. Calculate the best measure of spread.
4. The following data represent the time (in minutes) that each of 20 people waited in line at a local book store topurchase the latest Harry Potter book:
15 8 5 10 14 17 21 23 6 19 31 34 30 31
3 22 17 25 5 16
a. Contruct a box plot for this data
b. Which measure of spread is better to use for this data?
c. Calculate the best measure of spread.
5. Firman’s Fitness Factory is a new gym that offers reasonably-priced family packages. The following tablerepresents the number of family packages sold during the opening month:
24 21 31 28 29
27 22 27 30 32
26 35 24 22 34
30 28 24 32 27
32 28 27 32 23
20 32 28 32 34
Construct a box-and-whisker plot for the data. Are the data symmetric or skewed?
Find the range, IQR and standard deviation for the following sets of data
6.
TABLE 4.5:
34 25 16 4312 34 39 4565 13 45 6734 41 30 15
7.
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TABLE 4.6:
2 5 95 7 37 9 1
Using the procedure outlined in this concept, check the following data sets for outliers:
8. 25, 33, 55, 32, 17, 19, 15, 18, 21
9. 149, 123, 126, 122, 129, 120
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CHAPTER 5 Unit 5 - Polynomials,Exponents and Radicals
Chapter Outline5.1 ADDITION AND SUBTRACTION OF POLYNOMIALS
5.2 EXPONENT PROPERTIES INVOLVING PRODUCTS
5.3 MULTIPLICATION OF POLYNOMIALS
5.4 SPECIAL PRODUCTS OF POLYNOMIALS
5.5 EXPONENT PROPERTIES INVOLVING QUOTIENTS
5.6 ZERO AND NEGATIVE EXPONENTS
5.7 DIVISION OF POLYNOMIALS BY MONOMIALS
5.8 DIVISION OF POLYNOMIALS BY POLYNOMIALS
5.9 SCIENTIFIC NOTATION
5.10 SQUARE ROOTS AND IRRATIONAL NUMBERS
5.11 PYTHAGOREAN THEOREM
5.12 MULTIPLICATION AND DIVISION OF RADICALS
5.13 ADDITION AND SUBTRACTION OF RADICALS
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5.1 Addition and Subtraction of Polynomials
What You Will Learn
• Write a polynomial expression in standard form.• Classify polynomial expression by degree• Simplifiying polynomials• Add and subtract polynomials
Polynomial Expression
A polynomial is made up of different terms that contain positive integer powers of the variables. Here is an exampleof a polynomial.
4x3 +2x2−3x+1
Each part of the polynomial that is added or subtracted is called a term of the polynomial. The example above is apolynomial with four terms.
The numbers appearing in each term in front of the variable are called the coefficients. The number appearing all byitself without a variable is called a constant.
In this case, the coefficient of x3 is 4, the coefficient of x2 is 2, the coefficient of x is -3 and the constant is 1.
Degrees of Polynomials and Standard Form
Each term in the polynomial has a degree. This is the power of the variable in that term.
4x3 Has a degree of 3 and is called a cubic term or 3rd order term.
2x2 Has a degree of 2 and is called the quadratic term or 2nd order term.
−3x Has a degree of 1 and is called the linear term or 1st order term.
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1 Has a degree of 0 and is called the constant.
By definition, the degree of the polynomial is the same as the degree of the term with the highest degree. Thisexample is a polynomial of degree 3, which is also called a “cubic” polynomial. (Why do you think it is called acubic?).
Polynomials can have more than one variable. Here is another example of a polynomial.
t4−6s3t2−12st +4s4−5
This is a polynomial because all exponents on the variables are positive integers. This polynomial has five terms.Let’s look at each term more closely. Note: The degree of a term is the sum of the powers on each variable in theterm.
t4 Has a degree of 4, so it’s a 4th order term
−6s3t2 Has a degree of 5, so it’s a 5th order term.
−12st Has a degree of 2, so it’s a 2nd order term
4s4 Has a degree of 4, so it’s a 4th order term
-5 Is a constant, so its degree is 0.
Since the highest degree of a term in this polynomial is 5, then this is polynomial of degree 5 or a 5th orderpolynomial.
A polynomial that has only one term has a special name. It is called a monomial (mono means one). A monomialcan be a constant, a variable, or a product of a constant and one or more variables. You can see that each term in apolynomial is a monomial. A polynomial is the sum of monomials. Here are some examples of monomials.
b2︸︷︷︸This is a monomial
8︸︷︷︸So is this
−2ab2︸ ︷︷ ︸and this
14
x4︸︷︷︸and this
−29xy︸ ︷︷ ︸and this
Example 1
For the following polynomials, identify the coefficient on each term, the degree of each term and the degree of thepolynomial.
a) x5−3x3 +4x2−5x+7
b) x4−3x3y2 +8x−12
Solution
a) x5−3x3 +4x2−5x+7
The coefficients of each term in order are 1, -3, 4, -5 and the constant is 7.
The degrees of each term are 5, 3, 2, 1, and 0. Therefore, the degree of the polynomial is 5.
b) x4−3x3y2 +8x−12
The coefficients of each term in order are 1, -3, 8 and the constant is -12.
The degrees of each term are 4, 5, 1, and 0. Therefore, the degree of the polynomial is 5.
Example 2
Identify the following expressions as polynomials or non-polynomials.
a) 5x2−2x
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b) 3x2−2x−2
c) x√
x−1
d) 5x3+1
e) 4x13
f) 4xy2−2x2y−3+ y3−3x3
Solution
(a) 5x2−2x This is a polynomial.
(b) 3x2−2x−2 This is not a polynomial because it has a negative exponent.
(c) x√
x−1 This is not a polynomial because is has a square root.
(d) 5x3+1 This is not a polynomial because the power of x appears in the denominator.
(e) 4x13 This is not a polynomial because it has a fractional exponent.
(f) 4xy2−2xy−3+ y3−3x3 This is a polynomial.
You saw that each term in a polynomial has a degree. The degree of the highest term is also the degree of thepolynomial. Often, we arrange the terms in a polynomial so that the term with the highest degree is first and it isfollowed by the other terms in order of decreasing power. This is called standard form.
The following polynomials are in standard form.
4x4−3x3 +2x2− x+1
a4b3−2a3b3 +3a4b−5ab2 +2
The first term of a polynomial in standard form is called the leading term and the coefficient of the leading term iscalled the leading coefficient.
The first polynomial above has a leading term of 4x4 and a leading coefficient of 4.
The second polynomial above has a leading term of a4b3 and a leading coefficient of 1.
Example 3
Rearrange the terms in the following polynomials so that they are in standard form. Indicate the leading term andleading coefficient of each polynomial.
(a) 7−3x3 +4x
(b) ab−a3 +2b
(c) −4b+4+b2
Solution
(a) 7−3x3 +4x is rearranged as −3x3 +4x+7. The leading term is −3x3 and the leading coefficient is -3.
(b) ab−a3 +2b is rearranged as −a3 +ab+2b. The leading term is −a3 and the leading coefficient is -1.
(c) −4b+4+b2 is rearranged as b2−4b+4. The leading term is b2 and the leading coefficient is 1.
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Simplifying Polynomials
A polynomial is simplified if it has no terms that are alike. Like terms are terms in the polynomial that have thesame variable(s) with the same exponents, but they can have different coefficients.
2x2y and 5x2y are like terms.
6x2y and 6xy2 are not like terms.
If we have a polynomial that has like terms, we simplify by combining them.
x2 +6xy−4xy+ y2
↗ ↖Like terms
This sample polynomial simplified by combining the like terms 6xy−4xy = 2xy. We write the simplified polynomialas
x2 +2xy+ y2
Example 4
Simplify the following polynomials by collecting like terms and combining them.
(a) 2x−4x2 +6+ x2−4+4x
(b) a3b3−5ab4 +2a3b−a3b3 +3ab4−a2b
Solution
(a) 2x−4x2 +6+ x2−4+4x
Rearrange the terms so that like terms are grouped together
= (−4x2 + x2)+(2x+4x)+(6−4)
Combine each set of like terms by adding or subtracting the coefficients
=−3x2 +6x+2
(b) a3b3−5ab4 +2a3b−a3b3 +3ab4−a2b
Rearrange the terms so that like terms are grouped together:
= (a3b3−a3b3)+(−5ab4 +3ab4)+2a3b−a2b
Combine each set of like terms:
= 0−2ab4 +2a3b−a2b=−2ab4 +2a3b−a2b
Add and Subtract Polynomials
Polynomial addition
To add two or more polynomials, write their sum and then simplify by combining like terms.
Example 5
Add and simplify the resulting polynomials.
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(a) Add 3x2−4x+7 and 2x3−4x2−6x+5.
(b) Add x2−2xy+ y2 and 2y2−4x2 and 10xy+ y3.
Solution:
(a) Add 3x2−4x+7 and 2x3−4x2−6x+5
= (3x2−4x+7)+(2x3−4x2−6x+5)
Group like terms = 2x3 +(3x2−4x2)+(−4x−6x)+(7+5)
Simplify = 2x3− x2−10x+12
(b) Add x2−2xy+ y2 and 2y2−3x2 and 10xy+ y3
= (x2−2xy+ y2)+(2y2−3x2)+(10xy+ y3)
Group like terms = (x2−3x2)+(y2 +2y2)+(−2xy+10xy)+ y3
Simplify = 2x2 +3y2 +8xy+ y3
Polynomial subtraction
To subtract one polynomial from another, add the opposite of each term of the polynomial you are subtracting.
Example 6
Subtract and simplify the resulting polynomials.
a) Subtract x3−3x2 +8x+12 from 4x2 +5x−9.
b) Subtract 5b2−2a2 from 4a2−8ab−9b2.
Solution
a)
(4x2 +5x−9)− (x3−3x2 +8x+12) = (4x2 +5x−9)+(−x3 +3x2−8x−12)
Group like terms =−x3− (4x2 +3x2)+(5x−8x)+(−9−12)
Simplify =−x3 +7x2−3x−21
b)
(4a2−8ab−9b2)− (5b2−2a2) = (4a2−8ab−9b2)+(−5b2 +2a2)
Group like terms = (4a2 +2a2)+(−9b2−5b2)−8ab
Simplify = 6a2−14b2−8ab
Note: An easy way to check your work after adding or subtracting polynomials is to substitute a convenient valuein for the variable, and check that your answer and the problem both give the same value. For example, in part (b)of Example 6, if we let a = 2 and b = 3, then we can check as follows.
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Given Solution
(4a2−8ab−9b2)− (5b2−2a2) 6a2−14b2−8ab
(4(2)2−8(2)(3)−9(3)2)− (5(3)2−2(2)2) 6(2)2−14(3)2−8(2)(3)
(4(4)−8(2)(3)−9(9))− (5(9)−2(4)) 6(4)−14(9)−8(2)(3)
(−113)−37 24−126−48
−150 −150
Since both expressions evaluate to the same number when we substitute in arbitrary values for the variables, we canbe reasonably sure that our answer is correct. Note, when you use this method, do not choose 0 or 1 for checkingsince these can lead to common problems.
Video Overview
MEDIAClick image to the left for more content.
Homework
Indicate which expressions are polynomials.
1. x2 +3x12
2. 13 x2y−9y2
3. 3x−3
Express each polynomial in standard form. Give the degree of each polynomial.
1. 3−2x2. 8−4x+3x3
3. −5+2x−5x2 +8x3
4. x2−9x4 +12
Add and simplify.
1. (x+8)+(−3x−5)2. (−2x2 +4x−12)+(7x+ x2)3. (2a2b−2a+9)+(5a2b−4b+5)4. (6.9a2−2.3b2 +2ab)+(3.1a−2.5b2 +b)
Subtract and simplify.
1. (−t +15t2)− (5t2 +2t−9)2. (−y2 +4y−5)− (5y2 +2y+7)3. (−5m2−m)− (3m2 +4m−5)4. (2a2b−3ab2 +5a2b2)− (2a2b2 +4a2b−5b2)
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5.2 Exponent Properties Involving Products
What You Will Learn
• Meaning of exponents• Use the product of a power property.• Use the power of a product property.• Simplify expressions involving product properties of exponents.
Meaning of Exponents
In this chapter, we will discuss exponents, we will start with what the word exponent means.
Consider the area of the square shown right. We know that the area is given by:
But we also know that for any rectangle, Area = (width) · (height), so we can see that:
Similarly, the volume of the cube is given by:
Volume = width ·depth ·height = x · x · x
But we also know that the volume of the cube is given by Volume = x3 so clearly
x3 = x · x · x
You probably know that the power (the small number to the top right of the x) tells you how many x’s to multiplytogether. In these examples the x is called the base and the power (or exponent) tells us how many factors of thebase there are in the full expression.
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x2 = x · x︸︷︷︸2 factors of x
x7 = x · x · x · x · x · x · x︸ ︷︷ ︸7 factors of x
x3 = x · x · x︸ ︷︷ ︸3 factors of x
xn = x · x · · · · · · · x︸ ︷︷ ︸n factors of x
Example 1
Write in exponential form.
(a) 2 ·2
(b) (−3)(−3)(−3)
(c) y · y · y · y · y
(d) (3a)(3a)(3a)(3a)
Solution
(a) 2 ·2 = 22 because we have 2 factors of 2
(b) (−3)(−3)(−3) = (−3)3 because we have 3 factors of (−3)
(c) y · y · y · y · y = y5 because we have 5 factors of y
(d) (3a)(3a)(3a)(3a) = (3a)4 because we have 4 factors of 3a
When we deal with numbers, we usually just simplify. We’d rather deal with 16 than with 24. However, withvariables, we need the exponents, because we’d rather deal with x7 than with x · x · x · x · x · x · x.
Let’s simplify Example 1 by evaluating the numbers.
Example 2
Simplify.
(a) 2 ·2
(b) (−3)(−3)(−3)
(c) y · y · y · y · y
(d) (3a)(3a)(3a)(3a)
Solution
(a) 2 ·2 = 22 = 4
(b) (−3)(−3)(−3) = (−3)3 =−27
(c) y · y · y · y · y = y5
(d) (3a)(3a)(3a)(3a) = (3a)4 = 34 ·a4 = 81a4
Note: You must be careful when taking powers of negative numbers. Remember these rules.
(negative number) · (positive number) = negative number
(negative number) · (negative number) = positive number
For even powers of negative numbers, the answer is always positive. Since we have an even number of factors, wemake pairs of negative numbers and all the negatives cancel out.
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(−2)6 = (−2)(−2)(−2)(−2)(−2)(−2) = (−2)(−2)︸ ︷︷ ︸+4
·(−2)(−2)︸ ︷︷ ︸+4
·(−2)(−2)︸ ︷︷ ︸+4
=+64
For odd powers of negative numbers, the answer is always negative. Since we have an odd number of factors, wecan make pairs of negative numbers to get positive numbers but there is always an unpaired negative factor, so theanswer is negative:
Ex: (−2)5 = (−2)(−2)(−2)(−2)(−2) = (−2)(−2)︸ ︷︷ ︸+4
·(−2)(−2)︸ ︷︷ ︸+4
·(−2)︸ ︷︷ ︸−2
=−32
Use the Product of Powers Property
What happens when we multiply one power of x by another? See what happens when we multiply x to the power 5by x cubed. To illustrate better we will use the full factored form for each:
(x · x · x · x · x)︸ ︷︷ ︸x5
·(x · x · x)︸ ︷︷ ︸x3
= (x · x · x · x · x · x · x · x)︸ ︷︷ ︸x8
So x5 · x3 = x8. You may already see the pattern to multiplying powers, but let’s confirm it with another example.We will multiply x squared by x to the power 4:
(x · x)︸ ︷︷ ︸x2
·(x · x · x · x)︸ ︷︷ ︸x4
= (x · x · x · x · x · x)︸ ︷︷ ︸x6
So x2 ·x4 = x6. Look carefully at the powers and how many factors there are in each calculation. 5 factors of x times3 factors of x equals (5+3) = 8 factors of x. 2 factors of x times 4 factors of x equals (2+4) = 6 factors of x.
You should see that when we take the product of two powers of x, the number of factors of x in the answer is thesum of factors in the terms you are multiplying. In other words the exponent of x in the answer is the sum of theexponents in the product.
Product rule for exponents: xn · xm = xn+m
Example 3
Multiply x4 · x5.
Solution
x4 · x5 = x4+5 = x9
When multiplying exponents of the same base, it is a simple case of adding the exponents. It is important that whenyou use the product rule you avoid easy-to-make mistakes. Consider the following.
Example 4
Multiply 22 ·23.
Solution
22 ·23 = 25 = 32
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Note that when you use the product rule you DO NOT MULTIPLY BASES. In other words, you must avoid thecommon error of writing���
���XXXXXX22 ·23 = 45. Try it with your calculator and check which is right!
Example 5
Multiply 22 ·33.
Solution
22 ·33 = 4 ·27 = 108
In this case, the bases are different. The product rule for powers ONLY APPLIES TO TERMS THAT HAVE THESAME BASE. Common mistakes with problems like this include���
���XXXXXX22 ·33 = 65.
Use the Power of a Product Property
We will now look at what happens when we raise a whole expression to a power. Let’s take x to the power 4 andcube it. Again we will us the full factored form for each.
(x4)3 = x4 · x4 · x4 3 factors of x to the power 4.
(x · x · x · x)︸ ︷︷ ︸x4
·(x · x · x · x)︸ ︷︷ ︸x4
·(x · x · x · x)︸ ︷︷ ︸x4
= (x · x · x · x · x · x · x · x · x · x · x · x)︸ ︷︷ ︸x12
So (x4)3 = x12. It is clear that when we raise a power of x to a new power, the powers multiply.
When we take an expression and raise it to a power, we are multiplying the existing powers of x by the power abovethe parenthesis.
Power rule for exponents: (xn)m = xn·m
Power of a product
If we have a product inside the parenthesis and a power on the parenthesis, then the power goes on each elementinside. So that, for example, (x2y)4 =)(x2)4 · (y)4 = x8y4. Watch how it works the long way.
(x · x · y)︸ ︷︷ ︸x2y
·(x · x · y)︸ ︷︷ ︸x2y
·(x · x · y)︸ ︷︷ ︸x2y
·(x · x · y)︸ ︷︷ ︸x2y
= (x · x · x · x · x · x · x · x · y · y · y · y)︸ ︷︷ ︸x8y4
Power rule for exponents: (xn)m = xnm and (xnym)p = xnpymp
WATCH OUT! This does NOT work if you have a sum or difference inside the parenthesis. For example, (x+y)2 6= x2 + y2. This is a commonly made mistake. It is easily avoidable if you remember what an exponent means(x+ y)2 = (x+ y)(x+ y). We will learn how to simplify this expression in a later chapter.
Let’s apply the rules we just learned to a few examples.
When we have numbers, we just evaluate and most of the time it is not really important to use the product rule andthe power rule.
Example 6
Simplify the following expressions.
(a) 34 ·37
(b) 26 ·2
(c) (42)3
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Solution
In each of the examples, we want to evaluate the numbers.
(a) Use the product rule first: 35 ·37 = 312
Then evaluate the result: 312 = 531,441
OR
We can evaluate each part separately and then multiply them. 35 ·37 = 243 ·2,187 = 531,441.
(b) Use the product rule first. 26 ·2 = 27
Then evaluate the result. 27 = 128
OR
We can evaluate each part separately and then multiply them. 26 ·2 = 64 ·2 = 128
(c) Use the power rule first. (42)3 = 46
Then evaluate the result. 46 = 4096
OR
We evaluate inside the parenthesis first. (42)3 = (16)3
Then apply the power outside the parenthesis. (16)3 = 4096
When we have just one variable in the expression then we just apply the rules.
Example 7
Simplify the following expressions.
(a) x2 · x7
(b) (y3)5
Solution
(a) Use the product rule. x2 · x7 = x2+7 = x9
(b) Use the power rule. (y3)5 = y3·5 = y15
When we have a mix of numbers and variables, we apply the rules to the numbers or to each variable separately.
Example 8
Simplify the following expressions.
(a) (3x2y3) · (4xy2)
(b) (4xyz) · (x2y3) · (2yz4)
(c) (2a3b3)2
Solution
(a) We group like terms together.
(3x2y3) · (4xy2) = (3 ·4) · (x2 · x) · (y3 · y2)
We multiply the numbers and apply the product rule on each grouping.
12x3y5
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(b) We groups like terms together.
(4xyz) · (x2y3) · (2yz4) = (4 ·2) · (x · x2) · (y · y3 · y) · (z · z4)
We multiply the numbers and apply the product rule on each grouping.
8x3y5z5
(c) We apply the power rule for each separate term in the parenthesis.
(2a3b3)2 = 22 · (a3)2 · (b3)2
We evaluate the numbers and apply the power rule for each term.
4a6b6
In problems that we need to apply the product and power rules together, we must keep in mind the order of operation.Exponent operations take precedence over multiplication.
Example 9
Simplify the following expressions.
(a) (x2)2 · x3
(b) (2x2y) · (3xy2)3
(c) (4a2b3)2 · (2ab4)3
Solution
(a) (x2)2 · x3
We apply the power rule first on the first parenthesis.
(x2)2 · x3 = x4 · x3
Then apply the product rule to combine the two terms.
x4 · x3 = x7
(b) (2x2y) · (3xy2)3
We must apply the power rule on the second parenthesis first.
(2x2y) · (3xy2)3 = (2x2y) · (27x3y6)
Then we can apply the product rule to combine the two parentheses.
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(2x2y) · (27x3y6) = 54x5y7
(c) (4a2b3)2 · (2ab4)3
We apply the power rule on each of the parentheses separately.
(4a2b3)2 · (2ab4)3 = (16a4b6) · (8a3b12)
Then we can apply the product rule to combine the two parentheses.
(16a4b6) · (8a3b12) = 128a7b18
Video Overview
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Homework
Write in exponential notation.
1. 4 ·4 ·4 ·4 ·42. 3x ·3x ·3x3. (−2a)(−2a)(−2a)(−2a)4. 6 ·6 ·6 · x · x · y · y · y · y
Find each number:
1. 54
2. (−2)6
3. (0.1)5
4. (−0.6)3
Multiply and simplify.
1. 63 ·66
2. 22 ·24 ·26
3. 32 ·43
4. x2 · x4
5. (−2y4)(−3y)6. (4a2)(−3a)(−5a4)
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Simplify.
1. (a3)4
2. (xy)2
3. (3a2b3)4
4. (−2xy4z2)5
5. (−8x)3(5x)2
6. (4a2)(−2a3)4
7. (12xy)(12xy)2
8. (2xy2)(−x2y)2(3x2y2)
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5.3 Multiplication of Polynomials
What You Will Learn
• Multiply a polynomial by a monomial• Multiply a polynomial by a binomial
Multiply a Polynomial by a Monomial
When multiplying polynomials we must remember the exponent rules that we learned in the previous sections.
The Product Rule xn · xm = xn+m
This says that if we multiply expressions that have the same base, we just add the exponents and keep the baseunchanged.
If the expressions we are multiplying have coefficients and more than one variable, we multiply the coefficients justas we would any number and we apply the product rule on each variable separately.
(2x2y3) · (3x2y) = (2 ·3) · (x2+2) · (y3+1) = 6x4y4
To multiply monomials, we need to multiply the coefficients separately and then apply the exponent rules to eachvariable separately. Let’s try some examples.
Example 1
Multiply the following monomials.
a) (2x2)(5x3)
b) (−3y4)(2y2)
c) (3xy5)(−6x4y2)
d) (−12a2b3c4)(−3a2b2)
Solution
a) (2x2)(5x3) = (2 ·5) · (x2 · x3) = 10x2+3 = 10x5
b) (−3y4)(2y2) = (−3 ·2) · (y4 · y2) =−6y4+2 =−6y6
c) (3xy5)(−6x4y2) = 18x1+4y5+2 =−18x5y7
d) (−12a2b3c4)(−3a2b2) = 36a2+2b3+2c4 = 36a4b5c4
To multiply a polynomial by a monomial, we use the Distributive Property.
Example 2
Multiply
a) 3(x2 +3x−5)
b) 4x(3x2−7)
c) −7y(4y2−2y+1)
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Solution
a) 3(x2 +3x−5) = 3(x2)+3(3x)−3(5) = 3x2 +9x−15
b) 4x(3x2−7) = (4x)(3x2)+(4x)(−7) = 12x3−28x
c) −7y(4y2−2y+1) = (−7y)(4y2)+(−7y)(−2y)+(−7y)(1) =−28y3 +14y2−7y
Notice that the use of the Distributive Property simplifies the problems to just multiplying monomials by monomialsand adding all the separate parts together.
Example 3
Multiply
a) 2x3(−3x4 +2x3−10x2 +7x+9)
b) −7a2bc3(5a2−3b2−9c2)
Solution
a)
2x3(−3x4 +2x3−10x2 +7x+9) = (2x3)(−3x4)+(2x3)(2x3)+(2x3)(−10x2)+(2x3)(7x)+(2x3)(9)
=−6x7 +4x6−20x5 +14x4 +18x3
b)
−7a2bc3(5a2−3b2−9c2) = (−7a2bc3)(5a2)+(−7a2bc3)(−3b2)+(−7a2bc3)(−9c2)
=−35a4bc3 +21a2b3c3 +63a2bc5
Multiply a Polynomial by a Binomial
Let’s start by multiplying two binomials together. A binomial is a polynomial with two terms, so a product of twobinomials will take the form.
(a+b)(c+d)
The Distributive Property also applies in this situation. Let’s think of the first parenthesis as one term. TheDistributive Property says that the term in front of the parenthesis multiplies with each term inside the parenthesisseparately. Then, we add the results of the products.
(a+b)(c+d) = a · (c+d)+b · (c+d)
We see that we can apply the distributive property on each of the parenthesis in turn.
a · (c+d)+b · (c+d) = a · c+a ·d +b · c+b ·d (or ac+ad +bc+bd)
What you should notice is that when multiplying any two polynomials, every term in one polynomial is multipliedby every term in the other polynomial.
A common phrase that describes this process is FOIL, this stands for First, Outside, Inside, Last. The terms thatcome first in each parenthesis are multiplied together, then the two outside terms, then the inside terms, then the lastterms in each parenthesis.
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Let’s look at some examples of multiplying polynomials.
Example 4
Multiply and simplify (2x+1)(x+3)
Solution
We must multiply each term in the first polynomial with each term in the second polynomial.
Let’s try to be systematic to make sure that we get all the products.
First, multiply the first term in the first parenthesis by all the terms in the second parenthesis.
We are now done with the first term.
Now we multiply the second term in the first parenthesis by all terms in the second parenthesis and add them to theprevious terms.
We are done with the multiplication and we can simplify.
(2x)(x)+(2x)(3)+(1)(x)+(1)(3) = 2x2 +6x+ x+3
= 2x2 +7x+3
This way of multiplying polynomials is called in-line multiplication or horizontal multiplication.
Example 5
Multiply and simplify
(a) (4x−5)(x−20)
(b) (3x−2)(3x+2)
(c) (2x−3)(3x2 +2x−5)
Solution
a)
(4x−5)(x−20) = (4x)(x)+(4x)(−20)+(−5)(x)+(−5)(−20)
= 4x2−80x−5x+100 = 4x2−85x+100
b)
(3x−2)(3x+2) = (3x)(3x)+(3x)(2)+(−2)(3x)+(−2)(2)
= 9x2 +6x−6x−4 = 9x2−4
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(c)
(2x−3)(3x2 +2x−5) = (2x)(3x2)+(2x)(2x)+(2x)(−5)+(−3)(3x2)+(−3)(2x)+(−3)(−5)
= 6x3 +4x2−10x−9x2−6x+15 = 6x3−5x2−16x+15
Video Overview
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Homework
Multiply the following monomials.
1. (2x)(−7x)2. (−5a2b)(−12a3b3)3. (3xy2z2)(15x2yz3)
Multiply and simplify.
1. 2x(4x−5)2. 9x3(3x2−2x+7)3. −3a2b(9a2−4b2)4. (x−3)(x+2)5. (a2 +2)(3a2−4)6. (7x−2)(9x−5)7. (2x−1)(2x2− x+3)8. (3x+2)(9x2−6x+4)9. (a2 +2a−3)(a2−3a+4)
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5.4 Special Products of Polynomials
What You Will Learn
• Find the square of a binomial• Find the product of binomials using sum and difference formula
Find the Square of a Binomial
A special binomial product is the square of a binomial. Consider the following multiplication.
(x+4)(x+4)
Since we are multiplying the same expression by itself that means that we are squaring the expression. This meansthat:
(x+4)(x+4) = (x+4)2
Lets multiply: (x+4)(x+4) = x2 +4x+4x+16
And combine like terms: = x2 +8x+16
Notice that the middle terms are the same. Is this a coincidence? In order to find that out, let’s square a generallinear binomial.
(a+b)2 = (a+b)(a+b) = a2 +ab+ab+b2
= a2 +2ab+b2
It looks like the middle terms are the same again. So far we have squared the sum of binomials. Let’s now square adifference of binomials.
(a−b)2 = (a−b)(a−b) = a2−ab−ab+b2
= a2−2ab+b2
We notice a pattern when squaring binomials. To square a binomial, add the square of the first term, add or subtracttwice the product of the terms, and the square of the second term. You should remember these formulas:
Square of a Binomial
(a+b)2 = a2 +2ab+b2 and (a−b)2 = a2−2ab−b2
Remember! A polynomial that is raised to an exponent means that we multiply the polynomial by itself howevermany times the exponent indicates. For instance
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(a+b)2 = (a+b)(a+b)
Don’t make the common mistake (a+b)2 = a2 +b2. To see why (a+b)2 6= a2 +b2 try substituting numbers for aand b into the equation (for example, a = 4 and b = 3), and you will see that it is not a true statement. The middleterm, 2ab, is needed to make the equation work.
We can apply the formulas for squaring binomials to any number of problems.
Example 1
Square each binomial and simplify.
(a) (x+10)2
(b) (2x−3)2
(c) (x2 +4)2
(d) (5x−2y)2
Solution
Let’s use the square of a binomial formula to multiply each expression.
a) (x+10)2
If we let a = x and b = 10, then
(a2 + b) = a2 + 2a b + b2
↓ ↓ ↓ ↓ ↓ ↓(x + 10)2 = (x)2 +2(x)(10)+(10)2
= x2 +20x+100
b) (2x−3)2
If we let a = 2x and b = 3, then
(a−b)2 = a2−2ab+b2
(2x−3)2 = (2x)2−2(2x)(3)+(3)2
= 4x2−12x+9
c) (x+4)2
If we let a = x2 and b = 4, then
(x2 +4)2 = (x2)2 +2(x2)(4)+(4)2
= x4 +8x2 +16
d) (5x−2y)2
If we let a = 5x and b = 2y, then
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(5x−2y)2 = (5x)2−2(5x)(2y)+(2y)2
= 25x2−20xy+4y2
Find the Product of Binomials Using Sum and Difference Patterns
Another special binomial product is the product of a sum and a difference of terms. For example, let’s multiply thefollowing binomials.
(x+4)(x−4) = x2−4x+4x−16
= x2−16
Notice that the middle terms are opposites of each other, so they cancel out when we collect like terms. This is nota coincidence. This always happens when we multiply a sum and difference of the same terms.
(a+b)(a−b) = a2−ab−b2
= a2−b2
When multiplying a sum and difference of the same two terms, the middle terms cancel out. We get the square ofthe first term minus the square of the second term. You should remember this formula.
Sum and Difference Formula (also called Difference of the Squares or DOTS)
(a+b)(a−b) = a2−b2
Let’s apply this formula to a few examples.
Example 2
Multiply the following binomials and simplify.
(a) (x+3)(x−3)
(b) (5x+9)(5x−9)
(c) (2x3 +7)(2x3−7)
(d) (4x+5y)(4x−5y)
Solution
(a) Let a = x and b = 3, then
(a+b)(a−b) = a2 − b2
↓ ↓ ↓ ↓ ↓ ↓(x+3)(x−3) = (x)2− (3)2
= x2−9
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(b) Let a = 5x and b = 9, then
(a + b)(a − b) = a2 − b2
↓ ↓ ↓ ↓ ↓ ↓(5x+9)(5x−9) = (5x)2− (9)2
= 25x2−81
(c) Let a = 2x3 and b = 7, then
(2x3 +7)(2x3−7) = (2x3)2− (7)2
= 4x6−49
(d) Let a = 4x and b = 5y, then
(4x+5y)(4x−5y) = (4x)2− (5y)2
= 16x2−25y2
Video Overview
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Homework
Use the special product for squaring binomials to multiply these expressions.
1. (x+9)2
2. (3x−7)2
3. (4x2 + y2)2
4. (8x−3)2
Use the special product of a sum and difference to multiply these expressions.
1. (2x−1)(2x+1)2. (x−12)(x+12)3. (5a−2b)(5a+2b)4. (ab−1)(ab+1)
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5.5 Exponent Properties Involving Quotients
What You Will Learn
• Use the quotient of powers property.• Use the power of a quotient property.• Simplify expressions involving quotient properties of exponents.
Use the Quotient of Powers Property
You saw in a previous section that we can use exponent rules to simplify products of numbers and variables. Inthis section, you will learn that there are similar rules you can use to simplify quotients. Let’s take an example of aquotient, x7 divided by x4.
x7
x4 = �x ·�x ·�x ·�x · x · x · x�x ·�x ·�x ·�x
=x · x · x
1= x3
You should see that when we divide two powers of x, the number of factors of x in the solution is the differencebetween the factors in the numerator of the fraction, and the factors in the denominator. In other words, whendividing expressions with the same base, keep the base and subtract the exponent in the denominator from theexponent in the numerator.
Quotient Rule for Exponents: xn
xm = xn−m
When we have problems with different bases, we apply the quotient rule separately for each base.
x5y3
x3y2 = �x ·�x ·�x · x · x�x ·�x ·�x
· �y ·�y · y�y ·�y
=x · x1· y
1= x2y OR
x5y3
x3y2 = x5−3 · y3−2 = x2y
Example 1
Simplify each of the following expressions using the quotient rule.
(a) x10
x5
(b) a6
a
(c) a5b4
a3b2
Solution
Apply the quotient rule.
(a) x10
x5 = x10−5 = x5
(b) a6
a = a6−1 = a5
(c) a5b4
a3b2 = a5−3 ·b4−2 = a2b2
Now let’s see what happens if the exponent on the denominator is bigger than the exponent in the numerator.
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Example 2
Divide. x4÷ x7 Apply the quotient rule.
x4
x7 = x4−7 = x−3
A negative exponent!? What does that mean?
Let’s do the division longhand by writing each term in factored form.
x4
x6 = �x ·�x ·�x ·�x�x ·�x ·�x ·�x · x · x
=1
x · x=
1x2
We see that when the exponent in the denominator is bigger than the exponent in the numerator, we still subtract thepowers. This time we subtract the smaller power from the bigger power and we leave the x’s in the denominator.
When you simplify quotients, to get answers with positive exponents you subtract the smaller exponent from thebigger exponent and leave the variable where the bigger power was.
• We also discovered what a negative power means x−3 = 1x3 . We’ll learn more on this in the next section!
Example 3
Simplify the following expressions, leaving all powers positive.
(a) x2
x6
(b) a2b6
a5b
Solution
(a) Subtract the exponent in the numerator from the exponent in the denominator and leave the x’s in the denominator.
x2
x6 =1
x6−2 =1x4
(b) Apply the rule on each variable separately.
a2b6
a5b=
1a5−2 ·
b6−1
1=
b5
a3
The Power of a Quotient Property
When we apply a power to a quotient, we can learn another special rule. Here is an example.
(x3
y2
)4
=
(x3
y2
)·(
x3
y2
)·(
x3
y2
)·(
x3
y2
)=
(x · x · x) · (x · x · x) · (x · x · x) · (x · x · x)(y · y) · (y · y) · (y · y) · (y · y)
=x12
y8
Notice that the power on the outside of the parenthesis multiplies with the power of the x in the numerator and thepower of the y in the denominator. This is called the power of a quotient rule.
Power Rule for Quotients(
xn
ym
)p= xn·p
ym·p
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Simplifying Expressions Involving Quotient Properties of Exponents
Let’s apply the rules we just learned to a few examples.
• When we have numbers with exponents and not variables with exponents, we evaluate.
Example 4
Simplify the following expressions.
(a) 45
42
(b) 53
57
(c)(
34
52
)2
Solution
In each of the examples, we want to evaluate the numbers.
(a) Use the quotient rule first.
45
42 = 45−2 = 43
Then evaluate the result.
43 = 64
OR
We can evaluate each part separately and then divide.
102416
= 64
(b) Use the quotient rule first.
53
57 =1
57−3 =154
Then evaluate the result.
154 =
1625
OR
We can evaluate each part separately and then reduce.
53
57 =125
78125=
1625
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It makes more sense to apply the quotient rule first for examples (a) and (b). In this way the numbers we areevaluating are smaller because they are simplified first before applying the power.
(c) Use the power rule for quotients first.
(34
52
)2
=38
54
Then evaluate the result.
38
54 =6561625
OR
We evaluate inside the parenthesis first.
(34
52
)2
=
(8125
)2
Then apply the power outside the parenthesis.
(8125
)2
=6561625
When we have just one variable in the expression, then we apply the rules straightforwardly.
Example 5: Simplify the following expressions:
(a) x12
x5
(b)(
x4
x
)5
Solution:
(a) Use the quotient rule.
x12
x5 = x12−5 = x7
(b) Use the power rule for quotients first.
(x4
x
)5
=x20
x5
Then apply the quotient rule
x20
x5 = x15
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OR
Use the quotient rule inside the parenthesis first.
(x4
x
)5
= (x3)5
Then apply the power rule.
(x3)5 = x15
When we have a mix of numbers and variables, we apply the rules to each number or each variable separately.
Example 6
Simplify the following expressions.
(a) 6x2y3
2xy2
(b)(
2a3b3
8a7b
)2
Solution
(a) We group like terms together.
6x2y3
2xy2 =62· x
2
x· y
3
y2
We reduce the numbers and apply the quotient rule on each grouping.
3xy >
(b) We apply the quotient rule inside the parenthesis first.
(2a3b3
8a7b
)2
=
(b2
4a4
)2
Apply the power rule for quotients.
(b2
4a4
)2
=b4
16a8
In problems that we need to apply several rules together, we must keep in mind the order of operations.
Example 7
Simplify the following expressions.
(a) (x2)2 · x6
x4
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(b)(
16a2
4b5
)3· b2
a16
Solution
(a) We apply the power rule first on the first parenthesis.
(x2)2 · x6
x4 = x4 · x6
x4
Then apply the quotient rule to simplify the fraction.
x4 · x6
x4 = x4 · x4
Apply the product rule to simplify.
x4 · x2 = x6
(b) Simplify inside the first parenthesis by reducing the numbers.
(4a2
b5
)3
· b2
a16
Then we can apply the power rule on the first parenthesis.
(4a2
b5
)3
· b2
a16 =64a6
b15 ·b2
a16
Group like terms together.
64a6
b15 ·b2
a16 = 64 · a6
a16 ·b2
b15
Apply the quotient rule on each fraction.
64 · a6
a16 ·b2
b15 =64
a10b13
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Homework
Evaluate the following expressions.
1. 56
52
2. 67
63
3. 34
310
4.(
22
33
)3
Simplify the following expressions.
1. a3
a2
2. x5
x9
3.(
a3b4
a2b
)3
4. x6y2
x2y5
5. 6a3
2a2
6. 15x5
5x
7.(
18a4
15a10
)4
8. 25yx6
20y5x2
9.(
x6y2
x4y4
)3
10.(
6a2
4b4
)2· 5b
3a
11. (3ab)2(4a3b4)3
(6a2b)4
12. (2a2bc2)(6abc3)4ab2c
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5.6 Zero and Negative Exponents
What You Will Learn
• Simplify expressions with zero exponents.• Simplify expressions with negative exponents.• Evaluate exponential expressions with no variables.
Simplify Expressions with Exponents of Zero
Let us look again at the quotient rule for exponents (that xn
xm = xn−m) and consider what happens when n = m. Let’stake the example of x4 divided by x4.
x4
x4 = x(4−4) = x0
Now we arrived at the quotient rule by considering how the factors of x cancel in such a fraction. Let’s do that againwith our example of x4 divided by x4.
x4
x4 = �x ·�x ·�x ·�x�x ·�x ·�x ·�x
= 1
So x0 = 1.
This works for any value of the exponent, not just 4.
xn
xn = xn−n = x0
Since there is the same number of factors in the numerator as in the denominator, they cancel each other out and weobtain x0 = 1. The zero exponent rule says that any number raised to the power zero is one.
Zero Rule for Exponents: x0 = 1, x 6= 0
Simplify Expressions With Negative Exponents
Again we will look at the quotient rule for exponents (that xn
xm = xn−m) and this time consider what happens whenm > n. Let’s take the example of x4 divided by x6.x4
x6 = x(4−6) = x−2 for x 6= 0.
By the quotient rule our exponent for x is -2. But what does a negative exponent really mean? Let’s do the samecalculation long-hand by dividing the factors of x4 by the factors of x6.
x4
x6 = �x ·�x ·�x ·�x�x ·�x ·�x ·�x · x · x
=1
x · x=
1x2
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So we see that x to the power -2 is the same as one divided by x to the power +2. Here is the negative power rule forexponents.
Negative Power Rule for Exponents 1xn = x−n x 6= 0
You will also see negative powers applied to products and fractions. For example, here it is applied to a product.
(x3y)−2 = x−6y−2 using the power rule
x−6y−2 =1x6 ·
1y2 =
1x6y2 using the negative power rule separately on each variable
Here is an example of a negative power applied to a quotient.
(ab
)−3=
a−3
b−3 using the power rule for quotients
a−3
b−3 =a−3
1· 1
b−3 =1a3 ·
b3
1using the negative power rule on each variable separately
1a3 ·
b3
1=
b3
a3 simplifying the division of fractions
b3
a3 =
(ba
)3
using the power rule for quotients in reverse.
The last step is not necessary but it helps define another rule that will save us time. A fraction to a negative power is“flipped”.
Negative Power Rule for Fractions(
xy
)−n=( y
x
)n, x 6= 0,y 6= 0
In some instances, it is more useful to write expressions without fractions and that makes use of negative powers.
Example 1
Write the following expressions without fractions.
(a) 1x
(b) 2x2
(c) x2
y3
(d) 3xy
Solution
We apply the negative rule for exponents 1xn = x−n on all the terms in the denominator of the fractions.
(a) 1x = x−1
(b) 2x2 = 2x−2
(c) x2
y3 = x2y−3
(d) 3xy = 3x−1y−1
Sometimes, it is more useful to write expressions without negative exponents.
Example 2
Write the following expressions without negative exponents.
(a) 3x−3
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(b) a2b−3c−1
(c) 4x−1y3
(d) 2x−2
y−3
Solution
We apply the negative rule for exponents 1xn = x−n on all the terms that have negative exponents.
(a) 3x−3 = 3x3
(b) a2b−3c−1 = a2
b3c
(c) 4x−1y3 = 4y3
x
(d) 2x−2
y−3 = 2y3
x2
Example 3
Simplify the following expressions and write them without fractions.
(a) 4a2b3
2a5b
(b)(
x3y2
)3· x2y
4
Solution
(a) Reduce the numbers and apply quotient rule on each variable separately.
4a2b3
6a5b= 2 ·a2−5 ·b3−1 = 2a−3b2
(b) Apply the power rule for quotients first.
(2xy2
)3
· x2y4
=8x2
y6 ·x2y4
Then simplify the numbers, use product rule on the x’s and the quotient rule on the y’s.
8x3
y6 ·x2y4
= 2 · x3+2 · y1−6 = 2x5y−5
Example 4
Simplify the following expressions and write the answers without negative powers.
(a)(
ab−2
b3
)2
(b) x−3y2
x2y−2
Solution
(a) Apply the quotient rule inside the parenthesis.
(ab−2
b3
)2
= (ab−5)2
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Apply the power rule.
(ab−5)2 = a2b−10 =a2
b10
(b) Apply the quotient rule on each variable separately.
x−3y2
x2y−2 = x−3−2y2−(−2) = x−5y4 =y4
x5
Evaluate Exponential Expressions with No Variables
The same rules apply, don’t be tempted to multiply a base and it’s negative exponent as in example (d) below
Example 6
Evaluate the following expressions to a single number.
(a) 50
(b) 72
(c)(2
3
)3
(d) 3−3
Solution
(a) 50 = 1 Remember that a number raised to the power 0 is always 1.
(b) 72 = 7 ·7 = 49
(c)(2
3
)3= 23
33 =827
(d) 3−3 = 133 =
127 a common mistake on this problem is to multiply 3 by -3.
Video Overview
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Homework
Simplify the following expressions, be sure that there aren’t any negative exponents in the answer.
1. x−1 · y2
2. x−4
3. x−3
x−7
4. x−3y−5
z−7
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5.(a
b
)−2
6. (3a−2b2c3)3
7. x−3 · x3
8.
x4
x−5
9.
2x−4x4
10. 2−4
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5.7. Division of Polynomials by Monomials www.ck12.org
5.7 Division of Polynomials by Monomials
What You Will Learn
• Divide a polynomial by a monomial.
Divide a Polynomial by a Monomial
We’ll start by dividing a polynomial by a monomial. To do this, we divide each term of the polynomial by themonomial. When the numerator has more than one term, the monomial on the bottom of the fraction serves as thecommon denominator to all the terms in the numerator.
Example 1
Divide.
a) 8x2−4x+162
b) 3x2+6x−1x
c) −3x2−18x+69x
Solution
a) 8x2−4x+162 = 8x2
2 −4x2 + 16
2 = 4x2−2x+8
b) 3x3+6x−1x = 3x3
x + 6xx −
1x = 3x2 +6− 1
x
c) −3x2−18x+69x =−3x2
9x −18x9x + 6
9x =−x3 −2+ 2
3x
A common error is to cancel the denominator with just one term in the numerator.
Consider the quotient 3x+44 .
Remember that the denominator of 4 is common to both the terms in the numerator. In other words we are dividingboth of the terms in the numerator by the number 4.
The correct way to simplify is:
3x+44
=3x4+
44=
3x4+1
A common mistake is to cross out the number 4 from the numerator and the denominator, leaving just 3x. This isincorrect, because the entire numerator needs to be divided by 4.
Example 2
Divide 5x3−10x2+x−25−5x2 .
Solution
5x3−10x2 + x−25−5x2 =
5x3
−5x2 −10x2
−5x2 +x−5x2 −
25−5x2
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The negative sign in the denominator changes all the signs of the fractions:
−5x3
5x2 +10x2
5x2 −x
5x2 +255x2 =−x+2− 1
5x+
5x2
Video Overview of Dividing a Polynomial by a Monomial
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Homework
Divide the following polynomials:
1. 2x+42
2. x−4x
3. 5x−355x
4. x2+2x−5x
5. 4x2+12x−36−4x
6. 2x2+10x+72x2
7. x3−x−2x2
8. 5x4−93x
9. x3−12x2+3x−412x2
10. 3−6x+x3
−9x3
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5.8. Division of Polynomials by Polynomials www.ck12.org
5.8 Division of Polynomials by Polynomials
What You Will Learn
• To divide one polynomial by another using long division.
Long Division
We will be solving problems like the following:
2x−3 ) 2x3−3x2−8x+12
This question is asking 2x−3 times what would be 2x3−3x2−8x+12? This is done with a procedure very similarto long division with numbers.
The first question to ask yourself is how many times does 2x go into 2x3? The answer is x2 times.
x2
2x−3 ) 2x3−3x2−8x+122x3−3x2
0
Place x2 above the x2 term in the polynomial.
Multiply x2 by both terms in the divisor (2x and -3) and place them until their like terms. Subtract from the dividend(2x3−3x2−8x+12). Pull down the next two terms and repeat.
x2 −42x−3 ) 2x3−3x2−8x+12
2x3−3x2
−8x+12−8x+12
−8x divided by 2x =−4
After multiplying both terms in the divisor by -4, place that under the terms you brought down. When subtractingwe notice that everything cancels out. Therefore, just like we thought, x2−4 is a factor.
When dividing polynomials, not every divisor will go in evenly to the dividend. If there is a remainder, write it as afraction over the divisor.
Example A
(2x3−6x2 +5x−20)÷ (x2−5)
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Solution: Set up the problem using a long division bar.
x2−5 ) 2x3−6x2+5x−20
How many times does x2 go into 2x3? 2x times.
2xx2−5 ) 2x3 −6x2+5x−20
2x3−10x2
4x2+5x−20
Multiply 2x by the divisor. Subtract that from the dividend.
Repeat the previous steps. Now, how many times does x2 go into 4x2? 4 times.
2x +4x2−5 ) 2x3 −6x2+5x−20
2x3−10x2
4x2+5x−204x2 −20
5x
At this point, we are done. x2 cannot go into 5x because it has a higher degree. Therefore, 5x is a remainder. Thecomplete answer would be 2x+4+ 5x
x2−5 .
Example B
(3x4 + x3−17x2 +19x−6)÷ (x2−2x+1). Determine if x2−2x+1 goes evenly into 3x4 + x3−17x2 +19x−6. Ifso, try to factor the divisor and quotient further.
Solution: First, do the long division. If x2−2x+1 goes in evenly, then the remainder will be zero.
3x2 +7x −6x2−2x+1 ) 3x4 +x3−17x2+19x−6
3x4−6x3 3x2
7x3−20x2+19x7x3−14x2 +7x
−6x2+12x−6−6x2+12x−6
0
This means that x2−2x+1 and 3x2 +7x−6 both go evenly into 3x4 + x3−17x2 +19x−6.
Practice Exercises
1. (5x4 +6x3−12x2−3)÷ (x2 +3)
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2. Is (x+4) a factor of x3−2x2−51x−108? If so, find any other factors.
Answers
1. Make sure to put a placeholder in for the x−term.
5x2 +6x −27x2 +3 ) 5x4+6x3−12x2 +0x −3
5x4 +15x2
6x3−27x2 +0x6x3 +18x
−27x2−18x −3−27x2 −81
−18x+78
The final answer is 5x2 +6x−27− 18x−78x2+3 .
2. Divide (x+4) into x3−2x2−51x−108 and if the remainder is zero, it is a factor.
x2 −6x −27x+4 ) x3−2x2−51x−108
x3+4x2
−6x2−51x−6x2−24x
−27x−108−27x−108
0
x+4 is a factor.
Video Overview of Dividing a Polynomial by a Polynomial
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Divide the following polynomials using long division.
1. (2x3 +5x2−7x−6)÷ (x+1)2. (x4−10x3 +15x−30)÷ (x−5)3. (2x4−8x3 +4x2−11x−1)÷ (x2−1)4. (3x3−4x2 +5x−2)÷ (3x+2)
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5. (3x4−5x3−21x2−30x+8)÷ (x−4)6. (2x5−5x3 +6x2−15x+20)÷ (2x2 +3)
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5.9 Scientific Notation
What You Will Learn
• Write numbers in scientific notation.• Evaluate expressions in scientific notation.
Introduction – Powers of 10
Consider the number six hundred and forty three thousand, two hundred and ninety seven. We write it as 643,297and each digit’s position has a “value” assigned to it. You may have seen a table like this before.
hundred-thousands ten-thousands thousands hundreds tens units (ones)
6 4 3 2 9 7
We have seen that when we write an exponent above a number it means that we have to multiply a certain number offactors of that number together. We have also seen that a zero exponent always gives us one, and negative exponentsmake fractional answers. Look carefully at the table above. Do you notice that all the column headings are powersof ten? Here they are listed.
100,000 = 105
10,000 = 104
1,000 = 103
100 = 102
10 = 101
Even the “units” column is really just a power of ten. Unit means 1 and 1 = 100.
If we divide 643,297 by 100,000 we get 6.43297. If we multiply this by 100,000 we get back to our original number.But we have just seen that 100,000 is the same as 105, so if we multiply 6.43297 by 105 we should also get ouroriginal answer. In other words
6.43297×105 = 643,297
So we have found a new way of writing numbers! What do you think happens when we continue the powers of ten?Past the units column down to zero we get into decimals, here the exponent becomes negative.
Writing Numbers Greater Than One in Scientific Notation
Scientific notation numbers are always written in the following form.
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a×10b
Where 1 ≤ a < 10 and b, the exponent, is an integer. This notation is especially useful for numbers that are eithervery small or very large. When we use scientific notation to write numbers, the exponent on the 10 determines theposition of the decimal point.
Look at the following examples.
1.07×104 = 10,700
1.07×103 = 1,070
1.07×102 = 107
1.07×101 = 10.7
1.07×100 = 1.07
1.07×10−1 = 0.107
1.07×10−2 = 0.0107
1.07×10−3 = 0.00107
1.07×10−4 = 0.000107
Look at the first term of the list and examine the position of the decimal point in both expressions.
So the exponent on the ten acts to move the decimal point over to the right. An exponent of 4 moves it 4 places andan exponent of 3 would move it 3 places.
Example 1
Write the following numbers in scientific notation.
(a) 63
(b) 9,654
(c) 653,937,000
(d) 1,000,000,006
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(a) 63 = 6.3×10 = 6.3×101
(b) 9,654 = 9.654×1,000 = 9.654×103
(c) 653,937,000 = 6.53937000×100,000,000 = 6.53937×108
(d) 1,000,000,006 = 1.000000006×1,000,000,000 = 1.000000006×109
Example 2
The Sun is approximately 93 million miles for Earth. Write this distance in scientific notation.
This time we will simply write out the number long-hand (with a decimal point) and count decimal places.
Solution
93,000,000,0︸ ︷︷ ︸7 decimal places
= 9.3×107 miles
A Note on Significant Figures
We often combine scientific notation with rounding numbers. If you look at Example 2, the distance you are givenhas been rounded. It is unlikely that the distance is exactly 93 million miles! Looking back at the numbers inExample 1, if we round the final two answers to 2 significant figures (2 s.f.) they become:
1(c) 6.5×108
1(d) 1.0×109
Note that the zero after the decimal point has been left in for Example 1(d) to indicate that the result has beenrounded. It is important to know when it is OK to round and when it is not.
Writing Numbers Less Than One in Scientific Notation
We have seen how we can use scientific notation to express large numbers, but it is equally good at expressingextremely small numbers. Consider the following example.
Example 3
The time taken for a light beam to cross a football pitch is 0.0000004 seconds. Express this time in scientific notation.
We will proceed in a similar way as before.
0.0000004 = 4×0.0000001 = 4× 110,000,000
= 4× 1107 = 4×10−7
So...
Just as a positive exponent on the ten moves the decimal point that many places to the right, a negative exponentmoves the decimal place that many places to the left.
Example 4
Express the following numbers in scientific notation.
(a) 0.003
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(b) 0.000056
(c) 0.00005007
(d) 0.00000000000954
Let’s use the method of counting how many places we would move the decimal point before it is after the firstnon-zero number. This will give us the value for our negative exponent.
(a) 0.003︸ ︷︷ ︸3 decimal places
= 3×10−3
(b) 0.000056︸ ︷︷ ︸5 decimal places
= 5.6×10−5
(c) 0.00005007︸ ︷︷ ︸5 decimal places
= 5.007×10−5
(d) 0.0000000000954︸ ︷︷ ︸12 decimal places
= 9.54×10−12
Evaluating Expressions in Scientific Notation
When we are faced with products and quotients involving scientific notation, we need to remember the rules forexponents that we learned earlier. It is relatively straightforward to work with scientific notation problems if youremember to deal with all the powers of 10 together. The following examples illustrate this.
Example 5
Evaluate the following expressions and write your answer in scientific notation.
(a) (3.2×106) · (8.7×1011)
(b) (5.2×10−4) · (3.8×10−19)
(c) (1.7×106) · (2.7×10−11)
The key to evaluating expressions involving scientific notation is to keep the powers of 10 together and deal withthem separately. Remember that when we use scientific notation, the leading number must be between 1 and 10.We need to move the decimal point over one place to the left. See how this adds 1 to the exponent on the 10.
(a)
(3.2×106) · (8.7×1011) = 3.2×8.7︸ ︷︷ ︸27.84
×106×1011︸ ︷︷ ︸1017
(3.2×106) · (8.7×1011) = 2.784×101×1017
Solution
(3.2×106) · (8.7×1011) = 2.784×1018
(b)
(5.2×10−4) · (3.8×10−19) = 5.2×3.8︸ ︷︷ ︸19.76
×10−4×10−19︸ ︷︷ ︸10−23
= 1.976×101×10−23
Solution
(5.2×10−4) · (3.8×10−19) = 1.976×10−22
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(c)
(1.7×106) · (2.7×10−11) = 1.7×2.7︸ ︷︷ ︸4.59
×106×10−11︸ ︷︷ ︸10−5
Solution
(1.7×106) · (2.7×10−11) = 4.59×10−5
Example 6
Evaluate the following expressions. Round to 3 significant figures and write your answer in scientific notation.
(a) (3.2×106)÷ (8.7×1011)
(b) (5.2×10−4)÷ (3.8×10−19)
(c) (1.7×106)÷ (2.7×10−11)
It will be easier if we convert to fractions and THEN separate out the powers of 10.
(a)
(3.2×106)÷ (8.7×1011) =3.2×106
8.7×1011 Next we separate the powers of 10.
=3.28.7× 106
1011 Evaluate each fraction (round to 3 s.f.):
= 0.368×10(6−11) Remember how to write scientific notation!
= 3.68×10−1×10−5
Solution
(3.2×106)÷ (8.7×1011) = 3.86×10−6 (rounded to 3 significant figures)
(b)
(5.2×10−4)÷ (3.8×1019) =5.2×10−4
3.8×10−19 Separate the powers of 10.
=5.23.8× 10−4
10−19 Evaluate each fraction (round to 3 s.f.).
= 1.37×10((−4)− (−19))
= 1.37×1015
Solution
(5.2×10−4)÷ (3.8×10−19) = 1.37×1015 (rounded to 3 significant figures)
(c)
(1.7×106)÷ (2.7×10−11) =1.7×106
2.7×10−11 Next we separate the powers of 10.
=1.72.7× 106
10−11 Evaluate each fraction (round to 3 s.f.).
= 0.630×10(6−(−11)) Remember how to write scientific notation!
= 6.30×10−1×1017
Solution
(1.7×106)÷ (2.7×10−11) = 6.30×1016 (rounded to 3 significant figures)
Note that the final zero has been left in to indicate that the result has been rounded.
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Video Overview
MEDIAClick image to the left for more content.
Homework
1. Write the numerical value of the following.
a. 3.102×102
b. 7.4×104
c. 1.75×10−3
d. 2.9×10−5
e. 9.99×10−9
2. Write the following numbers in scientific notation.
a. 120,000b. 1,765,244c. 12d. 0.00281e. 0.000000027
3. Evaluate
(a) (4×102)∗ (2×103)(b) (8×109)∗ (4×10−3)(c) (3×102)÷ (2×10−2)
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5.10. Square Roots and Irrational Numbers www.ck12.org
5.10 Square Roots and Irrational Numbers
What You Will Learn
• Square Roots• Rational and Irrational Numbers• Simplifying Square Roots
Square Roots
The square root of a number n is any number such that s2 = n.
Every positive number has two square roots, the positive and the negative. The symbol used to represent the squareroot is
√x. It is assumed that this is the positive square root of x. To show both the positive and negative values, you
can use the symbol ±, read “plus or minus.”
For example:√
81 = 9 means the positive square root of 81.
−√
81 =−9 means the negative square root of 81.
±√
81 =±9 means the positive or negative square root of 81.
Example A
A square measures 324 square meters. How long is one side of the square?
Solution: The area of a square is s2 = Area. The value of Area can be replaced with 324.
s2 = 324
The value of s represents the square root of 324.
s =√
324 = 18
The square is 18 meters long by 18 meters wide.
Approximating Square Roots
When the square root of a number is a whole number, this number is called a perfect square. 9 is a perfect squarebecause
√9 = 3.
Not all square roots are whole numbers. Many square roots are irrational numbers, meaning there is no rationalnumber equivalent. For example, 2 is the square root of 4 because 2×2 = 4. The number 7 is the square root of 49because 7×7 = 49. What is the square root of 5?
There is no whole number multiplied by itself that equals five, so√
5 is not a whole number. To find the value of√5, we can use estimation.
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To estimate the square root of a number, look for the perfect integers less than and greater than the value, and thenestimate the decimal.
Example B
Estimate√
5.
Solution: The perfect square below 5 is 4 and the perfect square above 5 is 9. Therefore, 4 < 5 < 9. Therefore,√
5is between
√4 and
√9, or 2 <
√5 < 3. Because 5 is closer to 4 than 9, the decimal is a low value:
√5≈ 2.2.
For example, suppose you do not have a calculator and you need to find√
18. You know there is no whole numbersquared that equals 18, so
√18 is an irrational number. The value is between
√16 = 4 and
√25 = 5. However, we
need to find a more simplified value of√
18.
Begin by writing the prime factorization of√
18.√
18 =√
9×2 =√
9×√
2.√
9 = 3 but√
2 does not have awhole number value. Therefore, the simplified value of
√18 is 3
√2.
You can check your answer on a calculator by finding the decimal approximation for each square root.
Example C
Find the simplified value of√
75.
Solution:
√75 =
√25×3 =
√25×
√3 = 5
√3
Guided Practice
The area of a square is 50 square feet. What are the lengths of its sides?
Solution:
We know that the formula for the area of a square is a = s2. Using this formula:
a = s2
50 = s2
√50 =
√s2
√50 = s
Now we will simplify:√
50 =√
25 ·2 = 5√
2.
The length of each side of the square is 5√
2 feet.
Video Overview of Square Roots
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MEDIAClick image to the left for more content.
Video Overview of Simplifying Square Roots
MEDIAClick image to the left for more content.
Homework
Find the following square roots. Give your answer in the simplest form.
1.√
252.√
243.√
204.√
2005.√
2000
6.
√14
7.
√94
8.√
0.169.√
0.110.√
0.01
Use a calculator to find the following square roots. Round to two decimal places.
1.√
132.√
993.√
1234.√
25.√
20006.√
0.257.√
1.358.√
0.379.√
0.710.√
0.01
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www.ck12.org Chapter 5. Unit 5 - Polynomials, Exponents and Radicals
5.11 Pythagorean Theorem
What You Will Learn
• Pythagorean Theorem
Pythagorean Theorem
One of the most important theorems in history is Pythagorean’s Theorem. Put it words, it states, “The sum of thesquare of each leg of a right triangle is equal to the square of the hypotenuse.”
Let’s review basic right triangle anatomy.
The two segments forming the right angle (90◦) are called the legs of the right triangle. The segment opposite theright angle is called the hypotenuse.
So, the Pythagorean Theorem states, (leg1)2 +(leg2)
2 = (hypotenuse)2:
a2 +b2 = c2
Or, to find the hypotenuse, c =√
a2 +b2.
This theorem will give you practice in using square roots
Homework
Find the missing length of each right triangle.
1. a = 12, b = 16, c =?2. a =? , b = 20, c = 303. a = 4, b =? , c = 11
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4.
5.
6.7. One leg of a right triangle is 4 feet less than the hypotenuse. The other leg is 12 feet. Find the lengths of the
three sides of the triangle.8. One leg of a right triangle is 3 more than twice the length of the other. The hypotenuse is 3 times the length
of the short leg. Find the lengths of the three legs of the triangle.
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5.12 Multiplication and Division of Radicals
What You Will Learn
• Multiply radicals• Divide radicals
Multiply Radicals
Example A
Simplify√
3 ·√
12.
Solution:√
3 ·√
12 =√
36 = 6
Example B
√3(√
6+5) =√
18+5√
3
which simplifies to
3√
2+5√
3
Divide Radicals
Dividing radicals is more complicated. A radical in the denominator of a fraction is not considered simplified bymathematicians. In order to simplify the fraction, you must rationalize the denominator.
To rationalize the denominator means to remove any radical signs from the denominator of the fraction usingmultiplication.
Remember:√
a×√
a =√
a2 = a
Example B
Simplify 2√3
.
Solution:
We must clear the denominator of its radical using the property above. Remember, what you do to one piece of afraction, you must do to all pieces of the fraction.
2√3×√
3√3=
2√
3√32
=2√
33
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5.12. Multiplication and Division of Radicals www.ck12.org
Video Overview or Rationalizing the Denominator
MEDIAClick image to the left for more content.
Homework
Multiply the following expressions.
1.√
6(√
10+√
8)
2.(√
a−√
b)(√
a+√
b)
3.(2√
x+5)(
2√
x+5)
Rationalize the denominator.
1. 7√15
2. 9√10
3. 2x√5x
4.√
5√3y
5. The volume of a spherical balloon is 950cm3. Find the radius of the balloon. (Volume of a sphere = 43 πR3)
6. A rectangular picture is 9 inches wide and 12 inches long. The picture has a frame of uniform width. If thecombined area of picture and frame is 180in2, what is the width of the frame?
7. The volume of a soda can is 355 cm3. The height of the can is four times the radius of the base. Find theradius of the base of the cylinder.
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5.13 Addition and Subtraction of Radicals
What You Will Learn
• Add and Subtract Radicals
Add and Subtract Radiclas
To add or subtract radicals, they must have the same square root associated with them
a n√x+b n√x = (a+b) n√x
Example A
Add: 3√
5+6√
5.
Solution:
The value “√
5” is considered a like term. Using the rule above:
3√
5+6√
5 = (3+6)√
5 = 9√
5
Example B
Simplify 2√
13+6√
12.
Solution:
The square roots are not like terms, so there can be no further simplification.
In some cases, the radical may need to be reduced before addition/subtraction is possible.
Example C
Simplify 4√
3+2√
12.
Solution:√
12 simplifies to 2√
3.
4√
3+2√
12→ 4√
3+2(
2√
3)
4√
3+4√
3 = 8√
3
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5.13. Addition and Subtraction of Radicals www.ck12.org
Video Overview
MEDIAClick image to the left for more content.
Homework
Simplify the following expressions as much as possible.
1. 3√
8−6√
322.√
180+6√
4053.√
6−√
27+2√
54+3√
484.√
8−4√
985.√
48+√
27
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www.ck12.org Chapter 6. Unit 6 - Quadratic Functions
CHAPTER 6 Unit 6 - Quadratic FunctionsChapter Outline
6.1 MONOMIAL FACTORS OF POLYNOMIALS
6.2 FACTORING QUADRATIC EXPRESSIONS
6.3 FACTORING POLYNOMIALS WHERE A DOES NOT EQUAL 1
6.4 FACTORING THE DIFFERENCE OF TWO SQUARES (DOTS)
6.5 SOLVE QUADRATIC EQUATIONS BY FACTORING
6.6 SOLVE QUADRATIC EQUATIONS BY SQUARE ROOTS
6.7 SOLVE QUADRATIC EQUATIONS BY COMPLETING THE SQUARE
6.8 SOLVING QUADRATIC EQUATIONS BY THE QUADRATIC FORMULA
6.9 GRAPHS OF QUADRATIC FUNCTIONS
6.10 APPLICATIONS OF QUADRATIC EQUATIONS
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6.1 Monomial Factors of Polynomials
What You Will Learn
• Greatest Common Factor (GCF)• Factoring out the GCF from a polynomial
Greatest Common Factor (GCF)
The GCF is the highest number that divides exactly into two or more numbers . For example, the GCF of 12 and 18is 6 because that is the largest number that divides exactly into both numbers. If you are looking for the GCF of aterm that also includes variables, you need to also factor out the highest common power of the variable.
Find the GCF of 8x2 and 24x3
Solution
The GCF is 8x2 because 8 is the largest number that divides evenly into both numbers and x2 is the highest commonpower of the variable.
Factoring Out the GCF From a Polynomial
When a GCF is factored from a polynomial, you divide each term by the common factor. What is left over remainsin parentheses.
Example B
Factor:
a. 15x−25b. 3a+9b+6
Solution:
1. We see that the factor of 5 divides evenly from all terms.
15x−25 = 5(3x−5)
2. We see that the factor of 3 divides evenly from all terms.
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3a+9b+6 = 3(a+3b+2)
Now we will use examples where variables also need to be factored out.
Example C
Find the greatest common factor.
a3−3a2 +4a
Solution:
Notice that the factor a appears in all terms of a3−3a2 +4a but each term has a different power of a. The commonfactor is the lowest power that appears in the expression. In this case the factor is a.
Let’s rewrite: a3−3a2 +4a = a(a2)+a(−3a)+a(4)
Factor a to get a(a2−3a+4).
Example C
Find the greatest common factor.
5x3y−15x2y2 +25xy3
Solution:
The common factor is 5xy.
When we factor 5xy, we obtain 5xy(x2−3xy+5y2).
Practice Exercises
Find the greatest common factor.
16x2y3z2 +4x3yz+8x2y4z5
Solution:
First, look at the coefficients to see if they share any common factors. They do: 4.
Next, look for the lowest power of each variable, because that is the most you can factor out. The lowest power of xis x2. The lowest powers of y and z are to the first power.
This means we can factor out 4x2yz. Now, we have to determine what is left in each term after we factor out 4x2yz:
16x2y3z2 +4x3yz+8x2y4z5 = 4x2yz(4y2z+ x+2y3z4)
Video Overview
MEDIAClick image to the left for more content.
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Homework
Factor the common factor from the following polynomials.
1. 36a2 +9a3−6a7
2. yx3y2 +12x+16y3. 3x3−21x4. 5x6 +15x4
5. 4x3 +10x2−2x6. −10x6 +12x5−4x4
7. 12xy+24xy2 +36xy3
8. 5a3−7a9. 45y12 +30y10
10. 16xy2z+4x3y
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6.2 Factoring Quadratic Expressions
What You Will Learn
• Factor quadratic trinomials in standard form.
Write Quadratic Expressions in Standard Form
Quadratic polynomials are polynomials of 2nd degree. The standard form of a quadratic polynomial is written as
ax2 +bx+ c
Here a,b, and c stand for constant numbers. Factoring these polynomials depends on the values of these constants.
Factor when a = 1, b is Positive, and c is Positive
Let’s first consider the case where a = 1,b is positive and c is positive. The quadratic trinomials will take thefollowing form.
x2 +bx+ c
You know from multiplying binomials that when you multiply two factors (x+m)(x+ n) you obtain a quadraticpolynomial. Let’s try an example and see what happens. We use The Distributive Property.
(x+2)(x+5) = x2 +5x+2x+10
To simplify this polynomial we would combine the like terms in the middle by adding them.
(x+2)(x+5) = x2 +7x+10
To factor we need to do this process in reverse. This means that we start with x2 +7x+10 and we need to find twonumbers that multiply to 10 and add to 7.
Let’s try some specific examples.
Example 1
Factor x2 +5x+6
Solution We are looking for an answer that is a product of two binomials in parentheses.
(x+ )(x+ )
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To fill in the blanks, we want two numbers m and n that multiply to 6 and add to 5. A good strategy is to list thepossible ways we can multiply two numbers to give us 6 and then see which of these pairs of numbers add to 5. Thenumber six can be written as the product of.
6 = 1 ·6 and 1+6 = 7
6 = 2 ·3 and 2+3 = 5 ← This is the correct choice.
So the answer is (x+2)(x+3).
We can check to see if this is correct by multiplying (x+2)(x+3).
x + 2
x + 3
3x + 6
x2 + 2x
x2 + 5x + 9
The answer checks out.
Example 2
Factor x2 +7x+12
Solution
We are looking for an answer that is a product of two parentheses (x+ )(x+ ).
The number 12 can be written as the product of the following numbers.
12 = 1 ·12 and 1+12 = 13
12 = 2 ·6 and 2+6 = 8
12 = 3 ·4 and 3+4 = 7 ← This is the correct choice.
The answer is (x+3)(x+4).
Example 3
Factor x2 +8x+12.
Solution
We are looking for an answer that is a product of the two parentheses (x+ )(x+ ).
The number 12 can be written as the product of the following numbers.
12 = 1 ·12 and 1+12 = 13
12 = 2 ·6 and 2+6 = 8 ← This is the correct choice.
12 = 3 ·4 and 3+4 = 7
The answer is (x+2)(x+6).
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Example 4
Factor x2 +12x+36.
Solution
We are looking for an answer that is a product of the two parentheses (x+ )(x+ ).
The number 36 can be written as the product of the following numbers.
36 = 1 ·36 and 1+36 = 37
36 = 2 ·18 and 2+18 = 20
36 = 3 ·12 and 3+12 = 15
36 = 4 ·9 and 4+9 = 13
36 = 6 ·6 and 6+6 = 12 ← This is the correct choice
The answer is (x+6)(x+6).
Factor when a = 1, b is Negative and c is Positive
Now let’s see how this method works if the middle coefficient (b) is negative.
Example 5
Factor x2−6x+8.
Solution
We are looking for an answer that is a product of the two parentheses (x+ )(x+ ).
The number 8 can be written as the product of the following numbers.
8 = 1 ·8 and 1+8 = 9 Notice that these are two different choices.
But also,
8 = (−1) · (−8) and −1+(−8) =−9 Notice that these are two different choices.
8 = 2 ·4 and 2+4 = 6
But also,
8 = (−2) · (−4) and −2+(−4) =−6 ← This is the correct choice.
The answer is (x−2)(x−4)
We can check to see if this is correct by multiplying (x−2)(x−4).
x − 2
x − 4
− 4x + 8
x2 − 2x
x2 − 6x + 8
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The answer checks out.
Example 6
Factor x−17x+16
Solution
We are looking for an answer that is a product of two parentheses: (x± )(x± ).
The number 16 can be written as the product of the following numbers:
16 = 1 ·16 and 1+16 = 17
16 = (−1) · (−16) and −1+(−16) =−17 ← This is the correct choice.
16 = 2 ·8 and 2+8 = 10
16 = (−2) · (−8) and −2+(−8) =−10
16 = 4 ·4 and 4+4 = 8
16 = (−4) · (−4) and −4+(−4) =−8
The answer is (x−1)(x−16).
Factor when a = 1 and c is Negative
Now let’s see how this method works if the constant term is negative.
Example 7
Factor x2 +2x−15
Solution
We are looking for an answer that is a product of two parentheses (x± )(x± ).
In this case, we must take the negative sign into account. The number -15 can be written as the product of thefollowing numbers.
−15 =−1 ·15 and −1+15 = 14 Notice that these are two different choices.
And also,
−15 = 1 · (−15) and 1+(−15) =−14 Notice that these are two different choices.
−15 =−3 ·5 and −3+5 = 2 ← This is the correct choice.
−15 = 3 · (−5) and 3+(−5) =−2
The answer is (x−3)(x+5).
We can check to see if this is correct by multiplying (x−3)(x+5).
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x − 3
x + 5
5x − 15
x2 − 3x
x2 + 2x − 15
The answer checks out.
Example 8
Factor x2−10x−24
Solution
We are looking for an answer that is a product of two parentheses (x± )(x± ).
The number -24 can be written as the product of the following numbers.
−24 =−1 ·24 and −1+24 = 23
−24 = 1 · (−24) and 1+(−24) =−23
−24 =−2 ·12 and −2+12 = 10
−24 = 2 · (−12) and 2+(−12) =−10 ← This is the correct choice.
−24 =−3 ·8 and −3+8 = 5
−24 = 3 · (−8) and 3+(−8) =−5
−24 =−4 ·6 and −4+6 = 2
−24 = 4 · (−6) and 4+(−6) =−2
The answer is (x−12)(x+2).
Example 9
Factor x2 +34x−35
Solution
We are looking for an answer that is a product of two parentheses (x± )(x± )
The number -35 can be written as the product of the following numbers:
−35 =−1 ·35 and −1+35 = 34 ← This is the correct choice.
−35 = 1 · (−35) and 1+(−35) =−34
−35 =−5 ·7 and −5+7 = 2
−35 = 5 · (−7) and 5+(−7) =−2
The answer is (x−1)(x+35).
Video Overview
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6.2. Factoring Quadratic Expressions www.ck12.org
MEDIAClick image to the left for more content.
Homework
Factor the following quadratic polynomials.
1. x2 +10x+92. x2 +15x+503. x2 +10x+214. x2 +16x+485. x2−11x+246. x2−13x+427. x2−14x+338. x2−9x+209. x2 +5x−14
10. x2 +6x−2711. x2 +7x−7812. x2 +4x−3213. x2−12x−4514. x2−5x−5015. x2−3x−4016. −x2−5x+24
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6.3 Factoring Polynomials where a does notequal 1
What You Will Learn
• Factor a quadratic trinomial where a 6= 1
Factor Quadratic Trinomials Where a 6= 1
Factoring by grouping is a very useful method for factoring quadratic trinomials where a 6= 1.
To factor a quadratic polynomial where a 6= 1, we follow the following steps.
a. We find the product ac.b. We look for two numbers that multiply to give ac and add to give b.c. We rewrite the middle term using the two numbers we just found.d. We factor the expression by grouping.
Let’s apply this method to the following examples.
Example 6
Factor the following quadratic trinomials by grouping.
a) 3x2 +8x+4
b) 6x2−11x+4
c) 5x2−6x+1
Solution
Let’s follow the steps outlined above.
a) 3x2 +8x+4
Step 1 ac = 3 ·4 = 12
Step 2 The number 12 can be written as a product of two numbers in any of these ways:
12 = 1 ·12 and 1+12 = 13
12 = 2 ·6 and 2+6 = 8 This is the correct choice.
12 = 3 ·4 and 3+4 = 7
Step 3 Re-write the middle term as: 8x = 2x+6x, so the problem becomes the following.
3x2 +8x+4 = 3x2 +2x+6x+4
Step 4: Factor an x from the first two terms and 2 from the last two terms.
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6.3. Factoring Polynomials where a does not equal 1 www.ck12.org
x(3x+2)+2(3x+2)
Now factor the common binomial (3x+2).
(3x+2)(x+2)
Our answer is (3x+2)(x+2).
To check if this is correct we multiply (3x+2)(x+2).
3x+ 2
x + 2
6x+ 4
3x2 + 2x
3x2 + 8x + 4
The answer checks out.
b) 6x2−11x+4
Step 1 ac = 6 ·4 = 24
Step 2 The number 24 can be written as a product of two numbers in any of these ways.
24 = 1 ·24 and 1+24 = 25
24 =−1 · (−24) and −1+(−24) =−25
24 = 2 ·12 and 2+12 = 14
24 =−2 · (−12) and −2+(−12) =−14
24 = 3 ·8 and 3+8 = 11
24 =−3 · (−8) and −3+(−8) =−11 ← This is the correct choice.
24 = 4 ·6 and 4+6 = 10
24 =−4 · (−6) and −4+(−6) =−10
Step 3 Re-write the middle term as −11x =−3x−8x, so the problem becomes
6x2−11x+4 = 6x2−3x−8x+4
Step 4 Factor by grouping. Factor a 3x from the first two terms and factor -4 from the last two terms.
3x(2x−1)−4(2x−1)
Now factor the common binomial (2x−1).
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(2x−1)(3x−4)
Our answer is (2x−1)(3x−4).
c) 5x2−6x+1
Step 1 ac = 5 ·1 = 5
Step 2 The number 5 can be written as a product of two numbers in any of these ways.
5 = 1 ·5 and 1+5 = 6
5 =−1 · (−5) and −1+(−5) =−6 ← This is the correct choice
Step 3 Rewrite the middle term as −6x =−x−5x. The problem becomes
5x2−6x+1 = 5x2− x−5x+1
Step 4 Factor by grouping: factor an x from the first two terms and a factor of -1 from the last two terms
x(5x−1)−1(5x−1)
Now factor the common binomial (5x−1).
(5x−1)(x−1).
Our answer is (5x−1)(x−1).
Video Overview using Grouping
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Video Overview using Guess and Check
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Homework
Factor completely.
1. 2x2 +16x+302. −x3 +17x2−70x3. 2x2−5124. 12x3 +12x2 +3x
Factor the following quadratic binomials by grouping.
1. 4x2 +25x−212. 6x2 +7x+13. 4x2 +8x−54. 3x2 +16x+21
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6.4 Factoring the Difference of Two Squares(DOTS)
What You Will Learn
• Factor the difference of two squares.
Factor the Difference of Two Squares
We use the sum and difference formula to factor a difference of two squares. A difference of two squares can be aquadratic polynomial in this form.
a2−b2
Both terms in the polynomial are perfect squares. In a case like this, the polynomial factors into the sum anddifference of the square root of each term.
a2−b2 = (a+b)(a−b)
In these problems, the key is figuring out what the a and b terms are. Let’s do some examples of this type.
Example 1
Factor the difference of squares.
a) x2−9
b) x2−100
c) x2−1
Solution
a) Rewrite as x2−9 as x2−32. Now it is obvious that it is a difference of squares.
The difference of squares formula is a2−b2 = (a+b)(a−b)
Lets see how our problem matches with the formula x2−32 = (x+3)(x−3)
The answer is x2−9 = (x+3)(x−3).
We can check to see if this is correct by multiplying (x+3)(x−3).
x + 3
x − 3
−3x − 9
x2 +3x
x2 +0x − 9
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The answer checks out.
We could factor this polynomial without recognizing that it is a difference of squares. With the methods we learnedin the last section we know that a quadratic polynomial factors into the product of two binomials.
(x± )(x± )
We need to find two numbers that multiply to -9 and add to 0, since the middle term is missing.
We can write -9 as the following products
−9 =−1 ·9 and −1+9 = 8
−9 = 1 · (−9) and 1+(−9) =−8
−9 = 3 · (−3) and 3+(−3) = 0 ← This is the correct choice
We can factor x2−9 as (x+3)(x−3), which is the same answer as before.
You can always factor using methods for factoring trinomials, but it is faster if you can recognize special productssuch as the difference of squares.
b) Rewrite x2−100 as x2−102. This factors as (x+10)(x−10).
c) Rewrite x2−1 as x2−12. This factors as (x+1)(x−1).
Example 2
Factor the difference of squares.
a) 16x2−25
b) 4x2−81
c) 49x2−64
Solution
a) Rewrite 16x2−25 as (4x)2−52. This factors as (4x+5)(4x−5).
b) Rewrite 4x2−81 as (2x)2−92. This factors as (2x+9)(2x−9).
c) Rewrite 49x2−64 as (7x)2−82. This factors as (7x+8)(7x−8).
Example 3
Factor the difference of squares:
a) x2− y2
b) 9x2−4y2
c) x2y2−1
Solution
a) x2− y2 factors as (x+ y)(x− y).
b) Rewrite 9x2−4y2 as (3x)2− (2y)2. This factors as (3x+2y)(3x−2y).
c) Rewrite as x2y2−1 as (xy)2−12. This factors as (xy+1)(xy−1).
Example 4
Factor the difference of squares.
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a) x4−25
b) 16x4− y2
c) x2y8−64z2
Solution
a) Rewrite x4−25 as (x2)2−52. This factors as (x2 +5)(x2−5).
b) Rewrite 16x4− y2 as (4x2)2− y2. This factors as (x2 +5)(x2−5).
c) Rewrite x2y8−64z2 as (xy2)2− (8z). This factors as (xy2 +8z)(xy2−8z).
Video Overview
MEDIAClick image to the left for more content.
Homework
Factor the following difference of squares.
1. x2−42. x2−363. −x2 +1004. x2−4005. 9x2−46. 25x2−497. −36x2 +258. 16x2−81y2
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6.5. Solve Quadratic Equations by Factoring www.ck12.org
6.5 Solve Quadratic Equations by Factoring
What You Will Learn
• Solving a Quadratic Polynomial Equation by Factoring
Solve Quadratic Polynomial Equations by Factoring
With the methods we learned in the last two sections, we can factor many kinds of quadratic polynomials. This isvery helpful when we want to solve polynomial equations such as
ax2 +bx+ c = 0
To solve a quadratic equation like this, perform the following steps:
Step 1
If necessary, rewrite the equation in standard form so that
Quadratic expression = 0.
Step 2
Factor the polynomial completely.
Step 3
Use the Zero-Product rule to set each factor equal to zero.
Step 4
Solve each equation from Step 3.
Step 5
Check your answers by substituting your solutions into the original equation.
We will do a few examples that show how to solve quadratic polynomials using the factoring methods we justlearned.
Example 1
Solve the following polynomial equations.
a) x2 +7x+6 = 0
b) x2−8x =−12
c) x2 = 2x+15
Solution
a) x2 +7x+6 = 0
Rewrite This is not necessary since the equation is in the correct form already.
Factor We can write 6 as a product of the following numbers.
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6 = 1 ·6 and 1+6 = 7 ← This is the correct choice.
6 = 2 ·3 and 2+3 = 5
x2 +7x+6 = 0 factors as (x+1)(x+6) = 0
Set each factor equal to zero
x+1 = 0 or x+6 = 0
Solve
x =−1 or x =−6
Check Substitute each solution back into the original equation.
x =−1 (−1)2 +7(−1)+6 = 1−7+6 = 0 Checks out.
x =−6 (−6)2 +7(−6)+6 = 36−42+6 = 0 Checks out.
b) x2−8x =−12
Rewrite x2−8x =−12 is rewritten as x2−8x+12 = 0.
Factor We can write 12 as a product of the following numbers.
12 = 1 ·12 and 1+12 = 13
12 =−1 · (−12) and −1+(−12) =−13
12 = 2 ·6 and 2+6 = 8
12 =−2 · (−6) and −2+(−6) =−8 ← This is the correct choice.
12 = 3 ·4 and 3+4 = 7
12 =−3 · (−4) and −3+(−4) =−7
x2−8x+12 = 0 factors as (x−2)(x−6) = 0
Set each factor equal to zero.
x−2 = 0 or x−6 = 0
Solve.
x = 2 or x = 6
Check Substitute each solution back into the original equation.
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x = 2 (2)2−8(2) = 4−16 =−12 Checks out.
x = 2 (6)2−8(6) = 36−48 =−12 Checks out.
c) x2 = 2x+15
Rewrite x2 = 2x+15 is re-written as x2−2x−15 = 0.
Factor We can write -15 as a product of the following numbers.
−15 = 1 · (−15) and 1+(−15) =−14
−15 =−1 · (15) and −1+(15) = 14
−15 =−3 ·5 and −3+5 = 2
−15 = 3 · (−5) and 3+(−5) =−2 ← This is the correct choice.
x2−2x−15 = 0 factors as (x+3)(x−5) = 0.
Set each factor equal to zero
x+3 = 0 or x−5 = 0
Solve
x =−3 or x = 5
Check Substitute each solution back into the original equation.
x =−3 (−3)2 = 2(−3)+15⇒ 9 = 0 Checks out.
x = 5 (5)2 = 2(5)+15⇒ 25 = 25 Checks out.
Example 2
Let’s try this procedure with one of the more complicated factoring examples from a previous section.
6x2−11x+4
Step 1 ac = 6 ·4 = 24
Step 2 The number 24 can be written as a product of two numbers in any of these ways.
24 = 1 ·24 and 1+24 = 25
24 =−1 · (−24) and −1+(−24) =−25
24 = 2 ·12 and 2+12 = 14
24 =−2 · (−12) and −2+(−12) =−14
24 = 3 ·8 and 3+8 = 11
24 =−3 · (−8) and −3+(−8) =−11 ← This is the correct choice.
24 = 4 ·6 and 4+6 = 10
24 =−4 · (−6) and −4+(−6) =−10
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Step 3 Re-write the middle term as −11x =−3x−8x, so the problem becomes
6x2−11x+4 = 6x2−3x−8x+4
Step 4 Factor by grouping. Factor a 3x from the first two terms and factor -4 from the last two terms.
3x(2x−1)−4(2x−1)
Now factor the common binomial (2x−1).
(2x−1)(3x−4)
This gives us the factored form of the equation, now we set each factor equal to zero and solve for x
2x−1 = 0, adding one to each side gives 2x = 1, dividing each side by 2 gives x = 12
3x−4 = 0, adding four to each side gives 3x = 4, dividing each side by 3 gives x = 43
So the answers are 12 and 4
3
Video Overview
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Homework
Solve the following quadratic equations using factoring.
1. x2−11x+30 = 02. x2 +4x = 213. x2 +49 = 14x4. x2−64 = 05. x2−24x+144 = 06. 4x2−25 = 07. x2 +26x =−1698. 4x2−8x−5 = 09. 3x2 +19x =−6
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6.6 Solve Quadratic Equations by SquareRoots
What You Will Learn
• Solve quadratic equations involving perfect squares.• Solve quadratic equations with irrational answers
Solve Quadratic Equations Involving Perfect Squares
So far you know how to solve quadratic equations by factoring. However, this method works only if a quadraticpolynomial can be factored. Unfortunately, in practice, most quadratic polynomials are not factorable. In thissection you will continue learning new methods that can be used in solving quadratic equations. In particular, wewill examine equations in which we can take the square root of both sides of the equation in order to arrive at theresult.
Let’s first examine quadratic equation of the type
x2− c = 0
We can solve this equation by isolating the x2 term: x2 = c
Once the x2 term is isolated we can take the square root of both sides of the equation. Remember that when we takethe square root we get two answers: the positive square root and the negative square root:
x =√
c and x =−√
c
Often this is written as x =±√
c.
Example 1
Solve the following quadratic equations
a) x2−4 = 0
b) x2−25 = 0
Solution
a) x2−4 = 0
Isolate the x2 x2 = 4
Take the square root of both sides x =√
4 and x =−√
4
Answer x = 2 and x =−2
b) x2−25 = 0
Isolate the x2 x2 = 25
Take the square root of both sides x =√
25 and x =−√
25
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Answer x = 5 and x =−5
Another type of equation where we can find the solution using the square root is
ax2− c = 0
We can solve this equation by isolating the x2 term
ax2 = c
x2 =ca
Now we can take the square root of both sides of the equation.
x =√
ca
and x =−√
ca
Often this is written as x =±√
ca
.
Example 2
Solve the following quadratic equations.
a) 9x2−16 = 0
b) 81x2−1 = 0
Solution
a) 9x2−16 = 0
Isolate the x2. 9x2 = 16
x2 =169
Take the square root of both sides. x =
√169
and x =−√
169
Answer: x = 43 and x =−4
3
b) 81x2−1 = 0
Isolate the x2 81x2 = 1
x2 =1
81
Take the square root of both sides x =
√1
81and x =−
√181
Answer x = 19 and x =−1
9
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As you have seen previously, some quadratic equations have no real solutions.
Example 3
Solve the following quadratic equations.
a) x2 +1 = 0
b) 4x2 +9 = 0
Solution
a) x2 +1 = 0
Isolate the x2 x2 =−1
Take the square root of both sides: x =√−1 and x =−
√−1
Answer Square roots of negative numbers do not give real number results, so there are no real solutions to thisequation.
b) 4x2 +9 = 0
Isolate the x2 4x2 =−9
x2 =−94
Take the square root of both sides x =
√−9
4and x =−
√−9
4Answer There are no real solutions.
We can also use the square root function in some quadratic equations where one side of the equation is a perfectsquare. This is true if an equation is of the form
(x−2)2 = 9
Both sides of the equation are perfect squares. We take the square root of both sides.
x−2 = 3 and x−2 =−3
Solve both equations
Answer x = 5 and x =−1
Example 4
Solve the following quadratic equations.
a) (x−1)2 = 4
b) (x+3)2 = 1
Solution
a) (x−1)2 = 4
Take the square root of both sides. x−1 = 2 and x−1 =−2
Solve each equation. x = 3 and x =−1
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Answer x = 3 and x =−1
b) (x+3)2 = 1
Take the square root of both sides. x+3 = 1 and x+3 =−1
Solve each equation. x =−2 and x =−4
It might be necessary to factor the right hand side of the equation as a perfect square before applying the methodoutlined above.
Example 5
Solve the following quadratic equations.
a) x2 +8x+16 = 25
b) 4x2−40x+25 = 9
Solution
a) x2 +8x+16 = 25
Factor the right hand side. x2 +8x+16 = (x+4)2 so (x+4)2 = 25
Take the square root of both sides. x+4 = 5 and x+4 =−5
Solve each equation. x = 1 and x =−9
Answer x = 1 and x =−9
b) 4x2−20x+25 = 9
Factor the right hand side. 4x2−20x+25 = (2x−5)2 so (2x−5)2 = 9
Take the square root of both sides. 2x−5 = 3 and 2x−5 =−3
Solve each equation. 2x = 8 and 2x = 2
Answer x = 4 and x = 1
Solving Quadratic Equations with Irrational Answers
We use the methods we learned so far in this section to find approximate solutions to quadratic equations. We canget approximate solutions when taking the square root does not give an exact answer.
Example 6
Solve the following quadratic equations.
a) x2−3 = 0
b) 2x2−9 = 0
Solution
a)
Isolate the x2. x2 = 3
Take the square root of both sides. x =√
3 and x =−√
3
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Answer x≈ 1.73 and x≈−1.73
b)
Isolate the x2. 2x2 = 9 so x2 =92
Take the square root of both sides. x =
√92
and x =−√
92
Answer x≈ 2.12 and x≈−2.12
Example 7
Solve the following quadratic equations.
a) (2x+5)2 = 10
b) x2−2x+1 = 5
Solution.
a)
Take the square root of both sides. 2x+5 =√
10 and 2x+5 =−√
10
Solve both equations. x =−5+
√10
2and x =
−5−√
102
Answer x≈−0.92 and x≈−4.08
b)
Factor the right hand side. (x−1)2 = 5
Take the square root of both sides. x−1 =√
5 and x−1 =−√
5
Solve each equation. x = 1+√
5 and x = 1−√
5
Answer x≈ 3.24 and x≈−1.24
Video Overview
MEDIAClick image to the left for more content.
Homework
Solve the following quadratic equations.
1. x2−1 = 02. x2−100 = 03. x2 +16 = 04. 9x2−1 = 05. 4x2−49 = 0
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6. 64x2−9 = 07. x2−81 = 08. 25x2−36 = 09. x2 +9 = 0
10. x2−16 = 011. x2−36 = 012. 16x2−49 = 013. (x−2)2 = 114. (x+5)2 = 1615. (2x−1)2−4 = 016. (3x+4)2 = 917. (x−3)2 +25 = 018. x2−6 = 019. x2−20 = 020. 3x2 +14 = 021. (x−6)2 = 522. (4x+1)2−8 = 023. x2−10x+25 = 924. x2 +18x+81 = 125. 4x2−12x+9 = 1626. (x+10)2 = 227. x2 +14x+49 = 328. 2(x+3)2 = 829. Susan drops her camera in the river from a bridge that is 400 feet high. How long is it before she hears the
splash?30. It takes a rock 5.3 seconds to splash in the water when it is dropped from the top of a cliff. How high is the
cliff in meters?31. Nisha drops a rock from the roof of a building 50 feet high. Ashaan drops a quarter from the top story window,
40 feet high, exactly half a second after Nisha drops the rock. Which hits the ground first?
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6.7. Solve Quadratic Equations by Completing the Square www.ck12.org
6.7 Solve Quadratic Equations by Completingthe Square
What You Will Learn
• Solve quadratic equations by completing the square
Solve Quadratic Equations by completing the Square
Completing the square is another technique used to solve quadratic equations. When completing the square, the goalis to make a perfect square trinomial and factor it.
Example A
Solve x2−8x−1 = 10.
Solution:
1. Write the polynomial so that x2 and x are on the left side of the equation and the constants on the right. This isonly for organizational purposes, but it really helps. Leave a little space after the x−term.
x2−8x = 11
2. Now, “complete the square.” Determine what number would make a perfect square trinomial with x2−8x+c. Todo this, divide the x−term by 2 and square that number, or
(b2
)2.
(b2
)2
=
(82
)2
= 42 = 16
3. Add this number to both sides in order to keep the equation balanced.
x2−8x+16 = 11+16
4. Factor the left side to the square of a binomial and simplify the right.
(x−4)2 = 27
5. Solve by using square roots.
x−4 =±3√
3
x = 4±3√
3
Completing the square enables you to solve any quadratic equation using square roots. Through this process, we canmake an unfactorable quadratic equation solvable, like the one above. It can also be used with quadratic equationsthat have imaginary solutions.
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Example B
Solve x2 +12x+37 = 0
Solution: First, this is not a factorable quadratic equation. Therefore, the only way we know to solve this equationis to complete the square. Follow the steps from Example A.
1. Organize the polynomial, x’s on the left, constant on the right.
x2 +12x =−37
2. Find(b
2
)2and add it to both sides.
(b2
)2
=
(122
)2
= 62 = 36
x2 +12x+36 =−37+36
3. Factor the left side and solve.
(x+6)2 =−1
x+6 =±i
x =−6± i
Example C
Solve x2−11x−15 = 0.
Solution: This is not a factorable equation. Use completing the square.
1. Organize the polynomial, x’s on the left, constant on the right.
x2−11x = 15
2. Find(b
2
)2and add it to both sides.
(b2
)2
=
(112
)2
=1214
x2−11x+1214
= 15+1214
3. Factor the left side and solve.
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(x− 11
2
)2
=604+
1214(
x− 112
)2
=1814
x− 112
=±√
1812
x =112±√
1812
Practice Exercies
1. Find the value of c that would make x2−2x+ c a perfect square trinomial. Then, factor the trinomial.
Solve the following quadratic equations by completing the square.
2. x2 +10x+21 = 0
3. x−5x = 12
Answers
1. c =(b
2
)2=(2
2
)2= 12 = 1. The factors of x2−2x+1 are (x−1)(x−1) or (x−1)2.
2. Use the steps from the examples above.
x2 +10x+21 = 0
x2 +10x =−21
x2 +10x+(
102
)2
=−21+(
102
)2
x2 +10x+25 =−21+25
(x+5)2 = 4
x+5 =±2
x =−5±2
x =−7,−3
3. Use the steps from the examples above.
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x2−5x = 12
x2−5x+(
52
)2
= 12+(
52
)2
x2−5x+254
=484+
254(
x− 52
)2
=734
x− 52=±
√732
x =52±√
732
Video Overview
MEDIAClick image to the left for more content.
Homework
Determine the value of c that would complete the perfect square trinomial.
1. x2 +4x+ c2. x2−2x+ c3. x2 +16x+ c
Rewrite the perfect square trinomial as a square of a binomial.
4. x2 +6x+95. x2−7x+ 49
46. x2− 1
2 x+ 116
Solve the following quadratic equations by completing the square.
7. x2 +6x−15 = 08. x2 +10x+29 = 09. x2−14x+9 =−60
10. x2 +3x+18 =−211. x2−9x−5 = 2312. x2−20x = 60
Solve the following quadratic equations by factoring, square roots, or completing the square.
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13. x2 + x−30 = 014. x2−18x+90 = 015. x2 +15x+56 = 016. x2 +3x−24 = 1217. (x−2)2−20 =−4518. x2 +24x+44 =−1919. Solve x2 +7x−44 = 0 by factoring and completing the square. Which method do you prefer?20. Challenge Solve x2 + 17
8 x−2 =−9.
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www.ck12.org Chapter 6. Unit 6 - Quadratic Functions
6.8 Solving Quadratic Equations by theQuadratic Formula
What You Will Learn
• Deriving the quadratic formula• Solve quadratic equations using the quadratic formula.• Identify and choose methods for solving quadratic equations.
Deriving the Quadratic Formula
In this section, you will solve quadratic equations using the Quadratic Formula. Most of you are already familiarwith this formula from previous mathematics courses. It is probably the most used method for solving quadraticequations. For a quadratic equation in standard form
ax2 +bx+ c = 0
The solutions are found using the following formula.
x =−b±
√b2−4ac
2a
We will start by explaining where this formula comes from and then show how it is applied. This formula is derivedby solving a general quadratic equation using the method of completing the square that you learned in the previoussection.
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6.8. Solving Quadratic Equations by the Quadratic Formula www.ck12.org
We start with a general quadratic equation. ax2 +bx+ c = 0
Subtract the constant term from both sides. ax2 +bx =−c
Divide by the coefficient of the x2 term. x2 +ba
x =− ca
Rewrite. x2 +2(
b2a
)x =− c
a
Add the constant(
b2a
)2
to both sides. x2 +2(
b2a
)x+(
b2a
)2
=− ca+
b2
4a2
Factor the perfect square trinomial.(
x+b2a
)2
=−4ac4a2 +
b2
4a2
Simplify.(
x+b2a
)2
=b2−4ac
4a2
Take the square root of both sides. x+b
2a=
√b2−4ac
4a2 and x+b2a
=−√
b2−4ac4a2
Simplify. x+b
2a=
√b2−4ac
2aand x+
b2a
=−√
b2−4ac2a
x =− b2a
+
√b2−4ac
2a
x =− b2a−√
b2−4ac2a
x =−b+
√b2−4ac
2a
x =−b−
√b2−4ac
2a
This can be written more compactly as x = −b±√
b2−4ac2a .
You can see that the familiar formula comes directly from applying the method of completing the square. Applyingthe method of completing the square to solve quadratic equations can be tedious. The quadratic formula is a morestraightforward way of finding the solutions.
Solve Quadratic Equations Using the Quadratic Formula
Applying the quadratic formula basically amounts to plugging the values of a,b and c into the quadratic formula.
Example 1
Solve the following quadratic equation using the quadratic formula.
a) 2x2 +3x+1 = 0
b) x2−6x+5 = 0
c) −4x2 + x+1 = 0
Solution
Start with the quadratic formula and plug in the values of a,b and c.
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a)
Quadratic formula x =−b±
√b2−4ac
2a
Plug in the values a = 2,b = 3,c = 1. x =−3±
√(3)2−4(2)(1)2(2)
Simplify. x =−3±
√9−8
4=−3±
√1
4
Separate the two options. x =−3+1
4and x =
−3−14
Solve. x =−24
=−12
and x =−44
=−1
Answer x =−12 and x =−1
b)
Quadratic formula. x =−b±
√b2−4ac
2a
Plug in the values a = 1,b =−6,c = 5. x =−(−6)±
√(−6)2−4(1)(5)2(1)
Simplify. x =6±√
36−202
=6±√
162
Separate the two options. x =6+4
2and x =
6−42
Solve x =102
= 5 and x =22= 1
Answer x = 5 and x = 1
c)
Quadratic formula. x =−b±
√b2−4ac
2a
Plug in the values a =−4,b = 1,c = 1. x =−1±
√(1)2−4(−4)(1)2(−4)
Simplify. x =−1±
√1+16
−8=−1±
√17
−8
Separate the two options. x =−1+
√17
−8and x =
−1−√
17−8
Solve. x≈−.39 and x≈ .64
Answer x≈−.39 and x≈ .64
Often when we plug the values of the coefficients into the quadratic formula, we obtain a negative number inside thesquare root. Since the square root of a negative number does not give real answers, we say that the equation has noreal solutions. In more advanced mathematics classes, you will learn how to work with “complex” (or “imaginary”)solutions to quadratic equations.
Example 2
Solve the following quadratic equation using the quadratic formula x2 +2x+7 = 0
Solution:
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6.8. Solving Quadratic Equations by the Quadratic Formula www.ck12.org
a)
Quadratic formula. x =b±√
b2−4ac2a
Plug in the values a = 1,b = 2,c = 7. x =−2±
√(2)2−4(1)(7)2(1)
Simplify. x =−2±
√4−28
2=−2±
√−24
2
Answer There are no real solutions.
To apply the quadratic formula, we must make sure that the equation is written in standard form. For some problems,we must rewrite the equation before we apply the quadratic formula.
Example 3
Solve the following quadratic equation using the quadratic formula.
a) x2−6x = 10
b) 8x2 = 5x+6
Solution:
a)
Rewrite the equation in standard form. x2−6x−10 = 0
Quadratic formula x =−b±
√b2−4ac
2a
Plug in the values a = 1,b =−6,c =−10. x =−(−6)±
√(−6)2−4(1)(−10)
2(1)
Simplify. x =6±√
36+402
=6±√
762
Separate the two options. x =6+√
762
and x =6−√
762
Solve. x≈ 7.36 and x≈−1.36
Answer x≈ 7.36 and x≈−1.36
b)
Rewrite the equation in standard form. 8x2 +5x+6 = 0
Quadratic formula x =−b±
√b2−4ac
2a
Plug in the values a = 8,b = 5,c = 6. x =−5±
√(5)2−4(8)(6)2(8)
Simplify. x =−5±
√25−192
16=−5±
√−167
16
Answer no real solutions
Identify and Choose Methods for Solving Quadratic Equations.
In mathematics, you will need to solve quadratic equations that describe application problems or that are part ofmore complicated problems. You learned four ways of solving a quadratic equation.
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• Factoring.• Taking the square root.• Completing the square• Quadratic formula.
Usually you will not be told which method to use. You will have to make that decision yourself. However, here aresome guidelines to which methods are better in different situations.
Factoring is always best if the quadratic expression is easily factorable. It is always worthwhile to check if you canfactor because this is the fastest method. Many expressions are not factorable so this method is not used very oftenin practice.
Taking the square root is best used when there is no x term in the equation.
Completing the square isn’t typically used unless you are fond of it. It is most useful for deriving the quadraticformula.
Quadratic formula is the method that is used most often for solving a quadratic equation. When solving directlyby taking square root and factoring does not work, this is the method that most people prefer to use.
If you are using factoring or the quadratic formula make sure that the equation is in standard form.
Example 4
Solve each quadratic equation
a) x2−4x−5 = 0
b) x2 = 8
c) −4x2 + x = 2
d) 25x2−9 = 0
e) 3x2 = 8x
Solution
a) This expression if easily factorable so we can factor and apply the zero-product property:
Factor. (x−5)(x+1) = 0
Apply zero-product property. x−5 = 0 and x+1 = 0
Solve. x = 5 and x =−1
Answer x = 5 and x =−1
b) Since the expression is missing the x term we can take the square root:
Take the square root of both sides. x =√
8 and x =−√
8
Answer x = 2.83 and x =−2.83
c) Rewrite the equation in standard form.
It is not apparent right away if the expression is factorable, so we will use the quadratic formula.
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Quadratic formula x =−b±
√b2−4ac
2a
Plug in the values a =−4,b = 1,c =−2. x =−1±
√12−4(−4)(−2)2(−4)
Simplify. x =−1±
√1−32
−8=−1±
√−31
−8
Answer no real solution
d) This problem can be solved easily either with factoring or taking the square root. Let’s take the square root in thiscase.
Add 9 to both sides of the equation. 25x2 = 9
Divide both sides by 25. x2 =9
25
Take the square root of both sides. x =
√925
and x =−√
925
Simplify. x =35
and x =−35
Answer x = 35 and x =−3
5
e)
Rewrite the equation in standard form 3x2−8x = 0
Factor out common x term. x(3x−8) = 0
Set both terms to zero. x = 0 and 3x = 8
Solve. x = 0 and x =83= 2.67
Answer x = 0 and x = 2.67
Video Overview
MEDIAClick image to the left for more content.
Homework
Solve the following quadratic equations using the quadratic formula.
1. x2 +4x−21 = 0
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2. x2−6x = 123. 3x2− 1
2 x = 38
4. 2x2 + x−3 = 05. −x2−7x+12 = 06. −3x2 +5x = 07. 4x2 = 08. x2 +2x+6 = 0
Solve the following quadratic equations using the method of your choice.
9. x2− x = 610. x2−12 = 011. −2x2 +5x−3 = 012. x2 +7x−18 = 013. 3x2 +6x =−1014. −4x2 +4000x = 015. −3x2 +12x+1 = 016. x2 +6x+9 = 017. 81x2 +1 = 018. −4x2 +4x = 919. 36x2−21 = 020. x2−2x−3 = 0
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6.9 Graphs of Quadratic Functions
What You Will Learn
• Graphing quadratic functions.• Compare graphs of quadratic functions
Graphing Quadratic Functions
The graphs of quadratic functions are curved lines called parabolas. You don’t have to look hard to find parabolicshapes around you. Here are a few examples.
• The path that a ball or a rocket takes through the air.• Water flowing out of a drinking fountain.• The shape of a satellite dish.• The shape of the mirror in car headlights or a flashlight.
Let’s see what a parabola looks like by graphing the simplest quadratic function, y = x2.
We will graph this function by making a table of values. Since the graph will be curved we need to make sure thatwe pick enough points to get an accurate graph.
TABLE 6.1:
x y = x2
−3 (−3)2 = 9-2 (−2)2 = 4-1 (−1)2 = 10 (0)2 = 01 (1)2 = 12 (2)2 = 43 (3)2 = 9
We plot these points on a coordinate graph.
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To draw the parabola, draw a smooth curve through all the points. (Do not connect the points with straight lines).
Let’s graph a few more examples.
Example 1
Graph the following parabolas.
a) y = 2x2
b) y =−x2
c) y = x2−2x+3
Solution
a) y = 2x2 +4x+1
Make a table of values.
TABLE 6.2:
x y = 2x2 +4x+1−3 2(−3)2 +4(−3)+1 = 7-2 2(−2)2 +4(−2)+1 = 1-1 2(−1)2 +4(−1)+1 =−10 2(0)2 +4(0)+1 = 11 2(1)2 +4(1)+1 = 72 2(2)2 +4(2)+1 = 173 2(3)2 +4(3)+1 = 31
Notice that the last two points have large y−values. We will not graph them since that will make our y−scale toobig. Now plot the remaining points and join them with a smooth curve.
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b) y =−x2 +3
Make a table of values.
TABLE 6.3:
x y =−x2 +3−3 −(−3)2 +3 =−6-2 −(−2)2 +3 =−1-1 −(−1)2 +3 = 20 −(0)2 +3 = 31 −(1)2 +3 = 22 −(2)2 +3 =−13 −(3)2 +3 =−6
Plot the points and join them with a smooth curve.
Notice that it makes an “upside down” parabola. Our equation has a negative sign in front of the x2 term. The signof the coefficient of the x2 term determines whether the parabola turns up or down.
If the coefficient of x2; is positive, then the parabola turns up.
If the coefficient of x2; is negative, then the parabola turns down.
c) y = x2−8x+3
Make a table of values.
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TABLE 6.4:
x y = x2−8x+3−3 (−3)2−8(−3)+3 = 36-2 (−2)2−8(−2)+3 = 23-1 (−1)2−8(−1)+3 = 120 (0)2−8(0)+3 = 31 (1)2−8(1)+3 =−42 (2)2−8(2)+3 =−93 (3)2−8(3)+3 =−12
Let’s not graph the first two points in the table since the values are very big. Plot the points and join them with asmooth curve.
This does not look like the graph of a parabola. What is happening here? If it is not clear what the graph looks likechoose more points to graph until you can see a familiar curve. For negative values of x it looks like the values of yare getting bigger and bigger. Let’s pick more positive values of x beyond x = 3.
TABLE 6.5:
x y = x2−8x+3−1 (−1)2−8(−1)+3 = 120 (0)2−8(0)+3 = 31 (1)2−8(1)+3 =−40 (0)2−8(0)+3 = 31 (1)2−8(1)+3 =−42 (2)2−8(2)+3 =−93 (3)2−8(3)+3 =−124 (4)2−8(4)+3 =−135 (5)2−8(5)+3 =−126 (6)2−8(6)+3 =−97 (7)2−8(7)+3 =−48 (8)2−8(8)+3 = 3
Plot the points again and join them with a smooth curve.
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We now see the familiar parabolic shape. We could have graphed fewer points if we knew where the middle of thecurve, or the vertex, is. To find that, use the equation x = −b
2a In this example, b is -8 and a is 1, so the number inthe middle of the table should be 8
2·1 or 4, which it is.
Compare Graphs of Quadratic Functions
The general form (or standard form) of a quadratic function is:
y = ax2 +bx+ c
Here a,b and c are the coefficients. Remember a coefficient is just a number (i.e. a constant term) that goes beforea variable or it can be alone. You should know that if you have a quadratic function, its graph is always a parabola.While the graph of a quadratic is always the same basic shape, we have different situations where the graph couldbe upside down. It could be shifted to different locations or it could be “fatter” or “skinnier”. These situations aredetermined by the values of the coefficients. Let’s see how changing the coefficient changes the orientation, locationor shape of the parabola.
Orientation
Does the parabola open up or down?
The answer to that question is pretty simple:
• If a is positive, the parabola opens up.• If a is negative, the parabola opens down.
The following plot shows the graphs of y = x2 and y = −x2. You see that the parabola has the same shape in bothgraphs, but the graph of y = x2 is right-side-up and the graph of y =−x2 is upside-down.
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Dilation
Changing the value of the coefficient a makes the graph “fatter” or “skinnier”. Let’s look at how graphs compare fordifferent positive values of a. The plot on the left shows the graphs of y = −x2 and y = 3x2. The plot on the rightshows the graphs of y =−x2 and y =
(13
)x2.
Notice that the larger the value of a is, the skinnier the graph is. For example, in the first plot, the graph of y = 3x2
is skinnier than the graph of y = x2. Also, the smaller a is (i.e. the closer to 0), the fatter the graph is. For example,in the second plot, the graph of y =
(13
)x2 is fatter than the graph of y = x2. This might seem counter-intuitive, but
if you think about it, it should make sense. Let’s look at a table of values of these graphs and see if we can explainwhy this happens.
TABLE 6.6:
x y = x2 y = 3x2 y = 13 x2
−3 (−3)2 = 9 3(−3)2 = 27 (−3)23 = 3
-2 (−2)2 = 4 3(−2)2 = 12 (−2)23 = 4
3-1 (−1)2 = 1 3(−1)2 = 3 (−1)
23 = 1
30 (0)2 = 0 3(0)2 = 0 (0)
23 = 0
1 (1)2 = 1 3(1)2 = 3 (1)23 = 1
32 (2)2 = 4 3(2)2 = 12 (2)
23 = 4
33 (3)2 = 9 3(3)2 = 27 (3)
23 = 3
From the table, you can see that the values of y = 3x2 are bigger than the values of y = x2. This is because each value
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of y gets multiplied by 3. As a result, the parabola will be skinnier because it grows three times faster than y = x2.On the other hand, you can see that the values of y =
(13
)x2 are smaller than the values of y = x2. This is because
each value of y gets divided by 3. As a result, the parabola will be fatter because it grows at one third the rate ofy = x2.
Video Overview
MEDIAClick image to the left for more content.
Homework
Graph the following functions by making a table of values. Use the vertex to help you pick values for the table.
1. y = 4x2−42. y =−x2 + x+123. y = 2x2 +10x+84. y = 1
2 x2−2x5. y = x−2x2
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6.10 Applications of Quadratic Equations
What You Will Learn
• Solve real world problems using quadratic equations
Solve Real World Problems using Quadratic Equations
Now that we know most of the problem solving strategies for quadratic equations, we can see how these methodsapply to solving real world problems.
Example A - Pythagorean Theorem
One leg of a right triangle is 3 feet longer than the other leg. The hypotenuse is 15 feet. Find the dimensions of theright triangle.
Solution
Let x = the length of one leg of the triangle, then the other leg will measure x+3.
Let’s draw a diagram.
Use the Pythagorean Theorem (leg1)2 +(leg2)
2 = (hypotenuse)2 or a2 +b2 = c2.
Here a and b are the lengths of the legs and c is the length of the hypotenuse.
Let’s substitute the values from the diagram.
a2 +b2 = c2
x2 +(x+3)2 = 152
In order to solve, we need to get the polynomial in standard form. We must first distribute, collect like terms andre-write in the form polynomial = 0.
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x2 + x2 +6x+9 = 225
2x2 +6x+9 = 225
2x2 +6x−216 = 0
Factor the common monomial 2(x+3x−108) = 0.
To factor the trinomial inside the parenthesis we need to numbers that multiply to -108 and add to 3. It would take along time to go through all the options so let’s try some of the bigger factors.
−108 =−12· and −12+9 =−3
−108 = 12 · (−9) and 12+(−9) = 3 ← This is the correct choice.
We factor as: 2(x−9)(x+12) = 0.
Set each term equal to zero and solve
x−9 = 0 x+12 = 0
or
x = 9 x =−12
It makes no sense to have a negative answer for the length of a side of the triangle, so the answer must be thefollowing.
Answer x = 9 for one leg, and x+3 = 12 for the other leg.
Check 92 +122 = 81+144 = 225 = 152 so the answer checks.
Example B - Number Problems
The product of two positive numbers is 60. Find the two numbers if one of the numbers is 4 more than the other.
Solution
Let x = one of the numbers and x+4 equals the other number.
The product of these two numbers equals 60. We can write the equation.
x(x+4) = 60
In order to solve we must write the polynomial in standard form. Distribute, collect like terms and re-write in theform polynomial = 0.
x2 +4x = 60
x2 +4x−60 = 0
Factor by finding two numbers that multiply to -60 and add to 4. List some numbers that multiply to -60:
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−60 =−4 ·15 and −4+15 = 11
−60 = 4 · (−15) and 4+(−15) =−11
−60 =−5 ·12 and −5+12 = 7
−60 = 5 · (−12) and 5+(−12) =−7
−60 =−6 ·10 and −6+10 = 4 ← This is the correct choice
−60 = 6 · (−10) and 6+(−10) =−4
The expression factors as (x+10)(x−6) = 0.
Set each term equal to zero and solve.
x+10 = 0 x−6 = 0
or
x =−10 x = 6
Since we are looking for positive numbers, the answer must be the following.
Answer x = 6 for one number, and x+4 = 10 for the other number.
Check 6 ·10 = 60 so the answer checks.
Example C - Area of a rectangle
A rectangle has sides of x+5 and x−3. What value of x gives and area of 48?
Solution:
Make a sketch of this situation.
Area of the rectangle = length×width
(x+5)(x−3) = 48
In order to solve, we must write the polynomial in standard form. Distribute, collect like terms and rewrite in theform polynomial = 0.
x2 +2x−15 = 48
x2 +2x−63 = 0
Factor by finding two numbers that multiply to -63 and add to 2. List some numbers that multiply to -63.
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−63 =−7 ·9 and −7+9 = 2 ← This is the correct choice
−63 = 7 · (−9) and 7+(−9) =−2
The expression factors as (x+9)(x−7) = 0.
Set each term equal to zero and solve.
x+9 = 0 x−7 = 0
or
x =−9 x = 7
Since we are looking for positive numbers the answer must be x = 7.
Answer The width is x−3 = 4 and the length is x+5 = 12.
Check 4 ·12 = 48 so the answer checks out.
Homework
Solve the following application problems:
1. One leg of a right triangle is 7 feet longer than the other leg. The hypotenuse is 13 feet. Find the dimensionsof the right triangle.
2. A rectangle has sides of x+2 and x−1. What value of x gives an area of 108?3. The product of two positive numbers is 120. Find the two numbers if one numbers is 7 more than the other.4. Framing Warehouse offers a picture framing service. The cost for framing a picture is made up of two parts.
The cost of glass is $1 per square foot. The cost of the frame is $2 per linear foot. If the frame is a square,what size picture can you get framed for $20?
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CHAPTER 7 Unit 7 - ExponentialFunctions, Sequences and Modeling
Chapter Outline7.1 EXPONENTIAL GROWTH
7.2 EXPONENTIAL DECAY
7.3 ARITHMETIC SEQUENCES AND LINEAR FUNCTIONS
7.4 GEOMETRIC SEQUENCES AND EXPONENTIAL FUNCTIONS
7.5 DESCRIBING THE PATTERN AND WRITING A RECURSIVE RULE FOR A SE-QUENCE
7.6 LINEAR, EXPONENTIAL, AND QUADRATIC MODELS
7.7 REFERENCES
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7.1 Exponential Growth
What You Will Learn
• Graph an exponential growth function.• Compare graphs of exponential growth functions.• Solve real-world problems involving exponential growth.
Introduction
Exponential functions are different than other functions you have seen before because now the variable appears asthe exponent (or power) instead of the base. In this section, we will be working with functions where the base is aconstant number and the exponent is the variable. Here is an example.
y = 2x
This particular function describes something that doubles each time x increases by one. Lets look at a particularsituation where this might occur.
A colony of bacteria has a population of three thousand at noon on Sunday. During the next week, the colonyspopulation doubles every day. What is the population of the bacteria colony at noon on Saturday?
Lets make a table of values and calculate the population each day.
Day 0(Sunday) 1(Monday) 2(Tuesday) 3(Wednesday) 4(Thursday) 5(Friday) 6(Saturday)
Population (in thousands) 3 6 12 24 48 96 192
To get the population of bacteria for the next day we simply multiply the current days population by 2.
We start with a population of 3 (thousand): P = 3
To find the population on Monday we double P = 3 ·2The population on Tuesday will be double that on Monday P = 3 ·2 ·2
The population on Wednesday will be double that on Tuesday P = 3 ·2 ·2 ·2Thursday is double that on Wednesday P = 3 ·2 ·2 ·2 ·2
Friday is double that on Thursday P = 3 ·2 ·2 ·2 ·2 ·2Saturday is double that on Friday P = 3 ·2 ·2 ·2 ·2 ·2 ·2
You can see that this function describes a population that is multiplied by 2 each time a day passes.
If we define x as the number of days since Sunday at noon, then we can write the following.
P = 3.2x This is a formula that we can use to calculate the population on any day.
For instance, the population on Saturday at noon will be P = 3.26 = 3.64 = 192 (thousand) bacteria.
We used x = 6, since Saturday at noon is six days after Sunday at noon.
In general exponential function takes the form:
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y = A ·bx where A is the initial amount and b is the factor that the amount gets multiplied by each time x is increasedby one.
Graph Exponential Functions
Lets start this section by graphing some exponential functions. Since we dont yet know any special properties ofexponential functions, we will graph using a table of values.
Example 1
Graph the equation using a table of values y = 2x.
Solution
Lets make a table of values that includes both negative and positive values of x.
To evaluate the positive values of x, we just plug into the function and evaluate.
x = 1, y = 21 = 2
x = 2, y = 22 = 2 ·2 = 4
x = 3, y = 23 = 2 ·2 ·2 = 8
TABLE 7.1:
x y−3 1
8-2 1
4-1 1
20 11 22 43 8
For x = 0, we must remember that a number to the power 0 is always 1.
x = 0, y = 20 = 1
To evaluate the negative values of x, we must remember that x to a negative power means one over x to the samepositive power.
x =−1, y = 2−1 =121 =
12
x =−2, y = 2−2 =122 =
14
x =−3, y = 2−3 =123 =
18
When we plot the points on the coordinate axes we get the graph below. Exponentials always have this basic shape.That is, they start very small and then, once they start growing, they grow faster and faster, and soon they becomeextremely big!
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You may have heard people say that something is growing exponentially. This implies that the growth is very quick.An exponential function actually starts slow, but then grows faster and faster all the time. Specifically, our functiony above doubled each time we increased x by one.
This is the definition of exponential growth. There is a consistent fixed period during which the function will doubleor triple, or quadruple. The change is always a fixed proportion.
Compare Graphs of Exponential Growth Functions
Lets graph a few more exponential functions and see what happens as we change the constants in the functions. Thebasic shape of the exponential function should stay the same. But, it may become steeper or shallower depending onthe constants we are using.
We mentioned that the general form of the exponential function is y = A · bx where A is the initial amount and b isthe factor that the amount gets multiplied by each time x is increased by one. Lets see what happens for differentvalues of A.
Example 2
Graph the exponential function y = 3 ·2x and compare with the graph of y = 2x.
Solution
Lets make a table of values for y = 3 ·2x.
TABLE 7.2:
x y = 3 ·2x
−2 y = 3 ·2−2 = 3 · 122 =
34
-1 y = 3 ·2−1 = 3 · 121 =
32
0 y = 3 ·20 = 31 y = 3 ·21 = 62 y = 3 ·22 = 3 ·4 = 123 y = 3 ·23 = 3 ·8 = 24
Now let’s use this table to graph the function.
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We can see that the function y = 3 ·2x is bigger than function y = 2x. In both functions, the value of y doubled everytime x increases by one. However, y = 3 ·2x starts with a value of 3, while y = 2x starts with a value of 1, so it makessense that y = 3 ·2x would be bigger as its values of y keep getting doubled.
You might think that if the initial value A is less than one, then the corresponding exponential function would be lessthan y = 2x. This is indeed correct. Lets see how the graphs compare for A = 1
3 .
Example 3
Graph the exponential function y = 13 ·2
x and compare with the graph of y = 2x.
Solution
Lets make a table of values for y = 13 ·2
x.
TABLE 7.3:
x y = 13 ·2
x
−2 y = 13 ·2
−2 = 13 ·
122 =
112
-1 y = 13 ·2
−1 = 13 ·
121 =
16
0 y = 13 ·2
0 = 13
1 y = 13 ·2
1 = 23
2 y = 13 ·2
2 = 13 ·4 = 4
33 y = 1
3 ·23 = 1
3 ·8 = 83
Now let’s use this table to graph the function.
As expected, the exponential function y = 13 ·2
x is smaller that the exponential function y = 2x.
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Now, lets compare exponential functions whose bases are different.
The function y = 2x has a base of 2. That means that the value of y doubles every time x is increased by 1.
The function y = 3x has a base of 3. That means that the value of y triples every time x is increased by 1.
The function y = 5x has a base of 5. That means that the value of y gets multiplies by a factor of 5 every time x isincreased by 1.
The function y = 10x has a base of 10. That means that the value of y gets multiplied by a factor of 10 every time xis increased by 1.
What do you think will happen as the base number is increased? Lets find out.
Example 4
Graph the following exponential functions of the same graph y = 2x,y = 3x,y = 5x,y = 10x.
Solution
To graph these functions we should start by making a table of values for each of them.
TABLE 7.4:
x y = 2x y = 3x y = 5x y = 10x
−2 14
19
125
1100
-1 12
13
15
110
0 1 1 1 11 2 3 5 102 4 9 25 1003 8 27 125 1000
Now let’s graph these functions.
Notice that for x = 0 the values for all the functions are equal to 1. This means that the initial value of the functionsis the same and equal to 1. Even though all the functions start at the same value, they increase at different rates. Wecan see that the bigger the base is the faster the values of y will increase. It makes sense that something that tripleseach time will increase faster than something that just doubles each time.
Finally, lets examine what the graph of an exponential looks like if the value of A is negative.
Example 5
Graph the exponential function y =−5 ·2x.
Solution
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Lets make a table of values.
TABLE 7.5:
x y =−5 ·2x
−2 −54
-1 −52
0 -51 -102 -203 -40
Now let’s graph the function.
This result should not be unexpected. Since the initial value is negative and doubles with time, it makes sense thatthe value of y increasess, but in a negative direction. Notice that the shape of the graph remains that of a typicalexponential function, but is now a mirror image about the horizontal axis (i.e. upside down).
Solve Real-World Problems Involving Exponential Growth
We will now examine some real-world problems where exponential growth occurs.
Example 6
The population of a town is estimated to increase by 15% per year. The population today is 20 thousand. Make agraph of the population function and find out what the population will be ten years from now.
Solution
First, we need to write a function that describes the population of the town. The general form of an exponentialfunction is.
y = A ·bx
Define y as the population of the town.
Define x as the number of years from now.
A is the initial population, so A = 20 (thousand)
Finally, we must find what b is. We are told that the population increases by 15% each year.
To calculate percents, it is necessary to change them into decimals. 15% is equivalent to 0.15.
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15% of A is equal to 0.15A. This represents the increase in population from one year to the next.
In order to get the total population for the following year we must add the current population to the increase inpopulation. In other words A+0.15A = 1.15A. We see from this that the population must be multiplied by a factorof 1.15 each year.
This means that the base of the exponential is b = 1.15.
The formula that describes this problem is y = 20 · (1.15)x
Now lets make a table of values.
TABLE 7.6:
x y = 20 · (1.15)x
−10 4.9-5 9.90 205 40.210 80.9
Now let’s graph the function.
Notice that we used negative values of x in our table of values. Does it make sense to think of negative time? In thiscase x =−5 represents what the population was five years ago, so it can be useful information. The question askedin the problem was What will be the population of the town ten years from now?
To find the population exactly, we use x = 10 in the formula. We found y = 20 · (1.15)10 = 89,911.
Example 7
Peter earned $1500 last summer. If he deposited the money in a bank account that earns 5% interest compoundedyearly, how much money will he have after five years?
Solution
This problem deals with interest which is compounded yearly. This means that each year the interest is calculated onthe amount of money you have in the bank. That interest is added to the original amount and next year the interest iscalculated on this new amount. In this way, you get paid interest on the interest.
Lets write a function that describes the amount of money in the bank. The general form of an exponential functionis
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y = A ·bx
Define y as the amount of money in the bank.
Define x as the number of years from now.
A is the initial amount, so A = 1500.
Now we must find what b is.
We are told that the interest is 5% each year.
Change percents into decimals 5% is equivalent to 0.05.
5% of A is equal to 0.05A This represents the interest earned per year.
In order to get the total amount of money for the following year, we must add the interest earned to the initial amount.
A+0.05A = 1.05A
We see from this that the amount of money must be multiplied by a factor of 1.05 each year.
This means that the base of the exponential is b = 1.05
The formula that describes this problem is y = 1500 · (1.05)x
To find the total amount of money in the bank at the end of five years, we simply use x = 5 in our formula.
Answer y = 1500 · (1.05)5 = $1914.42
Video Overview
MEDIAClick image to the left for more content.
Homework
Graph the following exponential functions by making a table of values.
1. y = 3x
2. y = 5 ·3x
3. y = 40 ·4x
4. y = 3 ·10x
Solve the following problems.
5. A chain letter is sent out to 10 people telling everyone to make 10 copies of the letter and send each one to anew person. Assume that everyone who receives the letter sends it to ten new people and that it takes a weekfor each cycle. How many people receive the letter on the sixth week?
6. Nadia received $200 for her 10th birthday. If she saves it in a bank with a 7.5% interest compounded yearly,how much money will she have in the bank by her 21st birthday?
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7.2 Exponential Decay
What You Will Learn
• Graph an exponential decay function.• Compare graphs of exponential decay functions.• Solve real-world problems involving exponential decay.
Introduction
In the last section, we looked at graphs of exponential functions. We saw that exponentials functions describe aquantity that doubles, triples, quadruples, or simply gets multiplied by the same factor. All the functions we lookedat in the last section were exponentially increasing functions. They started small and then became large very fast. Inthis section, we are going to look at exponentially decreasing functions. An example of such a function is a quantitythat gets decreased by one half each time. Lets look at a specific example.
For her fifth birthday, Nadias grandmother gave her a full bag of candy. Nadia counted her candy and found outthat there were 160 pieces in the bag. As you might suspect Nadia loves candy so she ate half the candy on the firstday. Her mother told her that if she eats it at that rate it will be all gone the next day and she will not have anymoreuntil her next birthday. Nadia devised a clever plan. She will always eat half of the candy that is left in the bag eachday. She thinks that she will get candy every day and her candy will never run out. How much candy does Nadiahave at the end of the week? Would the candy really last forever?
Lets make a table of values for this problem.
Day 0 1 2 3 4 5 6 7
No. of Candies 160 80 40 20 10 5 2.5 1.25
You can see that if Nadia eats half the candies each day, then by the end of the week she only has 1.25 candies leftin her bag.
Lets write an equation for this exponential function.
Nadia started with 160 pieces. y = 160
After the first she has12
of that amount. y = 160 · 12
After the second day she has12
of the last amount. y = 160 · 12· 1
2
You see that in order to get the amount of candy left at the end of each day we keep multiplying by 12 .
We can write the exponential function as
y = 160 · 12
x
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Notice that this is the same general form as the exponential functions in the last section.
y = A ·bx
Here A = 160 is the initial amount and b = 12 is the factor that the quantity gets multiplied by each time. The
difference is that now b is a fraction that is less than one, instead of a number that is greater than one.
This is a good rule to remember for exponential functions.
If b is greater than one, then the exponential function increased, but
If b is less than one (but still positive), then the exponential function decreased
Lets now graph the candy problem function. The resulting graph is shown below.
So, will Nadias candy last forever? We saw that by the end of the week she has 1.25 candies left so there doesntseem to be much hope for that. But if you look at the graph you will see that the graph never really gets to zero.
Theoretically there will always be some candy left, but she will be eating very tiny fractions of a candy every dayafter the first week!
This is a fundamental feature of an exponential decay function. Its value get smaller and smaller and approacheszero but it never quite gets there. In mathematics we say that the function asymptotes to the value zero. This meansthat it approaches that value closer and closer without ever actually getting there.
Graph an Exponential Decay Function
The graph of an exponential decay function will always take the same basic shape as the one in the previous figure.Lets graph another example by making a table of values.
Example 1
Graph the exponential function y = 5 ·(1
2
)x
Solution
Lets start by making a table of values.
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TABLE 7.7:
x y = 5 ·(1
2
)x
−3 y = 5 ·(1
2
)−3= 5.23 = 40
-2 y = 5 ·(1
2
)−2= 5.22 = 20
-1 y = 5 ·(1
2
)−1= 5.21 = 10
0 y = 5 ·(1
2
)0= 5.1 = 5
1 y = 5 ·(1
2
)1= 5
2
2 y = 5 ·(1
2
)2= 5
4
Now let’s graph the function.
Remember that a fraction to a negative power is equivalent to its reciprocal to the same positive power.
We said that an exponential decay function has the same general form as an exponentially increasing function, butthat the base b is a positive number less than one. When b can be written as a fraction, we can use the Property ofNegative Exponents that we discussed in Section 8.3 to write the function in a different form.
For instance, y = 5 ·(1
2
)x is equivalent to 5 ·2−x.
These two forms are both commonly used so it is important to know that they are equivalent.
Example 2
Graph the exponential function y = 8 ·3−x.
Solution
Here is our table of values and the graph of the function.
TABLE 7.8:
x y = 8 ·3−x
−3 y = 8.3−(−3) = 8 ·33 = 216-2 y = 8.3−(−2) = 8 ·32 = 72-1 y = 8.3−(−1) = 8 ·31 = 240 y = 8 ·30 = 81 y = 8 ·3−1 = 8
32 y = 8 ·3−2 = 8
9
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Compare Graphs of Exponential Decay Functions
You might have noticed that an exponentially decaying function is very similar to an exponentially increasingfunction. The two types of functions behave similarly, but they are backwards from each other.
The increasing function starts very small and increases very quickly and ends up very, very big. While the decreasingfunction starts very big and decreases very quickly to soon become very, very small. Lets graph two such functionstogether on the same graph and compare them.
Example 3
Graph the functions y = 4x and y = 4−x on the same coordinate axes.
Solution
Here is the table of values and the graph of the two functions.
Looking at the values in the table we see that the two functions are backwards of each other in the sense that thevalues for the two functions are reciprocals.
TABLE 7.9:
x y = 4x y = 4−x
−3 y = 4−3 = 164 y = 4−(−3) = 64
-2 y = 4−2 = 116 y = 4−(−2) = 16
-1 y = 4−1 = 14 y = 4−(−1) = 4
0 y = 40 = 1 y = 40 = 11 y = 41 = 4 y = 4−1 = 1
42 y = 42 = 16 y = 4−2 = 1
163 y = 43 = 64 y = 4−3 = 1
64
Here is the graph of the two functions. Notice that the two functions are mirror images of each others if the mirroris placed vertically on the y−axis.
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Solve Real-World Problems Involving Exponential Decay
Exponential decay problems appear in several application problems. Some examples of these are half-life problems,and depreciation problems. Lets solve an example of each of these problems.
Example 4 Half-Life
A radioactive substance has a half-life of one week. In other words, at the end of every week the level of radioactivityis half of its value at the beginning of the week. The initial level of radioactivity is 20 counts per second.
a) Draw the graph of the amount of radioactivity against time in weeks.
b) Find the formula that gives the radioactivity in terms of time.
c) Find the radioactivity left after three weeks
Solution
Lets start by making a table of values and then draw the graph.
TABLE 7.10:
time radioactivity0 201 102 53 2.54 1.255 0.625
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Exponential decay fits the general formula
y = A ·bx
In this case
y is the amount of radioactivity
x is the time in weeks
A = 20 is the starting amount
b = 12 since the substance losses half its value each week
The formula for this problem is: y = 20 ·(1
2
)x or y = 20 ·2−x.
Finally, to find out how much radioactivity is left after three weeks, we use x = 3 in the formula we just found.
y = 20 ·(
12
)3
=208
= 2.5
Example 5 Depreciation
The cost of a new car is $32,000. It depreciates at a rate of 15% per year. This means that it looses 15% of eachvalue each year.
Draw the graph of the cars value against time in year.
Find the formula that gives the value of the car in terms of time.
Find the value of the car when it is four years old.
Solution
Lets start by making a table of values. To fill in the values we start with 32,000 at time t = 0. Then we multiply thevalue of the car by 85% for each passing year. (Since the car looses 15% of its value, that means that it keeps 85%of its value). Remember that 85% means that we multiply by the decimal 0.85.
TABLE 7.11:
Time Value(Thousands)0 321 27.22 23.1
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TABLE 7.11: (continued)
Time Value(Thousands)3 19.74 16.75 14.2
Now draw the graph
Lets start with the general formula
y = A ·bx
In this case:
y is the value of the car
x is the time in years
A = 32 is the starting amount in thousands
b = 0.85 since we multiply the amount by this factor to get the value of the car next year
The formula for this problem is y = 32 · (0.85)x.
Finally, to find the value of the car when it is four years old, we use x = 4 in the formula we just found.
y = 32 · (0.85)4 = 16.7 thousand dollars or $16,704 if we don’t round.
Video Overview
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Homework
Graph the following exponential decay functions.
1. y = 15
x
2. y = 4 ·(2
3
)x
3. y = 3−x
4. y = 34 ·6
−x
Solve the following application problems.
5. The cost of a new ATV (all-terrain vehicle) is $7200. It depreciates at 18% peryear.Draw the graph of the vehicles value against time in years.Find the formula that gives the value of the ATV in terms of time.Find the value of the ATV when it is ten year old.
6. A person is infected by a certain bacterial infection. When he goes to the doctor the population of bacteria is2 million. The doctor prescribes an antibiotic that reduces the bacteria population to 1
4 of its size each day.
(a) Draw the graph of the size of the bacteria population against time in days.(b) Find the formula that gives the size of the bacteria population in term of time.(c) Find the size of the bacteria population ten days after the drug was first taken.(d) Find the size of the bacteria population after 2 weeks (14 days)
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7.3 Arithmetic Sequences and Linear Func-tions
What You Will Learn
• Arithmetic Sequence• How it relates to a linear function
Arithmetic Sequence
Look at this sequence.
You probably saw a pattern right away. If there were another set of boxes, you’d probably guess at how many therewould be.
If you saw this same pattern in terms of numbers, it would look like this:
2,4,6,8,10
This set of numbers is called a sequence ; it is a series of numbers that follow a pattern.
If there was another set of boxes, you’d probably guess there would be 12, right? Just like if you added anothernumber to the sequence, you’d write 12. You noticed that there was a difference of 2 between each two numbers, orterms, in the sequence.
When we have a sequence with a fixed number between each of the terms, we call this sequence an arithmeticsequence .
Take a look at this one.
What is the common difference between each of the terms in the sequence?
The difference is 5 between each number.
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This is an arithmetic sequence. You can see that you have to be a bit of a detective to figure out the numberpatterns.
Finding the difference between two terms in a sequence is one way to look at sequences. We have used tables ofvalues for several types of equations and we have used those tables of values to create graphs. Graphs are helpfulbecause they are visual representations of the same numbers. When values rise, we can see them rise on a graph.Let’s use the same ideas, then, to graph arithmetic sequences.
Take a look at this one.
Graph the sequence 2, 5, 8, 11, 14, 17,...
First convert it into a table of values with independent values being the term number and the dependent values beingthe actual term.
Use this table to create a graph.
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You can see the pattern clearly in the graph. That is one of the wonderful things about graphing arithmetic sequences.
In the graph that we created in the example, each term was expressed as a single point. This is called discrete data—only the exact points are shown. This type of data is usually involves things that are counted in whole numberslike people or boxes. Depending on what type of situation you are graphing, you may choose to connect the pointswith a line. The line demonstrates that there are data points between the points that we have graphed. This is calledcontinuous data and usually involves things like temperature or length that can change fractionally.
We can write an equation to describe this data if we compare it to the linear functions we learned in a previoussection . To find the equation for the function, we start with the general form of a linear function: y = mx+b. Sincem is the slope of the line, it’s also the quantity by which y increases every time the value of x increases by one. Inthis case, the slope is 3. The y intercept is the value of y when x is 0. To find the y value when x is zero, we willneed to subtract 3 from 2, this will give us -1 for the y intercept. The constant b is the value of the function whenx = 0. Therefore, the function is y = 3x−1.
So, we can graph sequences and classify them as either discrete or continuous data. Yet another possibility iscontinuing a sequence in either direction by adding terms that follow the same pattern.
Identify the pattern in the following sequences.
Example A
3,7,11,15
Solution: Add four
Example B
18,8,−2
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Solution: Subtract 10
Example C
81,86,91,96
Solution: Add five
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Homework
Directions: (a) Write the common difference for each sequence. If there is not a pattern, indicate this in your answer.(b) If there is a common difference, write a linear equation for each sequence.
1. -9, -7, -5, -3, -12. 5.05, 5.1, 5.15, 5.2, 5.253. 3, 6, 10, 15, 21, 284. 17, 14, 11, 8, 5, 25. 10, 9, 8, 7, 66. 3, 5, 7, 9, 117. 3, 9, 278. 4, 8, 16, 329. 2, 3, 5, 9
10. 5, 11, 23, 4711. 16, 8, 4, 212. 5, 10, 15, 2013. 3, 6, 9, 12
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7.4 Geometric Sequences and ExponentialFunctions
What You Will Learn
• Geometric Sequence• How it relates to an exponential function
Geometric Sequence
Which would you prefer: being given one million dollars, or one penny the first day, double that penny the next day,and then double the previous day’s pennies and so on for a month?
At first glance you would think that a million dollars would be more. However, let’s use geometric sequences beforewe decide to see how the pennies add up. You start with a penny the first day and keep doubling each day until theend of the month (30 days).
1st Day 1 penny = 20
2nd Day 2 pennies = 21
3rd Day 4 pennies = 22
4th Day 8 pennies = 23
30th Day = 229
229 = 536,870,912 pennies or $5,368,709.12, which is well over five times greater than $1,000,000.00.
Geometric Sequence: a sequence of numbers in which each number in the sequence is found by multiplying theprevious number by a fixed amount called the common ratio.
nth term in a geometric sequence: an = a1rn−1 (a1 = first term, r = common ratio)
The Common Ratio of a Geometric Sequence
The common ratio, r, in any geometric sequence can be found by dividing any term by the preceding term. If weknow the common ratio in a sequence then we can find any term in the sequence.
Example A
Find the eighth term in the geometric sequence.
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1, 2, 4,...
Solution:
First we need to find the common ratio: r = 21 = 2.
The eighth term is given by the formula 2 = 1 ·27 = 128.
In other words, to get the eighth term we started with the first term, which is 1, and multiplied by two seven times.
Graphing Geometric Sequences
Exponential graphs and geometric sequence graphs look very much alike. Exponential graphs are continuous,however, and the sequence graphs are discrete with distinct points (1st term and 2nd term, etc). If we graph thesequence describing the doubline of pennies, it looks like this.
FIGURE 7.1
Each time the x−value increases by one, the y−value is multiplied by 2. Since the ratio is always the same, thefunction is exponential.
To find the equation for the function that represents these values, we start with the general form of an exponentialfunction, y = a ·bx.
Here b is the ratio between the values of y each time x is increased by one. The constant a is the value of the sequencewhen x =0, which is the first day in this example. Therefore, the function is y = 1 ·2x or y = 2x
Example B
A courtier presented the Indian king with a beautiful, hand-made chessboard. The king asked what he would like inreturn for his gift and the courtier surprised the king by asking for one grain of rice on the first square, two grainson the second, four grains on the third, etc. The king readily agreed and asked for the rice to be brought (fromMeadows et al. 1972, p.29 via Porritt 2005). How many grains of rice does the king have to put on the last square?
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Solution:
A chessboard is an 8×8 square grid, so it contains a total of 64 squares.
The courtier asked for one grain of rice on the first square, 2 grains of rice on the second square, 4 grains of rice onthe third square and so on. We can write this as a geometric sequence.
1, 2, 4,...
The numbers double each time, so the common ratio is r = 2.
The problem asks how many grains of rice the king needs to put on the last square. What we need to find is the 64th
term in the sequence. This means multiplying the starting term, 1, by the common ratio 64 times in a row. Let’s usethe formula.
an = a1rn−1, where an is the nth term, a1 Is the first term and r is the common ratio.
a64 = 1 ·263 = 9,223,372,036,854,775,808 grains of rice.
Practice Exercise
Find the next number in the geometric sequence: 0.01 , 0.06 , 0.36 , 2.16 , 12.96 , ...
Solution:
The common ratio is:0.060.01 = 6.
Since the first value in the sequence is 0.01, the geometric sequence is
an = 0.01(6)n−1
and since we are looking for the 6th term
a6 = 0.01(6)6−1 = 0.01(6)5 = 77.76.
The next number in the sequence is 77.76.
Video Overview
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Homework
1. Define a geometric sequence.2. Using the chessboard example, how many grains of rice should be placed on the:
a. Fourth square?b. Tenth square?c. Twenty-fifth square?d. 60th square?
3. Is the following an example of a geometric sequence? Explain your answer. –1, 1, –1, 1, –1, ...4. Use the Petri dish example from the Concept. At what time are 25,000 bacteria in the dish?
Determine the constant ratio, r, for each geometric sequence.
5. 8, 6,...6. 2, 4, 8, 16,...7. 9,3, 1
3 ,19 , ...
8. 2, –8, 32, –128
Determine the first five terms of each geometric sequence.
9. a1 = 2,r = 310. a1 = 90,r =−1
311. a1 = 6,r =−2
Find the missing terms in each geometric sequence.
12. 3, ____, 48, 192, ____13. 81, ____, ____, ____, 114. 9
4 , ____, ____, 23 , ____
Find the indicated term of each geometric sequence.
15. a1 = 4,r = 2 Find a6.16. a1 =−7,r =−3
4 Find a4.17. a1 =−10,r =−3 Find a10.
18. A ball is tossed from a height of four feet. Each bounce is 80% as high as the previous bounce.
a. Write an equation to represent the situation.b. How high is the ball after the fifth bounce?
19. An ant walks past several stacks of Lego blocks. There is one block in the first stack, three blocks in thesecond stack, and nine blocks in the third stack. In fact, in each successive stack there are triple the number ofblocks there were in the previous stack.
a. How many blocks are in the eighth stack?b. When is the stack 343 blocks high?
20. A super-ball has a 75% rebound ratio. When you drop it from a height of 20 feet, it bounces and bounces andbounces...
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a. How high does the ball bounce after it strikes the ground for the third time?b. How high does the ball bounce after it strikes the ground for the seventeenth time?
21. Anne goes bungee jumping off a bridge above water. On the initial jump, the bungee cord stretches by 120feet. On the next bounce, the stretch is 60% of the original jump and each additional bounce stretches the ropeby 60% of the previous stretch.
a. What will the rope stretch be on the third bounce?b. What will the rope stretch be on the 12th bounce?
22. A virus population doubles every 30 minutes. It begins with a population of 30.
a. How many viral cells will be present after 5 hours?b. When will it reach 1,000,000 cells? Round to the nearest half-hour.
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7.5 Describing the Pattern and Writing a Recur-sive Rule for a Sequence
What You Will Learn
• Recursive Rule• Use recursive rules to describe arithmetic and geometric sequences
Recursive Rule
A recursive rule for a sequence is a formula which tells us how to progress from one term to the next in a sequence.Generally, the variable n is used to represent the term number. In other words, n takes on the values 1 (first term), 2(second term), 3 (third term), etc. The variable, an represents the nth term and an−1 represents the term preceding an.
Example sequence: 4,7,11,16, . . . ,an−1,an
In the above sequence, a1 = 4, a2 = 7, a3 = 11 and a4 = 16.
Example A
Describe the pattern and write a recursive rule for the sequence: 9,11,13,15, . . .
Solution: First we need to determine what the pattern is in the sequence. If we subtract each term from the onefollowing it, we see that there is a common difference of 29. We can therefore use an−1 and an to write a recursiverule as follows: an = an−1 +29
Example B
Write a recursive rule for the sequence: 3,9,27,81, . . .
Solution: In this sequence, each term is multiplied by 3 to get the next term. We can write a recursive rule:an = 3an−1
Example C
Write a recursive rule for the sequence: 1,1,2,3,5,8, . . .
Solution: This is a special sequence called the Fibonacci sequence. In this sequence each term is the sum of theprevious two terms. We can write the recursive rule for this sequence as follows: an = an−2 +an−1.
Practice Exercises
Write the recursive rules for the following sequences.
1. 1,2,4,8, . . .
2. 1,−2,−5,−8, . . .
3. 1,2,4,7, . . .
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Answers
1. In this sequence each term is double the previous term so the recursive rule is: an = 2an−1
2. This time three is subtracted each time to get the next term: an = an−1−3.
3. This one is a little trickier to express. Try looking at each term as shown below:
a1 = 1
a2 = a1 +1
a3 = a2 +2
a4 = a3 +3...
an = an−1 +(n−1)
Video Overview for Arithmetic Sequences
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Video Overview for Geometric Sequence
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Homework
Describe the pattern and write a recursive rule for the following sequences.
1. 14 ,−
12 ,1,−2 . . .
2. 5,11,17,23, . . .3. 33,28,23,18, . . .4. 1,4,16,64, . . .5. 21,30,39,48, . . .6. 100,75,50,25, . . .7. 243,162,108,72, . . .8. 128,96,72,54, . . .
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9. 1,5,10,16,23, . . .10. 0,2,2,4,6, . . .11. 3,5,8,12, . . .12. 0,2,6,12, . . .13. 4,9,14,19, . . .14. 1
2 ,23 ,
34 ,
45 . . .
15. 4,5,9,14,23, . . .
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7.6 Linear, Exponential, and Quadratic Models
What You Will Learn
• Identify functions using differences and ratios• Write equations for functions• Perform exponential and quadratic regressions with a graphing calculator• Solve real world problems by comparing function models
Introduction
In this course we’ve learned about three types of functions, linear, quadratic and exponential.
• Linear functions take the form y = mx+b.• Quadratic functions take the form y = ax2 +bx+ c.• Exponential functions take the form y = a ·bx.
In real-world applications, the function that describes some physical situation is not given; it has to be found beforethe problem can be solved. For example, scientific data such as observations of planetary motion are often collectedas a set of measurements given in a table. Part of the scientist’s job is to figure out which function best fits the data.In this section, you’ll learn some methods that are used to identify which function describes the relationship betweenthe variables in a problem.
Identify Functions Using Differences or Ratios
One method for identifying functions is to look at the difference or the ratio of different values of the dependentvariable. For example, if the difference between values of the dependent variable is the same each time wechange the independent variable by the same amount, then the function is linear .
Example A
Determine if the function represented by the following table of values is linear.
TABLE 7.12:
x y−2 –4–1 –10 21 52 8
If we take the difference between consecutive y−values, we see that each time the x−value increases by one, they−value always increases by 3.
Since the difference is always the same, the function is linear. This might remind you of the arithmetic sequence
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discussion we had earlier in this section, a linear function is always having the same value added to go from term toterm.
When we look at the difference of the y−values, we have to make sure that we examine entries for which thex−values increase by the same amount.
For example, examine the values in this table:
TABLE 7.13:
x y0 51 103 204 256 35
At first glance this function might not look linear, because the difference in the y−values is not always the same. Butif we look closer, we can see that when the y−value increases by 10 instead of 5, it’s because the x−value increasedby 2 instead of 1. Whenever the x−value increases by the same amount, the y−value does too, so the function islinear.
Another way to think of this is in mathematical notation. We can say that a function is linear if y2−y1x2−x1
is always thesame for any two pairs of x− and y−values. Notice that the expression we used here is simply the definition of theslope of a line.
Differences can also be used to identify quadratic functions. For a quadratic function, when we increase thex−values by the same amount, the difference between y−values will not be the same. However, the differenceof the differences of the y−values will be the same.
Here are some examples of quadratic relationships represented by tables of values:
In this quadratic function, y = x2, when we increase the x−value by one, the value of y increases by different values.However, it increases at a constant rate, so the difference of the difference is always 2.
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In this quadratic function, y = 2x2− 3x+ 1, when we increase the x−value by one, the value of y increases bydifferent values. However, the increase is constant: the difference of the difference is always 4.
To identify exponential functions, we use ratios instead of differences. If the ratio between values of the dependentvariable is the same each time we change the independent variable by the same amount, then the function isexponential.
Example B
Determine if the function represented by each table of values is exponential.
a)
b)
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a) If we take the ratio of consecutive y−values, we see that each time the x−value increases by one, the y−value ismultiplied by 3. Since the ratio is always the same, the function is exponential.
b) If we take the ratio of consecutive y−values, we see that each time the x−value increases by one, the y−value ismultiplied by 1
2 . Since the ratio is always the same, the function is exponential.
Write Equations for Functions
Once we identify which type of function fits the given values, we can write an equation for the function by startingwith the general form for that type of function. In this section, we will do this using algebra. It can be tricky to findthe equation for a function by hand, in the next section, we will use the graphing calculator to find the equations todescribe the functions.
Example C
Determine what type of function represents the values in the following table.
TABLE 7.14:
x y0 51 12 -33 -74 -11
Solution
Let’s first check the difference of consecutive values of y.
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If we take the difference between consecutive y−values, we see that each time the x−value increases by one, they−value always decreases by 4. Since the difference is always the same, the function is linear.
To find the equation for the function, we start with the general form of a linear function: y = mx+b. Since m is theslope of the line, it’s also the quantity by which y increases every time the value of x increases by one. The constantb is the value of the function when x = 0. Therefore, the function is y =−4x+5.
Example D
Determine what type of function represents the values in the following table.
TABLE 7.15:
x y0 01 52 203 454 805 1256 180
Solution
Here, the difference between consecutive y−values isn’t constant, so the function is not linear. Let’s look at thosedifferences more closely.
TABLE 7.16:
x y0 01 5 5−0 = 52 20 20−5 = 153 45 45−20 = 254 80 80−45 = 355 125 125−80 = 45
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TABLE 7.16: (continued)
x y6 180 180−125 = 55
When the x−value increases by one, the difference between the y−values increases by 10 every time. Since thedifference of the differences is constant, the function describing this set of values is quadratic.
To find the equation for the function that represents these values, we start with the general form of a quadraticfunction: y = ax2 +bx+ c.
We need to use the values in the table to find the values of the constants: a,b and c.
The value of c represents the value of the function when x = 0, so c = 0.
Plug in the point (1,5) : 5 = a+b
Plug in the point (2,20) : 20 = 4a+2b⇒ 10 = 2a+b
To find a and b,we solve the system of equations: 5 = a+b
10 = 2a+b
Solve the first equation for b : 5 = a+b⇒ b = 5−a
Plug the first equation into the second: 10 = 2a+5−a
Solve for a and b a = 5 and b = 0
Therefore the equation of the quadratic function is y = 5x2.
It can be difficult to find the equation of the functions by hand, later on in this section, we will use the graphingcalculator to find the equations that fit experimental data
Example E
Determine what type of function represents the values in the following table.
TABLE 7.17:
x y0 4001 5002 253 6.254 1.5625
Solution
The differences between consecutive y−values aren’t the same, and the differences between those differences aren’tthe same either. So let’s check the ratios instead.
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Each time the x−value increases by one, the y−value is multiplied by 14 . Since the ratio is always the same, the
function is exponential.
To find the equation for the function that represents these values, we start with the general form of an exponentialfunction, y = a ·bx.
Here b is the ratio between the values of y each time x is increased by one. The constant a is the value of the functionwhen x = 0. Therefore, the function is y = 400
(14
)x.
Perform exponential and quadratic regressions with a graphing calculator
Earlier, you learned how to perform linear regression with a graphing calculator to find the equation of a straight linethat fits a linear data set. In this section, you’ll learn how to perform exponential and quadratic regression to findequations for curves that fit non-linear data sets.
Example A
The following table shows how many miles per gallon a car gets at different speeds.
TABLE 7.18:
Speed (mph) Miles per gallon30 1835 2040 2345 2550 2855 3060 2965 2570 25
Using a graphing calculator:
a) Draw the scatterplot of the data.
b) Find the quadratic function of best fit.
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c) Draw the quadratic function of best fit on the scatterplot.
d) Find the speed that maximizes the miles per gallon.
e) Predict the miles per gallon of the car if you drive at a speed of 48 mph.
Solution
Step 1: Input the data.
Press [STAT] and choose the [EDIT] option.
Input the values of x in the first column (L1) and the values of y in the second column (L2). ( Note: in order to cleara list, move the cursor to the top so that L1 or L2 is highlighted. Then press [CLEAR] and then [ENTER].)
Step 2: Draw the scatterplot.
First press [Y=] and clear any function on the screen by pressing [CLEAR] when the old function is highlighted.
Press [STATPLOT] [STAT] and [Y=] and choose option 1.
Choose the ON option; after TYPE, choose the first graph type (scatterplot) and make sure that the Xlist and Ylistnames match the names on top of the columns in the input table.
Press [GRAPH] and make sure that the window is set so you see all the points in the scatterplot. In this case, thesettings should be 30≤ x≤ 80 and 0≤ y≤ 40. You can set the window size by pressing the [WINDOW] key at thetop.
Step 3: Perform quadratic regression.
Press [STAT] and use the right arrow to choose [CALC].
Choose Option 5 (QuadReg) and press [ENTER]. You will see “QuadReg” on the screen.
Type in L1,L2 after ’QuadReg’ and press [ENTER]. The calculator shows the quadratic function: y = −0.017x2 +1.9x−25
Step 4: Graph the function.
Press [Y=] and input the function you just found.
Press [GRAPH] and you will see the curve fit drawn over the data points.
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To find the speed that maximizes the miles per gallon, use [TRACE] and move the cursor to the top of the parabola.You can also use [CALC] [2nd] [TRACE] and option 4:Maximum, for a more accurate answer. The speed thatmaximizes miles per gallon is 56 mph.
Finally, plug x = 48 into the equation you found: y =−0.017(48)2 +1.9(48)−25 = 27.032 miles per gallon.
Note: The image above shows our function plotted on the same graph as the data points from the table. One thingthat is clear from this graph is that predictions made with this function won’t make sense for all values of x. Forexample, if x < 15, this graph predicts that we will get negative mileage, which is impossible. Part of the skill ofusing regression on your calculator is being aware of the strengths and limitations of this method of fitting functionsto data.
Example B
The following table shows the amount of money an investor has in an account each year for 10 years.
TABLE 7.19:
Year Value of account1996 $50001997 $54001998 $58001999 $63002000 $68002001 $73002002 $79002003 $86002004 $93002005 $100002006 $11000
Using a graphing calculator:
a) Draw a scatterplot of the value of the account as the dependent variable, and the number of years since 1996 asthe independent variable.
b) Find the exponential function that fits the data.
c) Draw the exponential function on the scatterplot.
d) What will be the value of the account in 2020?
Solution
Step 1: Input the data.
Press [STAT] and choose the [EDIT] option.
Input the values of x in the first column (L1) and the values of y in the second column (L2).
Step 2: Draw the scatterplot.
First press [Y=] and clear any function on the screen.
Press [GRAPH] and choose Option 1.
Choose the ON option and make sure that the Xlist and Ylist names match the names on top of the columns in theinput table.
Press [GRAPH] make sure that the window is set so you see all the points in the scatterplot. In this case the settings
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should be 0≤ x≤ 10 and 0≤ y≤ 11000.
Step 3: Perform exponential regression.
Press [STAT] and use the right arrow to choose [CALC].
Choose Option 0 and press [ENTER]. You will see “ExpReg” on the screen.
Press [ENTER]. The calculator shows the exponential function: y = 4975.7(1.08)x
Step 4: Graph the function.
Press [Y=] and input the function you just found. Press [GRAPH].
Finally, plug x = 2020−1996 = 24 into the function: y = 4975.7(1.08)24 = $31551.81
In 2020, the account will have a value of $31551.81.
Note: The function above is the curve that comes closest to all the data points. It won’t return y−values that areexactly the same as in the data table, but they will be close. It is actually more accurate to use the curve fit valuesthan the data points.
Example C
The following table shows the number of students enrolled in public elementary schools in the US (source: USCensus Bureau). Make a scatterplot with the number of students as the dependent variable, and the number of yearssince 1990 as the independent variable. Find which curve fits this data the best and predict the school enrollment inthe year 2007.
TABLE 7.20:
Year Number of students (millions)1990 26.61991 26.61992 27.11993 27.71994 28.11995 28.41996 28.11997 29.11998 29.32003 32.5
Solution
We need to perform linear, quadratic and exponential regression on this data set to see which function represents thevalues in the table the best.
Step 1: Input the data.
Input the values of x in the first column (L1) and the values of y in the second column (L2).
Step 2: Draw the scatterplot.
Set the window size: 0≤ x≤ 10 and 20≤ y≤ 40.
Here is the scatterplot:
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Step 3: Perform Regression.
Linear Regression
The function of the line of best fit is y = 0.51x+26.1. Here is the graph of the function on the scatterplot:
Quadratic Regression
The quadratic function of best fit is y = 0.064x2− .067x+26.84. Here is the graph of the function on the scatterplot:
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Exponential Regression
The exponential function of best fit is y = 26.2(1.018)x. Here is the graph of the function on the scatterplot:
From the graphs, it looks like the quadratic function is the best fit for this data set. We’ll use this function to predictschool enrollment in 2007.
x = 2007−1990 = 17 so y = 0.064(17)2− .067(17)+26.84 which is 44.2 million students.
Video Overview
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MEDIAClick image to the left for more content.
Homework
Determine whether the data in the following tables can be represented by a linear function.
TABLE 7.21:
x y−4 10-3 7-2 4-1 10 -21 -5
TABLE 7.22:
x y−2 4-1 30 21 32 63 11
TABLE 7.23:
x y0 501 752 1003 1254 1505 175
Determine whether the data in the following tables can be represented by a quadratic function.
TABLE 7.24:
x y−10 10-5 2.50 05 2.5
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TABLE 7.24: (continued)
x y10 1015 22.5
TABLE 7.25:
x y1 42 63 64 45 06 -6
TABLE 7.26:
x y−3 -27-2 -8-1 -10 01 12 83 27
Determine whether the data in the following tables can be represented by an exponential function.
TABLE 7.27:
x y0 2001 3002 18003 83004 258005 62700
TABLE 7.28:
x y0 1201 1802 2703 4054 607.55 911.25
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TABLE 7.29:
x y0 40001 24002 14403 8644 518.45 311.04
Determine what type of function represents the values in the following tables and find the equation of each function.
TABLE 7.30:
x y0 4001 5002 6253 781.254 976.5625
TABLE 7.31:
x y−9 -3-7 -2-5 -1-3 0-1 11 2
TABLE 7.32:
x y−3 14-2 4-1 -20 -41 -22 43 14
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7.7 References
1. . . CC BY-NC-SA
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