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Algebra 1 - Essex High School

Lea Ann SmithAndrew Gloag, (AndrewG)

Anne Gloag, (AnneG)

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Printed: August 25, 2014

AUTHORSLea Ann SmithAndrew Gloag, (AndrewG)Anne Gloag, (AnneG)

iii

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 Creating Equations from Patterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 702.4 Graphing using Table of Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 762.5 Graphing Using Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 822.6 Slope and Rate of Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 862.7 Graphs Using Slope-Intercept Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 932.8 Writing Equations of Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1022.9 Parallel and Perpendicular Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1092.10 Graphing Linear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1132.11 Graphs of Absolute Value Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1192.12 Fitting a Line to Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1232.13 Problem Solving with Linear Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

3 Unit 3 - Systems of Linear Equations 1393.1 Solving Linear Systems by Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1403.2 Solving Linear Systems by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1493.3 Solving Linear Systems by Combination (Elimination) . . . . . . . . . . . . . . . . . . . . . . 1523.4 Special Types of Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1593.5 Applications of Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1633.6 Systems of Linear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

4 Unit 4 - Polynomials, Exponents and Radicals 1764.1 Addition and Subtraction of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

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www.ck12.org Contents

4.2 Exponent Properties Involving Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1834.3 Multiplication of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1914.4 Special Products of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1954.5 Exponent Properties Involving Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1994.6 Zero and Negative Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2064.7 Division of Polynomials by Monomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2114.8 Division of Polynomials by Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2134.9 Scientific Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2174.10 Square Roots and Irrational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2234.11 Pythagorean Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2264.12 Multiplication and Division of Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2284.13 Addition and Subtraction of Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230

5 Unit 5 - Quadratic Functions 2325.1 Monomial Factors of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2335.2 Factoring Quadratic Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2365.3 Factoring Polynomials where a does not equal 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 2425.4 Factoring the Difference of Two Squares (DOTS) . . . . . . . . . . . . . . . . . . . . . . . . . 2465.5 Solve Quadratic Equations by Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2495.6 Solve Quadratic Equations by Square Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2535.7 Solve Quadratic Equations by Completing the Square . . . . . . . . . . . . . . . . . . . . . . . 2595.8 Solving Quadratic Equations by the Quadratic Formula . . . . . . . . . . . . . . . . . . . . . . 2645.9 Graphs of Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2715.10 Vertex Form of a Quadratic Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2785.11 Applications of Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

6 Unit 6 - Exponential Functions, Sequences and Modeling 2886.1 Exponential Growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2896.2 Exponential Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2976.3 Arithmetic Sequences and Linear Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3056.4 Geometric Sequences and Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . 3096.5 Describing the Pattern and Writing a Recursive Rule for a Sequence . . . . . . . . . . . . . . . 3146.6 Linear, Exponential, and Quadratic Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3176.7 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332

7 Appendix A - One Variable Statistics 3337.1 Dot Plots, Frequency Tables, Histograms and Bar Graphs . . . . . . . . . . . . . . . . . . . . . 3347.2 Measures of Central Tendency (Mean, Median and Mode) . . . . . . . . . . . . . . . . . . . . . 3447.3 Box and Whisker Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3517.4 Shape, Center and Spread . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355

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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

1

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

mx3

Variable expressions are used to describe real-world situations when we dont 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 tofind the value of. We find the value of an expression.

Why dont 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 cant 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 10k44 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.

1204476

Our answer is 76.

Sometimes, you will also have an expression to evaluate that uses division.

2


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.

243 = 8Now we add 2.

8+2 = 10

Our answer is 10.

Practice Exercises

Now it is your turn. Evaluate the following single-variable expressions.

a. 4x9 if x is 20b. 5y+6 if y is 9c. a4 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+(x7)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.

24 = 8Next, we add two.

8+2 = 10

3

1.1. Variables and Expressions www.ck12.org

Our answer is 10.

Lets 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 91Now you multiply numerator numerator and denominator denominator.91 23Next, 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 12

Video Overview

MEDIAClick image to the left for use the URL below.URL: http://www.ck12.org/flx/render/embeddedobject/70317

Homework

Directions: Evaluate each expression if the given value of r = 9.

4


1. r32. 63 r3. 11r

4. 2r+7

5. 3r+ r

6. 4r2r7. r+5r

8. 12r1Directions: Evaluate each expression when h =12.9. 703h10. 6h+6

11. 4h912. 11+ h413. 3h+h

14. 2h+5h

15. 6h2hDirections: Evaluate each multi-variable expression.

16. Evaluate xy+9y if x = 3 and y = 1317. Evaluate 2x+3y if x = 25 and y =

14

18. Evaluate 4y+3y2z if y = 2 and z =419. Evaluate 6z2(z+ x) if x is 3 and z is 420. Evaluate 8a+3b2c if a is 5, b is 4 and c is 3Here 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.

5

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 (72)423Solution: First, we check for parentheses. Yes, there they are and must be done first.

(72)423 = (5)423

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.

5423 = 20232023 = 103

Finally, addition and subtraction are equally important and must be done from left to right.

103 = 7 This is our answer.Example 2: Use the Order of Operations to simplify the following expressions.

a) 3572b) 3 (57)2c) (35) (72)Solutions:

a) There are no parentheses and no exponents. Go directly to multiplication and division from left to right: 3572 = 1572 = 153.5Now subtract: 153.5 = 11.5b) Parentheses must be done first: 3 (2)2

6


There are no exponents, so multiplication and division come next and are done left to right: 3(2)2=62=3c) Parentheses must be done first: (35) (72) = 153.5There are no exponents, multiplication, division, or addition, so simplify:

153.5 = 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 [1977]

Simplify according to the Order of Operations:

8 [1977] = 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 = 2b)(a+2

b+4 1)+b when a = 3 and b = 1

c) 2(

w+(x2z)(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 = 54 1. Rewriting 1 as a fraction, the expression becomes:

54 4

4=

14

b) (3+2)(1+4) =55 = 1

(11)+b Substituting 1 for b, the expression becomes 0+1 = 1c) 2

([11+(32(2))]

[(1+2)2)] 1)= 2

((11+7)

32 1)= 2

(189 1

)Continue simplifying: 2

(189 99

)= 2

(99

)= 2(1) = 2

7


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(x21)2 x4+12]+5x31 when x =3.

Substitute the value x =3 into the expression.

[3((3)21)2 (3)4+12]+5(3)31

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: 3x24y2+x4(x+y)

12

for x =2,y = 1.

8


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 .88 or 89 .

Video Overview


Homework

Use the Order of Operations to simplify the following expressions.

1. 8 (19 (2+5)7)2. 2+7111233. (3+7) (712)4. 2(3+(21))4(6+2) (35)5. 8 5+626. 93723+77. 8+126+68. (7232)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. 4x9x23x+1 when x = 214. z

2

x+y +x2

xy when x = 1, y =2, and z = 4.15. 4xyzy2x2 when x = 3, y = 2, and z = 5

9


16. x2z2

xz2x(zx) 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. 52 64+2 = 523. 124+103 3+7 = 1124. 22325 36 = 3025. 1284 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+yzxy+yz+xz when x =

12 , y =

32 , and z =1

29. (x+y)2

4x2y2 when x = 3 and y =5d30. The formula to find the volume of a spherical object (like a ball) is V = 43(pi)r

3, where r = the radius of thesphere. Determine the volume for a grapefruit with a radius of 9 cm.

10


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 = 242Since both sides of the equation equal 22, these equations are equivalent.

15+7 = 22 and 242 = 22An 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(x7)+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.

11

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. Lets look at an example.

Example

x5 = 10This 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 lets check our answer. To do this we substitute our answer for x back into the equation and see if both sidesare equal.

155 = 1010 = 10

12


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. x3 = 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.

Lets 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 lets 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. x7 = 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

14


16. x4 = 1017. y8 = 2018. 5 y = 219. 22a = 1520. 18 y = 2Directions: 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.Lets 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 yHere 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


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: a12

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 n4. 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


Homework

Directions: List at least three possible values for the variable in each inequality

1. x < 132. y > 53. x < 24. y >3

18


5. a > 126. x 47. y 38. b39. a5

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

20


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


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) (34)(-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

22


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 = 10022

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 = 10022x+22 = 10022+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 x3 = 9Solution

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.

x3 = 9x3+3 = 9+3 The +3 and 3 on the left cancel each other. We evaluate 9+3

x = 12

Example 2

Solve x3 = 11Solution

To isolate x we need to add 3 to both sides of the equation. This time we will add vertically.

x3 = 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 z5 =7Solution

This time our variable is z, but dont let that worry you. Treat this variable like any other variable.

z5 =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.

24


x+6 = 26

6 =6x = 20

Example 5

Solve x+20 =11Solution

To isolate x we need to subtract 20 from both sides of the equation.

x+20 =1120 =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 shouldnt 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 = 1.5Our 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(

18 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 = 59x5 is equivalent to

95 x so x is being multiplied by 95 . To cancel, multiply by the reciprocal, 59 .

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1.6. Solving Equations Using One Step www.ck12.org

59

(9x5

)=

595

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. Lets see how this problem looks in algebra.

Example 8

Solve 5x = 325 To cancel the 5 we divide both sides by 5.

5x5=

3255

x = 65

Video Overview of Adding/Subtracting to solve equations


Video Overview of Multiplying/Dividing to solve equations


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.

26


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. x1.1 = 3.2c. 7x = 21d. 4x = 1e. 5x12 =

23

f. x+ 52 =23

g. x 56 = 382. Solve the following equations for the unknown variable.

a. q13 =13b. z+13 = 9c. 21s = 3d. t+ 12 =

13

e. 7 f11 =711

f. 34 =12 yg. 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

3x5 = 10We 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. Heres how you use algebra to solve it

3x5 = 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

28


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


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(4x7) = 21 Use the Distributive Property4x+812x+21 = 21 Combine like terms8x+29 = 21 Subtract 29 from each side29 =29

8x8 =

88 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 side7 =7

6x6=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


Video Overview of Solving Equations with the Distributive Property and Combining Like Terms


Homework

Solve the following equations for the unknown variable.

1. 4x3 = 92. 15 =5x+43. 5x (3x+2) = 14. 4(x+3) = 1

30


5. 45 = 94x6. 6x+123x = 77. 12+ 23 x = 88. 7(x3)4(x+1) = 159. 56 x 23 = 34

10. 3 47 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 3x4 = 2x Divide by 2

42=

2x2

Solution

x = 2

Example 2

Solve 9x = 45xThis time we will collect like terms (x terms) on the left of the equal sign.

9x = 45x+5x +5x Add 5x to both sides.

14x = 4

14x = 4 Divide by 14.14x14

=414

x =27

Solution

32


x = 27If there are like terms to combine, do that before you collect the variables on one side of the equal sign

Example 3

3x+56x = 13x+4 Combine like terms3x+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 Property8x12+7 = 4x Combine like terms8x5 = 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 isnt possible to get a numberfor x. This example will show how this can happen.

4x+5(x+1) = 3x3(2x+7) Use the Distributive Property4x+5x+5 = 3x+6x21 Combine like terms9x+5 = 9x21 Subtract 9x from each side9x =9x

5 =21

No matter what number is substituted in for x, the final result will be two numbers that dont 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 Property10x+68x = 9x+67x Combine like terms2x+6 = 2x+6 Subtract 2x from each side

2x =2x6 = 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


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 = 10x133. 3(x1) = 2(x+3)4. 8 (x+6) = 3x+65. 7(x+20) = x+56. 9(x2) = 3x+37. 5x (x3) = 2x+79x8. 35 x 110 x = 12 x+19. x5 =(5 x)

10. 9x+3x10 = 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 lets think about the important information. The lumber is divided into three pieces. We dont 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+m10 = 70Now 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


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

36


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

=LWW

The Ws 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

Lets look at a more complicated formula. The area of a triangle is given by

A = 12 bh

38


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.

21A = 1

2bh 2

1Multiply both sides by

21

to clear the fraction

2A = bh

2Ah=

bhh

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

Heres 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 sideP2L = 2W

P2L2

=2W2

Now divide each side by 2

P2L2

=W or W =P2L

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

39

1.10. Solving Equations for a Variable www.ck12.org

W =P2L

2

W =242 5

2

W =2410

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 = 2pir, solve for r

b. If C=100, what is r? (pi is a constant that is always equal to 3.14)

6. a. y = 2x5, solve for xb. If y =23, what is x?

40


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 =

43

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 =1412 inches

30 inches =16830 =

285

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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.

42


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

43


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 5 164= 4 5 20

55 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 =

9x

Cross multiply:

x 4 = 9 34x4=

274


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

44


Example 7

x+23

=x1

4Cross Multiply

4(x+2) = 3(x1)4x+8 = 3x33x 3xx+8 =38 =8

x =11

Video Overview of Ratios


Video Overview of Rates


Video Overview of Proportions and Cross Multiplication


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. 23 ,

32 miles1.4 gallons , and

x13 are all ratios.

45


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 =5x

b. 1.257 =3.6x

c. 619 =x

11d. 1x =

0.015

e. 3004 =x

99f.

3x9=

x57

g. 42x+4 =95

h. 4x105 =3x+8

3

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?

46


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:

47

1.12. Percent Problems www.ck12.org

0.2 = 0.2100(

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 its word name. For example, 0.35 is 35 hundredths. This means itcold be written as 35100 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 cansolve 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.

48


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 74 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:

49

1.12. Percent Problems www.ck12.org

25100

=x

80Cross multiply

100x = 25 80100x = 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 90x = 27

Video Overview of Percents, Decimals and Fractions


Video Overview of the Percent Equation

50



Lesson Summary

A percent is simply a ratio with a base unit of 100, i.e. 13% = 13100 . The percent equation is: percent100 =

iso 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. 16b. 524c. 67d. 117e. 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 : 6x5 < 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.

6x5+5 < 7+5

Simplify.

6x < 12

4. Isolate the variable using the Multiplication or Division Property.

6x6

4 on the coordinate plane.

Solution

First, lets 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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2.10. Graphing Linear Inequalities www.ck12.org

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 yintercept, using yintercept 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 2x3.Solution

Step 1

The inequality is already written in slope-intercept form y 2x3.Step 2

Graph the equation y = 2x3 using slope intercept form.Step 3

Graph the inequality. We shade the plane above the line because y is greater than. The value 2x3 defines the line.The line is solid because the equal sign is included.

Example 6

Graph the inequality 5x2y > 4.

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www.ck12.org Chapter 2. Unit 2 - Linear Functions

Solution

Step 1

Rewrite the inequality in slope-intercept form.

2y >5x+4y >

52

x2

Notice that the inequality sign changed direction due to division of negative sign.

Step 2

Graph the equation y > 52 x2 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.

115


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

Lets 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 y1.29x+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.

116


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 Lets 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

117



Homework

Graph the following inequalities on the coordinate plane.

1. x < 202. y53. y 4x+34. y > x2 65. 3x4y 126. x+7y < 57. 6x+5y > 18. y+54x+109. x 12 y 5

10. 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?

118


2.11 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|= 21 |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

119

2.11. Graphs of Absolute Value Equations www.ck12.org

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 = |x1|.Solution:

Make a table of values and plot the ordered pairs. In order to efficiently make the table, it is best to put the vertex inthe center of the table. The vertex will always be where the value inside the absolute value is zero, we can find thatby setting x1 equal to zero. We find that x = 1, so we put that number in the center of the table

TABLE 2.7:

x y = |x1|

1 |11|= 20 |01|= 11 |11|= 02 |21|= 13 |31|= 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. Itis also possible to graph an absolute value function by considering the translation of the vertex and graphing theequation using the parent function y = |x|.

Example C

Graph y = |x3| by determining the translation of the vertex.

120


Solution: The vertex in this case is translated three units to the right. Starting from the point (3,0), draw the graphof the graph y = |x|

Example D

Graph y = |x|5.Solution: Be careful! Here, the minus 5 is not inside the absolute value. If there is a constant value outside of theabsolute value sign, that tells you about translations that are up and down instead of left and right. In this case, thevertex will move down 5.

Example E

Graph y =3|x+2|1.Solution:

Start by finding the horizontal translation by looking at the situation inside the absolute value.

0 = |x+2| x =2This means that the vertex moves two units to the left.

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2.11. Graphs of Absolute Value Equations www.ck12.org

The constant value outside the absolute value is -1, this means the vertex shifts down 1.

There is a -3 in front of the absolute value. The negative sign flips the graph upside down so the vertex is on thetop. The slope of the line changes from 11 to

31 due to the "3"

Video Overview


Homework

In 1 11, graph the function.

1. y = |x+3|2. y = |x6|3. y = |4x+2|4. y = |x4|35. |x4|= y6. |x2|= y7. y = |x|28. y = 2|x|+39. y =4|x+2|+3

10. y =|x+1|+5

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2.12 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.

123

2.12. Fitting a Line to Data www.ck12.org

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 dont 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 cant 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).

124


Start with the slope-intercept form of a line y = mx+b.

Find the slope m = 114.531 =6.52 = 3.25

Then y = 3.25x+b

Plug (3, 11) into the equation. 11 = 3.25(3)+b b = 1.25The 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 cant 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 cal

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