MAC1105-College Algebra

Chapter 2- Functions,Equations, and Inequalities

Solving Linear Equations

I. Solving Linear Equations

A.The numbers that make an equation a true statement are called thesolutions of the equation.

B.The solution set of an equation is the set of real numbers that make theequation true.

Ex. Given equation: 2 6 10x + = solution: 2 solution set: {2}

C. Two or more equations with the same solution set are calledequivalent equations.

Ex. 2 3 9 2 6x x+ = =, Both of these equations will have a solution of

x = 3 and therefore are equivalent equations.

D. Steps in solving linear equations:

1. Clear fractions.

2. Simplify each side of equation separately.

3. Isolate the variable term on one side.

4. Solve for the variable.

5. Check the solution/answer.

*Ultimately, you want to combine variable terms on one side and constant(s) onthe other side of the equation.

Ex. Solve the equation: 1. 2 6 12x + =

2. 3 7 5 1x x- = +

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3. 4 3 1 2 1 4 3 5( . ) . ( ) .x x x- = - +

4. 2 8 3 2 3 5 12x x+ = - - -[ ( ) ]

5. 52

39- = -x

6. x x- = +23

44( )

7. 5 5

45

x +=

8. 1

24

1

3( )x x+ =

9. 1

3

1

210x x+ =

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II. Identify Conditional Equations, Contradictions, and Identities

Type of linear equation # of Solution(s)

Conditional equation One

Contradiction None (solution set :Æ)

Identity Infinite number (solution set:R )

Ex. 1. 2 1 5 1 3x x x+ = + -

2. 2 3 1 6 3( )x x+ = +

Problem Solving and Using Formulas

I. Formulas: Simple interest i prt= Compound interest A pr

nnt= +( )1

Ex. 1. Johnny invests $1000 in a Nations Bank savings account thatearns 4% interest compounded quarterly. How much will he have in his account atthe end of 1 year? How much interest has he gained in 1 year?

2. Joe invests $1000 in a simple interest account that earns 4% annual interest rate. How much will he have in his account at the end of 1 year? Howmuch interest has he gained in 1 year?

3. Solve the equation 3 5 10x y- = for y

4. Given perimeter, P l w= +2 2 , solve this formula for w or for l

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5. Solve 1

22 6x y+ = for y

6. Solve 2 3 4 5( ) ( )x y x y+ = - +

7. Solve y mx b= + for m

8. Solve V lwh=1

3, for l

9. Solve C F= -5

932( ) for F

10. Solve zx

=- m

s for m

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Applications of Algebra

I. Translate a verbal statement into an algebraic expression or equation

A.

Phrase Algebraic Expression

A number increased by 6 x + 6

Twice a number 2x

5 less than a number x - 5

One-sixth of a number x

6 or

1

6x

2 more than 3 times a number 3 2x +

4 less than 6 times a number 6 4x -

3 times the sum of a number and 5 3 5( )x +

A number and the number increased by 8% x x+ 0 08.

A number and the number decreased by 8% x x- 0 08.

B. Examples:

1. An 8% commission on sales of x dollars:

2. The cost of purchasing xshirts at $5 each:

3. The number of cents in n nickels:

4. The distance traveled in t hours at 60 miles per hour:

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II. Use the Problem-Solving Procedure

A. Procedures

1. Read through and understand the problem (what’s being asked to find).

2. Translate the problem into mathematical language,express in equation.

a. Choose a variable to represent one quantity .

b. Write an equation that represents the word problem.

3. Carry out the mathematical calculations (solve the equation).

4. Answer the question asked.

B. Examples

1. John Doe took his family to visit the Magic Kingdom at Disney World. They stayed for three night at the Holiday Inn in Kissimmee. When they madetheir hotel reservation, they were quoted a rate of $170 per night. Their total billwas $589.50 , which included room tax and a $3 charge for a candy bar (from thein-room bar). Determine the room tax.

2. Jenny Kasten is purchasing her first home and she is considering twobanks for a $60,000 mortgage. Citicorp is charging 8.50% interest with no pointsfor a 30-year loan. (A point is a one-time charge of 1% of the amount of themortgage.) The monthly mortgage payments for the Citicorp mortgage would be$461.40. Citicorp is also charging a $200 application fee. Bank AmericaCorporation is charging 8.00% interest with 2 points for a 30-year loan. Themonthly mortgage payments for Bank America would be $440.04 and the cost ofthe points that Liz would need to pay at the time of closing is 0.02($60,000)=$1200. Bank America has waived its application fee.

a. How long would it take for the total payments of the Citicorpmortgage to equal the total payments of the Bank America mortgage?

b. If Jenny plans to keep her house for 20 years, which mortgagewould result in the lower total cost?

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Quadratic Equations and Functions

I. Quadratic Equations

Quadratic equation is an equation equivalent to ax bx c2 0+ + = , a ¹ 0

such that the coefficients a, b, c are real numbers.

II. Quadratic Functions

Quadratic function f is a second-degree polynomial function,

f x ax bx c( ) = + +2 , a ¹ 0, such that a, b, and c are real numbers.

III. Solving Quadratic Functions

A. Solving by factoring:

Example:

1. x x2 5 6 0- - =

( )( )x x- + =6 1 0

so x - =6 0 or x + =1 0

so x = 6 or x = -1

2. x 2 9 0- =

3. x x2 6 8 0- + =

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B. Solving by Completing the Square

Example: Solve for the unknown

1. x x2 6 10 0- - =

x x2 6 10- =

x x2 2 266

210

6

2- + = +( ) ( )

x x2 6 9 10 9- + = +

x x2 6 9 19- + =

( )( )x x- - =6

2

6

219

( )x - =3 192

( )x - = ±3 19

x = ± +19 3 Or 3 19±

2. x x2 6 8 0- + =

3. x 2 9 0- =

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To solve a quadratic equation by completing the square:

1. Isolate the terms with variables on one side of the equation and

arrange them in descending order.

2. Divide by the coefficient of the squared term if that coefficient

is not 1.

3. Complete the square by taking half the coefficient of the first-

degree term and adding its square on both sides of the equation.

4. Express one side of the equation as the square of a binomial

5. Use the principle of square roots.

6. Solve for the variable.

Example: Solve for the Unknown

1. 2 1 32x x- =

2. x x2 4 5 0- - =

3. 2 4 1 02x x- - =

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C. Solve by Using the Quadratic Formula

Consider any quadratic equation in standard form:

ax bx c2 0+ + = , a ¹ 0

xb b ac

a=

- ± -2 4

2

*Discriminant is b ac2 4-

if b ac2 4 0- = --> one real number solution

if b ac2 4 0- > - - > two real-number solutions

if b ac2 4 0- < - - >two imaginary-number

solutions,complex conjugates

Example:

Solve: x x2 5 8 0+ + =

x =- ± -

=- ± -5 5 4 1 8

2 1

5 7

2

2 ( )( )

( )

xi

=- ±5 7

2

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Example: Solve by quadratic formula

1. 3 5 2 02x x+ - =

2. 2 3 2 02y y- - =

3. 5 3 22m m+ =

4. 3 4 52x x+ =

More solving problems: 1. 1

4

1

5

1+ =

t 2. x

x+ =

65 3.

x x+-

-=

2

4

1

515

4. 3

2

2 4 4

42m m

m

m++ =

-

- 5. 3 4 1x - = 6. x - + =4 1 5

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7. 6 7 2x x+ = + 8. x x- + + =3 2 5 9. 2 3 7 2m m- = + -

10. t1

4 3= 11. t1

3 2= - 12. t1

5 2= 13. m1

2 7= -

14. x = 7 15. x = -45. 16. x - =1 4 17. 2 1 5 3x - - = -

18. 12 6 5- + =x 19. 3 1 4 1x + - = - 20. x - + =4 3 9

Graphing Quadratic Functions of the Type f x a x h k( ) ( )= - +2 :

a. recalling horizontal and vertical translation and vertical stretching or shrinking

b. vertex of a parabola: of f x ax bx cb

af

b

a( ) ( , ( ))= + + ®®®

- -2

2 2

Given: f x x x( ) = - + -2 14 47 and ask:

1. find the vertex

2. is there a maximum or minimum value, of so, where is it?

3. find the range

4. at what interval is the function increasing or decreasing?

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c. Graphing quadratic functions of the type: f x ax bx c a( ) ,= + + ¹2 0

Transform the equation into the standard form: f x a x h k( ) ( )= - +2 , vertex: (h k, )

Example: f x x x( ) = + +2 10 23

x x

x x

x x

x

2

2

2

10 25 25 23

10 25 25 23

10 25 2

5

+ + - +

+ + - +

+ + -

+

( )

( )

( )2 2-

\ = - - + -

\ = - -

f x x

vertex

( ) ( )

( , )

5 2

5 2

2

since a > 0, parabola upward, axis of symmetry@x = -5, minimum function value:

Try: I. Given: f x x x( ) = - + -2 1023

22

Determine:a. find the vertex b. determine whether there is a max or min value

c. find the range d. on what interval is the function increasing/decreasing?

II. Given: f xx

x( ) = - +2

24 8

Determine:a. find the vertex b. determine whether there is a max or min value

c. find the range d. on what interval is the function increasing/decreasing?

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Applications of Quadratic Equations

I. Guidelines for Applying Mathematics

1. Summarize the situation. Pinpoint the unknown ( often represented by x).

2. Draw a sketch (if possible)

3. Decide on a strategy, and find an equation in x.

4. Solve the equation for x.

5. Interpret the result.

II. Examples

1. An object is propelled vertically from the ground level with velocity v. We have already seen that the height s at time t is given by the equation s t vt= - +16 2 . If the initial velocity is 256 ft per second, when will the object be at a height of 768 ft?

2. Find two positive integers that have sum of 10 and product of 21.

3. A particular school is planning on extending the building. If the currentbuilding is rectangular and measures 325 feet by 210 feet. The planned extensionwill double this area and is to be carried out uniformly on the north and est sides. How much should the building be extended?

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The Complex Number System

A. The complex number is any number that can be written as a bi+ ,

such that a and b are real numbers. (Note that a and b both can be 0).

Real number a is said to be the real part of a bi+ , and real number b

is said to be the imaginary part.

Example: 3 2+ i, 5 6+ i, 6, 6 8+ i, 4i, 0

B. An imaginary number is a number that can be written as a bi+ ,

such that a and b are real numbers and b ¹ 0.

Example: 7 3+ i, 2

34- i, 16i

C. A pure imaginary number is an imaginary number

that has no real part.

Example: 3i, -5i, 8i

D. i = -1 and i 2 1= -

Example: express each number in terms of i

1. -3 = - = - =( )1 3 1 3 3i

2. - = - = - =16 1 16 1 16 4( ) i

3. - -14 =

4. - - =64

5. - =48

6. - =49

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Example: Simplify these expressions without using the i

1. i7 =

2. i3 =

3. i- =5

4. (- =i)6

5. ( )2 4i =

E. Addition and Subtraction of Complex Numbers

Example: Add or Subtract

1. ( ) ( )4 5 3 4 7 9+ + + = +i i i

2. ( ) ( )4 5 3 4 1 1 1+ - + = + = +i i i i

3. ( ) ( )8 6 4 3+ + + =i i

4. ( ) ( )6 9 4 7+ - + =i i

F. Multiplication

Example: Multiply and Simplify

1. - × - = - × × - ×2 6 1 2 1 6

= ×i i2 6 = i2 12

= - 12

2. - × - =5 7

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3. - - =4 3 5i i( )

4. ( )( )2 3 4 5+ + =i i

5. ( )3 6 2- =i

6. ( )( )3 6 3 6- + =i i

G. Conjugate of a Complex Number

The conjugate of a complex number a bi+ is a bi- , and the

conjugate of a bi- is a bi+ .

Example: Find the conjugate of each expression

1. 3 7+ i

2. 3-7i

3. - +4 8i

4. - -4 8i

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Example: Multiply

1. ( )( )5 8 5 8+ - =i i

2. ( )( )8 8i i- =

3. ( )( )3 4 3 4+ - =i i

Example: Simplify

1. ( )( )3 16 2 9+ - - - =

2. 3

5 10-=

i

Solving Linear Inequalities

I. Properties used to solve inequalities

A. If a b> , then a c b c+ > + .

B . If a b> , then a c b c- > - .

C If a b> . and c > 0, then ac bc> .

D. If a b> , and c > 0, then a

c

b

c> .

E. If a b> , and c < 0, then ac bc< .

F. If a b> , and c < 0, then a

c

b

c< .

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*When both sides of the inequalities are multiplied or divided by anegative number, the direction of the inequality symbol reverses.

G. Examples

1. Solve: 4 3 2 9x x+ £ - +

2. Solve: 4 2 4 8( )x x- £ -

3. Solve: y

3

2

54+ £

II. Graphing solutions on a number line, interval notation, and solution sets

A. Examples:

1. x a> 2. x a³

3. x a< 4. x a£

5. a x b< < 6. a x b£ £

7. a x b< £ 8. a x b£ <

9. x ³ 5 10. x < 3

11. 2 6< £x 12. - £ £ -6 1x

13. Solve: 3 2 5 8( )x x- £ +

and give solution both on a number line and in interval notation.

14. Solve: 1

24 14 5 6 3( )x x x+ ³ - -

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*Generally, when writing a solution to an inequality we write the variable on theleft.

Ex. - <6 x®®® > -x 6 - > ®®® < -4 4x x a x x a< ®®® > a x x a> ®®® <

*Try: A telephone operator informs a customer in a phone booth that the charge for calling Denver, Colorado, is $4.25 for the first 3 minutes and 45 cents for eachadditional minute. Any additional part of a minute will be rounded up to thenearest minute. Find the maximum time the customer can talk if he has only$9.50.

*Try: To be eligible to continue her financial assistance for college, NikitaMaxwell can earn no more than $2000 during her 8-week summer employment. She already earns $90 per week as a day-care assistant. She is considering addingan evening job at a fast-food restaurant, where she will earn $6.25 per hour. Whatis the maximum number of hours she can work at the restaurant withoutjeopardizing her financial assistance?

* Try:A small single-engine airplane can carry a maximum weight of 1500 pounds. Millie Johnson, the pilot, has to transport boxes weighing 80.4 pounds.

A. Write an inequality that can be used to determine the maximum number ofboxes that Millie can safely place on her plane if she weighs 125 pounds.

B. Find the maximum number of boxes that Millie can transport.

*Try: Pamela Person recently accepted a sales position in Ohio. She can selectbetween two payment plans. Plan 1 is a salary of $300 per week plus a 10%

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commission on sales. Plan 2 is a salary of $400 per week plus an 8% commissionon sales. For what amount of weekly sales would Pamela earn more by plan 1?

III. Solve compound inequalities involving “And”

A compound inequality is formed by joining two inequalities with theword and or or.

Examples: 3 < x and x < 5

x + >4 3 or 2 3 6x - <

x + £2 5 and 2 4 2x - > -

IV. Solved continued inequalities

Inequality in the form a x b< < is called continued inequality.

Examples: Solve 1 5 7< + £x

Solve - <-

<24 3

58

x

Solve - < - £ -12 3 5 4x

Solve 1

23 4 6< + <x

Solve 3

5

5

36<

- -<

x

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V. Solve compound inequalities involving “or”

Example: Try

1. x > 3 or x < 5

2. x + £ -3 1 or - + < -4 3 5x

3. - + <x 3 0 or 2 5 3x - ³

4. 4 2- < -x or 3 1 1x - < -

Inequalities with Absolute Value

I. X = 4 is equivalent to X = 4 or X = -4

For a > 0, and an algebraic expression X :

II. X a> is equivalent to X a> or X a< -

III.. X a< is equivalent to X a< and X a> -

Similar statements hold for X a£ and X a³ .

IV. ax b cx d+ = - is equivalent to ax b cx d+ = - or ax b cx d+ = - -( )

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Example: 1. x < 5 is equivalent to -5 < <x 5

2. x ³ 2 is equivalent to x £ -2 or x ³ 2

3. 2 5 4x + £ is equivalent to - £ + £4 2 5 4x

4. 6 2 1- ³x

5. 3 2 5x + <

6. x + >6 10

7. 6 4 8- £x

Solve for x: 1. 1 2 2- >x

2. 4 1 10 5x + - < -

3. 7 2 7 2x + - = -

4. x £ 14

5. 5 2

13

x

x

+

-=

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6. 4 2 13- < -x

7. 3 2 2y - ³ -

8. 3 8 4 6a a+ = -

9. x + <3

4

1

4

10. 2 1

35

x +>

11. 5 11 4x + £ - or 5 11 4x + ³

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