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Page 1: Algebraic Expressions, Linear - Kirkwood … Algebraic Expressions, Linear Equations, and Applications 5.1 Simplifying and Evaluating Algebraic Expressions 5.2 Translating English

Every day you are required to make decisions. Some decisions, like what

you eat for breakfast or what you wear, will likely affect your life for only a day. However, other decisions, like how much to spend on your first

house and which job offer to take, can affect the rest of your life. The skills you learn while studying algebra give you reasoning and problem-solving skills that can increase your confidence as you make some of life’s big decisions, especially decisions relating to money. Consider the following situation.

Suppose that you are a traveling salesperson and your company has allowed you $65 per day for a car rental. You rent a car for $45 per day plus $0.05 per mile. How many miles can you drive for the $65 allowed?

Plan: Set up an equation relating the amount of money spent and the money allowed in your budget and solve the equation.

Solution: Let x = the number of miles driven.

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5Algebraic Expressions, Linear Equations, and Applications5.1 Simplifying and Evaluating Algebraic Expressions

5.2 Translating English Phrases and Algebraic Expressions

5.3 Solving Linear Equations: x + b = c and ax = c

5.4 Solving Linear Equations: ax + b = c

5.5 Solving Linear Equations: ax + b = cx + d

5.6 Applications: Number Problems and Consecutive Integers

5.7 Applications: Percent Problems

5.8 Working with Formulas

5.9 Applications: Distance-Rate-Time, Interest, Mean, Cost

Chapter 5: Index of Key Terms and Ideas

Chapter 5: Test

Cumulative Review: Chapters 1 - 5

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565 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

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5.1 Simplifying and Evaluating Algebraic Expressions

Objective A Simplifying Algebraic Expressions

A single number is called a constant. Any constant or variable or the indicated product and/or quotient of constants and variables is called a term. Examples of terms are

16 3 5 2 1 3 5 142 3, , . , . , , , .x xy x bxy

− − −and

Note: As discussed in Section 1.5, an expression with 2 as the exponent is read “squared” and an expression with 3 as the exponent is read “cubed.” Thus 7 2x is read “seven x squared” and −4 3y is read “negative four y cubed.”

The number written next to a variable is called the coefficient (or the numerical coefficient) of the variable. For example, in

5 2x , 5 is the coefficient of x 2.

Similarly, in 1.3xy, 1.3 is the coefficient of xy.

notesIf no number is written next to a variable, the coefficient is understood to be 1. If a negative sign −( ) is next to a variable, the coefficient is understood to be −1. For example,

x x= ⋅1 , a a3 31= ⋅ , − = ⋅−x x1 , and − = ⋅−y y5 51 .

Like terms (or similar terms) are terms that are constants or terms that contain the same variables raised to the same exponents.

Note: The sum of the exponents on the variables is the degree of the term.

Like Terms

Like Terms

−6 1 84 14534

, . , , are like terms because each term is a constant.

−3 15 2 623

a a a a, , . , are like terms because each term contains the same variable a, raised to the same exponent, 1. (Remember that a a= 1.) These terms are first-degree in a.

Simplify and evaluate algebraic expressions by combining like terms.

Evaluate expressions for given values of the variables.

A

B

Objectives

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Simplifying and Evaluating Algebraic Expressions Section 5.1 566

5 3 22 2xy xy and − . are like terms because each term contains the same two variables, x and y, with x first-degree in both terms and y second-degree in both terms.

Unlike Terms 8 9 2x x and − are unlike terms (not like terms) because

the variable x is not of the same power in both terms. 8x is first-degree in x and −9 2x is second-degree in x.

Example 1

Like Terms

From the following list of terms, pick out the like terms.

− − −7 2 4 1 3 5 62 2, , . , , , , ,x x x y x x y 0

Solution

−7, 4.1, and 0 are like terms. (All are constants.)2x, −x, and 5x are like terms.3 2x y and −6 2x y are like terms.

1. From the following list of terms, pick out the like terms: 4x, −3.9, 3xy2, 5, 0, −2x, −6xy2, x.

−3.9, 5, 0; 4x, −2x, x; 3xy2, −6xy2

Now work margin exercise 1.

An algebraic expression is a combination of variables and numbers using any of the operations of addition, subtraction, multiplication, or division, as well as exponents. Examples of algebraic expressions are

xxy

zy

xx

x x x23

2 3 21423

12 93 4

10 4 11 1− +−

−− − +, , , .and

To simplify expressions that contain like terms we want to combine like terms.

To combine like terms, add (or subtract) the coefficients and keep the common variable expression.

Combining Like Terms

The procedure for combining like terms uses the distributive property in the form b c b ca a a+ = +( ) .

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In particular, with numerical coefficients, we can combine like terms as follows.

9 6 9 6

15

x x x

x

+ = +( )=

3 5 3 5

2

2 2 2

2

x x x

x

− = −( )= −

xy xy xy

xy

− = −( )

= −

1 6 1 0 1 6

0 6

. . .

.

Example 2

Combining Like Terms

a. 8x + 10x

Solution

8 10 8 10 18x x x x+ = +( ) =

b. 6.5y − 2.3y

Solution

6 5 2 3 6 5 2 3 4 2. . . . .y y y y− = −( ) =

c. 2 32 2x a x a+ + −

Solution

2 3 2 3

2 1 3 1

3 2

2 2 2 2

2

2

x a x a x x a a

x a

x a

+ + − = + + −

= +( ) + −( )= +

d. 4 7 5 1n n−( ) + +( )

Solution

4 7 5 1 4 28 5 5

4 5 28 5

4 5 28 5

9 23

n n n n

n n

n

n

−( ) + +( ) = − + +

= + − +

= +( ) + − +( )= −

by the distributive property

by the distributive property

by the commutative property of addition

Note: + = +12 2

x x and−a = −1a

Simplify by using the distributive property twice.

by the commutative property of addition

by the distributive property

Combine like terms.

by the distributive property

Add the coefficients algebraically.

by the distributive property

Subtract the coefficients algebraically.

by the distributive property Note: xy = 1xy = 1.0xy

Subtract the coefficients algebraically.

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Simplifying and Evaluating Algebraic Expressions Section 5.1 568

e. x x

x+

+3

25

Solution A fraction bar is a grouping symbol, similar to parentheses. So combine like terms in the numerator first.

x xx

xx

x x

x x

x

++ = +

= ⋅ +

= +=

32

542

5

42

5

2 5

7

2. Combine like terms whenever possible.

a. 3y + 8y

b. 4.6x − 2.5x

c. 4x2 + 6z − 3x2 + 4z

d. 2(m + 4) + 9(m − 2)

e. 6 24

4x x

x+

+

a. 11y b. 2.1x c. x2 + 10z d. 11m − 10 e. 6x

Reduce the fraction.

Combine like terms.

Now work margin exercise 2.

Objective B Evaluating Algebraic Expressions

In most cases, if an expression is to be evaluated, like terms should be combined first and then the resulting expression evaluated by following the rules for order of operations.

Parentheses must be used around negative numbers when substituting.

Without parentheses, an evaluation can be dramatically changed and lead to wrong answers, particularly when even exponents are involved. We analyze with the exponent 2 as follows.

In general, except for x = 0,

1. −x 2 is negative − − ⋅ − 6 1 6 362 2= ⋅ = −1 36 =

2. −( )x2

is positive −( ) −( ) −( )

6 6 6 36

2 = =

3. − ≠ −( )x x2 2 − ≠[ ]36 36

1. Combine like terms, if possible.2. Substitute the values given for any variables.3. Follow the rules for order of operations.

To Evaluate an Algebraic Expression

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Example 3

Evaluating Algebraic Expressions

a. Evaluate x 2 for x = 3 and for x = −4.

Solution

For

For

x x

x x

= = ( ) =

= = ( ) =− −

3 3

4 4

9

16

2 2

2 2

, .

, .

b. Evaluate −x 2 for x = 3 and for x = −4.

Solution

For

For

x x

x x

= − = −( ) = − ( ) = −= − = −( ) = − ( ) = −− −

3 3

4 4

1 9 9

1 16 16

2 2

2 2

, .

, .

3. a. Evaluate x2 for x = 2 and for x = −3.

b. Evaluate –x2 for x = 2 and for x = −3.

a. 4; 9 b. −4; −9

Now work margin exercise 3.

Example 4

Simplifying and Evaluating Algebraic Expressions

Simplify each expression below by combining like terms. Then evaluate the resulting expression using the given values for the variables.

a. Simplify and evaluate 2x + 5 + 7x for x = −3.

Solution

Simplify first. 2x + 5 + 7x = 2x + 7x + 5 = 9x + 5

Now evaluate. 9x + 5 = 9 53−( ) +

= −27 + 5

= −22

b. Simplify and evaluate 3ab − 4ab + 6a − a for a = 2, b = −1.

Solution

Simplify first. 3ab − 4ab + 6a − a = −ab + 5a

Now evaluate. −ab + 5a = − ( )( ) + ( )−1 52 21 Note: −ab = −1ab

= 2 + 10 = 12

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Simplifying and Evaluating Algebraic Expressions Section 5.1 570

c. Simplify and evaluate 5 3

42 1 5

x xx x

++ +( ) for = .

Solution

Simplify first.

5 34

2 184

2 2

2 2 2

4 2

x xx

xx

x x

x

++ +( ) = + +

= + += +

Now evaluate.

4 2 4 2

20 2

22

5x + = ( ) += +=

4. a. Simplify and evaluate −3x + 5x + 4 for x = −5.

b. Simplify and evaluate 7ab − 3ab + 4a − a for a = 3 and for b = −2.

c. Simplify and evaluate 4 2

23 3

x xx

++ +( )

for x = 3.

a. −6 b. −15 c. 27

Now work margin exercise 4.

Simplify the following expressions by combining like terms.

1. −2x − 5x 2. 12y + 6 − y + 10

3. 5 1 4x x−( ) + 4. 2 2 2b a b a− + +

Simplify the expression. Then evaluate the resulting expression for x = 3 and y = −2.

5. 2 3 4x y x y+( ) + −( )

Practice Problems

Practice Problem Answers 1. −7x 2. 11y + 16 3. 9x − 5 4. 3 2b 5. 6x + 2y; 14

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571 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

Exercises 5.1

Pick out the like terms in each list of terms. See Example1.

1. −516

7 8 9 3, , , , ,x x y 2. − −2 13 5 14 102 3 2 2 3x x x x x, , , ,

−5, 1

6, and 8 are like terms;

7x and 9x are like terms. −2 52 2x x, , and 14 2x are like terms;

−13 3x and 10 3x are like terms.

3. 5 6 3 5 22 2 2 2xy x xy x y x y x, , , , ,− − 4. 3 8 9 10 122 2 2 2 2ab ab ab a b a b ab a, , , , , ,− −

5xy and −6xy are like terms; −x 2 and 2 2x are like terms; 3 2x y and 5 2x y are like terms.

3 2ab and −ab2 are like terms; 8ab and ab are like terms; 9 2a b and −10 2a b are like terms.

5. 24 8 3 1 5 1 4 6 5 22 2, . , . , . , , , ,xyz xyz xyz xy z xyz− − 6. − − −35 1 62 312

75 2 52 2 2y y y y y y, . , , , , , , .

24, 8.3, and −6 are like terms; 1.5xyz, −1.4xyz, and xyz are like terms.

−35y, −y, and 75y are like terms;

1.62 and 12

are like terms;

−y2 , 3 2y , and 2 5 2. y are like terms.

Find the value of each numerical expression. See Example 3.

7. −( )82

64 8. −82 −64 9. −112 −121 10. −( )62

36

Simplify each expression by combining like terms. See Example 2.

11. 8x + 7x 15x 12. 3y + 8y 11y 13. 5 2x x+ −( ) 3x 14. 7 3x x+ −( ) 4x

15. −n − n −2n 16. −x − x −2x 17. 6 2 2y y− 5 2y 18. 16 52 2z z− 11 2z

19. 23 112 2x x+ 34 2x 20. 18 73 3x x+ 25 3x 21. 4x + 2 + 3x 7x + 2 22. 3x − 1 + x 4x − 1

23. 2x − 3y − x − y 24. x + y + x − 2y 25. 2 2 5 62 2 2x y x x− + + 26. 4a + 2a − 3b − a

x − 4y 2x − y 13 22x y− 5a − 3b

27. 3 1n n+( ) − 28. 2 4 1n n−( ) + + 29. 5 2 3a b a b−( ) + −

2n + 3 3n − 7 7a − 8b

30. 4 3 2 2a b a b− + +( ) 31. 3 2 2x y x y+( ) + −( ) 32. 4 5 3 2 7x y x y+( ) + ( )−

6a + b 8x + y 10x − y

33. 2 3 32 2x x x x+ − − 34. 2 32 2y y y y+ −4 − 35. 2 6 1 4 8 32 2n n n n− + − + −

2 2x x− y y2 + − + −2 2 22n n

36. 3 2 5 42 2n n n n+ +− − − 37. 3 4 52 2x xy xy y+ +− 38. 2 5 11 32x xy xy− + +

2 3 92n n+ − 3 2 2x xy y− + 2 6 32x xy+ +

39. x x

x+

+5

6 2x 40.

2 35

2y y

y+

− −y 41. yy y

−+2 43

−y 42. zz z

−+3 54

−z

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Simplifying and Evaluating Algebraic Expressions Section 5.1 572

For the following expressions, a. simplify, and b. evaluate the simplified expression for x = 4, y = 3, a = −2, and b = −1. See Example 4.

43. 5x + 4 − 2x 44. 7x − 17 − x 45. 5 2 5 3x x x− + −( ) a. 3x + 4 b. 16 a. 6x − 17 b. 7 a. −7x + 30 b. 2

46. x − 10 − 3x + 2 47. 6 5 13a a a+ − + 48. 3 1 2 2y y−( )+ +( )

a. −2x − 8 b. −16 a. 10a + 13 b. −7 a. 5y + 1 b. 16

49. 4 3 5 2y y+( )+ −( ) 50. − +( ) + −( )5 2x y x y 51. − +( ) + −( )2 3a b b a

a. 9y + 2 b. 29 a. −3x − 7y b. −33 a. −5a + b b. 9

52. 8 3 5 7 22 2 2. .x x x− + + 53. 3 1 0 9 4 5 32 2 2. . .a a a a− + − 54. 5 22 3ab b ab b+ − +

a. 3 6 22. x + b. 59.6 a. − +3 1 42. a a b. −20.4 a. 3 2 3ab b b+ + b. 6

55. 5 2 32 2a ab ab a+ − + 56. 2 4 1 1 3 1. .x x+( ) + −( ) 57. 1 3 2 2 6 8. .y y+( ) − −( ) a. 8 2a ab− b. −14 a. 3.7x + 1.1 b. 15.9 a. 3.9y − 18.2 b. −6.5

58. 3 5

212

a aa

+−

+ 59. 85 4

9a

a a+

+ 60.

− −−

++4 2

32 5

7b b b b

a. 8a b. −16 a. 9a b. −18 a. 3b b. −3

61. 5 3

44

5b b b b+

+− −

− 62. 2 3 2 9x x x+ − +( )

a. 3b b. −3 a. −x − 54 b. −58

63. Explain the difference between −52 and −( )52

.

64. The text recommends simplifying an expression (combining like terms) before evaluating. Do you think this is necessary?

Evaluate the expression 4 5 2 3 10 22x x x x− +( ) + + + for x = 3:

a. by substituting and then evaluating.

b. by first simplifying and then evaluating.

Which method would you recommend? Why?

Writing & Thinking

63. −52 is the square of 5 multiplied by −1 by the order of operations; while

−( )52

is the square of −5. ( − = −5 252 and

−( ) =5 252 )

64. a. For xx x x x

=− +( ) + + + = ( ) − ( ) +( ) + ( ) + + ( )

= (

34 5 2 3 10 2 4 3 5 3 2 3 3 10 2 3

4 9

2 2

)) − ( ) + + += − + + +=

5 5 9 10 6

36 25 9 10 6

36

b. 4 5 2 3 10 2 4 5 10 3 10 2

4

3 4 4 3

2 2

2

2 2

x x x x x x x x

x

x x

− +( ) + + + = − − + + +

=

= = ( )For

== ( )=

4 9

36

The second method, as it requires fewer calculations.

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5.2 Translating English Phrases and Algebraic Expressions

Objective A Translating English Phrases into Algebraic Expressions

Algebra is a language of mathematicians, and to understand mathematics, you must understand the language. We want to be able to change English phrases into their “algebraic” equivalents and vice versa. So if a problem is stated in English, we can translate the phrases into algebraic symbols and proceed to solve the problem according to the rules and methods developed for algebra.

Certain words are the keys to the basic operations. Some of these words are listed in Table 1 and highlighted in boldface in Example 1.

Key Words to Look For When Translating Phrases

Addition Subtraction Multiplication DivisionExponent(Powers)

add subtract (from) multiply divide square of

sum difference product quotient cube of

plus minus times

more than less than twice

increased by

decreased by of (with fractionsand percent)

total less

Table 1

The following examples illustrate how these key words used in English phrases can be translated into algebraic expressions. Note that in each case “a number” or “the number” implies the use of a variable (an unknown quantity).

Write the meaning of word phrases in algebraic expressions.

Write the meaning of algebraic expressions in words.

A

B

Objectives

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Translating English Phrases and Algebraic Expressions Section 5.2 574

Example 1

Translating English Phrases into Algebraic Expressions

English Phrase Algebraic Expressiona.

the of 3 and

3

the number r

product

times

multiplied by

x

x

3 eepresented by x

x

3

b.

3 a number

the 3

3

3

of and

added to

sum

plus

more than

z

z

z

zz

z

increased by 3

+ 3

c.

twice sum

product sum

the of and 1

the of 2 with the of

x

x and 1

2 the quantity found by a

number to 1

times adding

+( )2 1x

d.

twice plus

sum twice

times increased

1

the of and 1

2

x

x

x bby 1

1 the of 2 and a numbermore than product

+2 1x

e.

the between 5 a number and 3

3 the of

difference times

less than product aa number and 5

5 a number 3

3 the product of

times minus

subtracted from aa number and 5

5 a number, 3multiplied by less

−5 3n

f.

the a number

a number

square of

squared

x 2

g. the cost of renting a truck for one 30 + 0.25x day and driving x miles if the rate is $30 per day plus $0.25 per mile

1. Write the algebraic expression for each of the following English phrases.

a. 5 times y

b. the sum of x and 1

c. the product of 3 with the sum of z and 2

d. 3 more than the product of 8 and a number

e. 4 subtracted from 6 times x

f. the cube of a number

g. The cost of playing golf at a country club for a year if the membership dues are $5000 a year plus $10 per round, if r rounds of golf are played in a year.

a. 5y b. x + 1 c. 3(z + 2) d. 3 + 8x e. 6x - 4 f. x3

g. 5000 + 10r

Now work margin exercise 1.

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notesIn Example 1b., the phrase “the sum of z and 3” was translated as z + 3. If the expression had been translated as 3 + z, there would have been no mathematical error because addition is commutative. That is, z + 3 = 3 + z. However, in part e., the phrase “3 less than the product of a number and 5” must be translated as it was because subtraction is not commutative. Thus

“3 less than 5 times a number” means 5n − 3while “5 times a number less than 3” means 3 − 5nand “3 less 5 times a number” means 3 − 5n.

Therefore, be very careful when writing and/or interpreting expressions indicating subtraction. Be sure that the subtraction is in the order indicated by the wording in the problem. The same is true with expressions involving division.

The words quotient and difference deserve special mention because their use implies that the numbers given are to be operated on in the order given. That is, division and subtraction are done with the values in the same order that they are given in the problem. For example:

the quotient of y and 5 y5

the quotient of 5 and y 5y

the difference between 6 and x 6 − x

the difference between x and 6 x − 6

If we did not have these agreements concerning subtraction and division, then the phrases just illustrated might have more than one interpretation and be considered ambiguous.

An ambiguous phrase is one whose meaning is not clear or for which there may be two or more interpretations. This is a common occurrence in everyday language, and misunderstandings occur frequently. Imagine the difficulties diplomats have in communicating ideas from one language to another trying to avoid ambiguities. Even the order of subjects, verbs, and adjectives may not be the same from one language to another. Translating grammatical phrases in any language into mathematical expressions is quite similar. To avoid ambiguous phrases in mathematics, we try to be precise in the use of terminology, to be careful with grammatical construction, and to follow the rules for order of operations.

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Translating English Phrases and Algebraic Expressions Section 5.2 576

Objective B Translating Algebraic Expressions into English Phrases

Consider the three expressions to be translated into English:

7 1 6 3 7 1n n n+( ) −( ) +, , .and

In the first two expressions, we indicate the parentheses with a phrase such as “the quantity” or “the sum of” or “the difference between.” Without the parentheses, we agree that the operations used in the expression are to be indicated in the order given. Thus

7 1n +( ) can be translated as “seven times the sum of a number and 1,” while 7 1n + can be translated as “seven times a number plus 1.” 6 3n −( ) can be translated as “six times the difference between a number and 3.”

Example 2

Translating Algebraic Expressions to Phrases

Write an English phrase that indicates the meaning of each algebraic expression.

Algebraic Expression Possible English Phrase a. 5x the product of 5 and a number b. 2n + 8 twice a number increased by 8 c. 3 2a −( ) three times the difference

between a number and 2

Now work margin exercises 2.

Change the following phrases to algebraic expressions.

1. 7 less than a number 2. the quotient of y and 5

3. 14 more than 3 times a number

Change the following algebraic expressions into English phrases. (There may be more than one correct translation.)

4. 10 − x 5. 2 3y −( ) 6. 5n + 3n

Practice Problems

2. Write an English phrase for each of the following algebraic expressions.

a. 3x

b. 7x + 6

c. 2(n − 7)

2. a. the product of 3 and a number

b. 7 times a number increased by 6

c. twice the difference between n and 7

Practice Problem Answers

1. x − 7 2. y5

3. 3y + 14

4. 10 decreased by a number 5. twice the difference between a number and 3 6. 5 times a number plus 3 times the same number

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577 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

Exercises 5.2

Translate each algebraic expression into an equivalent English phrase. (There may be more than one correct translation.) See Example 2.

1. 4x 4 times a number 2. -9x the product of a number and -9

3. x + 5 5 more than a number 4. 4x − 7 7 less than 4 times a number

5. 7 1 1x +( ). 7 times the sum of a number and 1.1 6. 3 2 2 5. .x +( ) 3.2 times the sum of a number and 2.5

7. − −( )2 8x −2 times the difference between a number and 8

8. 10 4x +( ) 10 times the sum of a number and 4

9. 61x −( )

6 divided by the difference between a number and 1

10. 9

3x +( ) 9 divided by the sum of a number and 3

11. 5 2 3x +( ) 5 times the sum of twice a number and 3

12. 3 4 5x −( ) 3 times the difference of 4 times a number and 5

Write each pair of expressions in words. Notice the differences between the algebraic expressions and the corresponding English phrases.

13. 3 7 3 7x x+ +( ); 14. 4 1 4 1x x− −( ); 3 times a number plus 7; 3 times the sum of a

number and 7 1 less than 4 times a number; 4 times the

difference of a number and 1

15. 7 3 7 3x x− −( ); 16. 5 6 5 6x x+( ) +; The product of 7 and a number minus 3; 7

times the difference between a number and 3 5 times the sum of a number and 6; the sum of

5 times a number and 6

Write the algebraic expressions described by the English phrases. Choose your own variable. See Example 1.

17. 6 added to a number x + 6 18. 7 more than a number x + 7

19. 4 less than a number x − 4 20. a number decreased by 13 x − 13

21. the quotient of twice a number and 10 210

x 22. the difference between a number and 3, all

divided by 7 x − 3

7

23. 5 subtracted from three times a number 3x − 5

24. the sum of twice a number and four times the number 2x + 4x

25. 8 minus twice a number 8 − 2x 26. the sum of a number and 9 times the number x + 9x

27. twenty decreased by 4.8 times a number 20 − 4.8x

28. the difference between three times a number and five times the same number 3x − 5x

29. 9 times the sum of a number and 2 9 2x +( ) 30. 3 times the difference between a number and 8 3 8x −( )

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Translating English Phrases and Algebraic Expressions Section 5.2 578

31. 13 less than the product of 4 with the sum of a number and 1 4 1 13x +( ) −

32. 4 more than the product of 8 with the difference between a number and 6 8 6 4x −( ) +

33. eight more than the product of 3 and the sum of a number and 6 3 6 8x +( ) +

34. six less than twice the difference between a number and 7 2 7 6x −( ) −

35. four less than 3 times the difference between 7 and a number 3 7 4−( ) −x

36. nine more than twice the sum of 17 and a number 2 17 9+( ) +x

37. a. 6 less than a number x − 6 b. 6 less a number 6 − x

38. a. 20 less than a number x − 20 b. 20 less a number 20 − x

39. a. 5 less than 3 times a number 3x − 5 b. 5 less 3 times a number 5 − 3x

40. a. 6 less than 4 times a number 4x − 6 b. 6 less 4 times a number 6 − 4x

Write the algebraic expressions described by the English phrases using the given variable. See Example 1.

41. Time: the number of hours in d days 24d 42. Graphing Calculators: the cost of x graphing calculators if one calculator costs $115 115x

43. Gas Prices: the cost of x gallons of gasoline if the cost of one gallon is $3.15 3.15x

44. Time: the number of seconds in m minutes 60m

45. Time: the number of days in y years (Assume 365 days in a year.) 365y

46. Candy: the cost of x pounds of candy at $4.95 a pound 4.95x

47. Time: the number of days in t weeks and 3 days 7t + 3

48. Time: the number of minutes in h hours and 20 minutes 60h + 20

49. Football: the points scored by a football team on t touchdowns (7 points) and 1 field goal (3 points) 7t + 3

50. Vacation Time: the amount of vacation days an employee has after w weeks if she gets 0.2 vacation days for every week she works 0.2w

51. Car Rentals: the cost of renting a car for one day and driving m miles if the rate is $20 per day plus 15 cents per mile 20 + 0.15m

52. Fishing: the cost of purchasing a fishing rod and reel if the rod costs x dollars and the reel costs $8 more than twice the cost of the rod x x x+ +( ) = +2 8 3 8

53. Rectangles: the perimeter of a rectangle if the width is w centimeters and the length is 3 cm less than twice the width 2 2 2 3 6 6w w w+ −( ) = −

54. Squares: the area of a square with side c centimeters c2

55. Discuss the meaning of the term “ambiguous phrase.” A phrase whose meaning is not clear or for which there may be two or more interpretations.

Writing & Thinking

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579 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

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5.3 Solving Linear Equations: x + b = c and ax = c

Objective A Linear Equations

In this section, we will discuss solving linear equations in the following two forms:

x b c ax c+ = = and .

In these equations we treat a, b, and c as constants and x as the unknown quantity.

In the following sections, we will combine the techniques developed here and discuss solving linear equations in the forms:

ax b c ax b cx d+ = + = +and .

An equation is a statement that two algebraic expressions are equal. That is, both expressions represent the same number. If an equation contains a variable, any number that gives a true statement when substituted for the variable is called a solution to the equation. The solutions to an equation form a solution set. The process of finding the solution set is called solving the equation.

If a, b, and c are constants and a ≠ 0 then a linear equation in x is an equation that can be written in the form

ax b c+ = .

Note: A linear equation in x is also called a first-degree equation in x because the variable x can be written with the exponent 1. That is, x x= 1.

Linear Equation in x

Objective B Determining Possible Solutions

We begin by determining whether or not a particular number satisfies an equation. A number is said to be a solution or to satisfy an equation if it gives a true statement when substituted for the variable.

Example 1

Determining a Possible Solution

Determine whether or not the given real number is a solution to the given equation by substituting for the variable and checking to see if the resulting equation is true or false.

Define the term linear equation.

Learn how to determine whether or not a number is a solution to a given equation.

Solve equations of the form x + b = c.

Solve equations of the form ax = c.

A

B

C

D

Objectives

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Solving Linear Equations: x + b = c and ax = c Section 5.3 580

a. x + = −5 2 given that x = −7

Solution

−( ) + = −7 5 2 is true, so −7 is a solution.

b. 1 4 0 5. .+ =z given that z = −1 1.

Solution

1 4 0 51 1. ..+ ( ) =− is false because 1 4 1 1 0 3 0 5. . . . .+ −( ) = ≠

So −1.1 is not a solution.

c. 5 6 2 9. .− =y given that y = 2 7.

Solution

5 6 2 92 7. ..− = is true. So, 2.7 is a solution.

d. z − = −14 3 given that z = −10

Solution

−( ) − = −10 14 3 is false because − − = − = − ≠ −10 14 10 14 4 3.

So, −10 is not a solution.

Note: Notice that parentheses were used around negative numbers in the substitutions. This should be done to keep operations properly separated, particularly when negative numbers are involved.

1. Determine whether or not the given real number is a solution to the given equation by substituting for the variable and checking to see if the resulting equation is true or false.

a. x − 2 = −9 given that x = −7

b. 2.8 + z = 1.3 given that z = −1.7

c. 6.7 − y = 3 given that y = 3.7

d. |z| − 7 = −4 given that x = −12

a. −7 is a solution b. −1.7 is not a solution c. 3.7 is a solution d. −12 is not a solution

Now work margin exercise 1.

Objective C Solving Equations of the Form x + b = c

To begin, we need the addition principle of equality.

If the same algebraic expression is added to both sides of an equation, the new equation has the same solutions as the original equation. Symbolically, if A, B, and C are algebraic expressions, then the equations

A = Band

A + C = B + C

have the same solutions.

Addition Principle of Equality

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Equations with the same solutions are said to be equivalent equations.

The objective of solving linear (or first-degree) equations is to get the variable by itself (with a coefficient of +1) on one side of the equation and any constants on the other side. The following procedure will help in solving linear equations such as

x − =3 7, − = +11 5y , 3 2 2 5 3 2 0 8z z− + = +. . . , and x − =25

310

.

1. Combine any like terms on each side of the equation.2. Use the addition principle of equality and add the opposite of the

constant b to both sides. The objective is to isolate the variable on one side of the equation (either the left side or the right side) with a coefficient of +1.

3. Check your answer by substituting it for the variable in the original equation.

Procedure for Solving Linear Equations that Simplify to the Form x + b = c

Every linear equation has exactly one solution. This means that once a solution has been found, there is no need to search for another solution.Note that, as illustrated in Examples 1 through 3, variables other than x may be used.

Example 2

Solving x + b = c

a. x − =3 7

Solution

x

x

− =− + += ( )

3 7

3 73 3

Write the equation.

Add the opposite of 3 to bot 3 − hh sides.

Simplify.

x = 10

Check:

x − =

− ==

( )3 7

3 7

7 7

10?

Substitute

true statement

10.x =

x − =

− ==

( )3 7

3 7

7 7

10?

Substitute

true statement

10.x =

b. − = +11 5y

Solution

− = +11 5y Write the equation. Note that thevariable can be on the righ

tt side.

Add the opposite of +5 to both sides.

S

5− = +− =− − ( )11 5

16

5 5y

y

iimplify.

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Solving Linear Equations: x + b = c and ax = c Section 5.3 582

Check:

− = +

− =( ) +− = −

11 5

11 5

11 11

16

y?

Substitute .

true statement

16y = −

− = +

− =( ) +− = −

11 5

11 5

11 11

16

y?

Substitute .

true statement

16y = −

c. x − =25

310

Solution

x

x

− =

− + = +

25

310

25

310

25

25

Write the equation.

Add the opposite of 2

5−− 2

5toboth sides.

(The common denominator is 10.

= +x3

104

10))

Simplify.x =7

10

Check:

x − =

− =

− =

25

310

25

310

710

410

310

710

?

?

Substitute = .

The co

7

10x

mmmon denominator is 1

true statement

0.

310

=3

10

Solve each of the following linear equations.

2. a. x − 5 = 12

b. −8 = x + 4

c. x − =38

34

3. 5z − 3z + 1.6 = 4.1 + 0.9

2. a. x = 17 b. x = −12

c. x =98

3. z = 1.7

Example 3

Simplfying and Solving Equations

3 2 2 5 3 2 0 8z z− + = +. . .

Solution

3 2 2 5 3 2 0 8

2 5 4 0

z z

z

− + = ++ =

. . .

. .

Write the equation.

Combine like termss onboth sides of the equation.

Add the opp2.5

z + =− −2 5 4 02 5 2 5. .. . − oosite of +2.5

to both side

Simplify.

s

( )

=.

z 1 5.

Check:

3 2 2 5 3 2 0 8

3 2 2 5 3 2 0 8

4 5 3 0 2 5 3

1 5 1 5

z z− + = +

( ) − ( ) + = +

− + =

. . .

. . .

. . . .

?

?. .

22 0 8

4 0 4 0

+=

.

. .

Substitute z = 1.5.

Simplify.

true statement

Now work margin exercises 2 and 3.

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Objective D Solving Equations of the Form ax = c

To solve equations of the form ax = c, where a ≠ 0, we can use the idea of the reciprocal of the coefficient a. For example, as we studied earlier, the

reciprocal of 34

is 43

and 34

43

1⋅ = . Also, we need the multiplication (or

division) principle of equality as stated below.

If both sides of an equation are multiplied by (or divided by) the same nonzero constant, the new equation has the same solutions as the original equation. Symbolically, if A and B are algebraic expressions and C is any nonzero constant, then the equations

A = B,

AC = BC where C ≠ 0 ,

and AC

= BC

where C ≠ 0

have the same solutions. We say that the equations are equivalent.

Multiplication (or Division) Principle of Equality

Remember that the objective is to get the variable by itself on one side of the equation. That is, we want the variable to have +1 as its coefficient. The following procedure will accomplish this.

1. Combine any like terms on each side of the equation.

2. Use the multiplication (or division) principle of equality and multiply both sides of the equation by the reciprocal of the coefficient of the variable. (Note: This is the same as dividing both sides of the equation by the coefficient.) Thus the coefficient of the variable will become +1.

3. Check your answer by substituting it for the variable in the original equation.

Procedure for Solving Linear Equations that Simplify to the Form ax = c

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Solving Linear Equations: x + b = c and ax = c Section 5.3 584

Example 4

Solving ax = c

a. 5x = 20

Solution

5 20

5 2015

15

x

x

=

⋅( ) = ⋅

Write the equation.

Multiply by , the recipro1

5ccal of 5.

Use the associative property of multi15

515

201

= ⋅x pplication

Simplify.

.

1 4

4

⋅ ==

x

x

Multiplying by the reciprocal of the coefficient is the same as dividing

by the coefficient itself. So, we can multiply both sides by 15

, as we

did, or we can divide both sides by 5. In either case, the coefficient of x becomes +1.

5 20

5 20

4

x

x

x

=

=

=5 5

Check:

5 20

5 20

20 20

4

x =

⋅( ) ==

?

b. 1.1x + 0.2x = 12.2 − 3.1

Solution

When decimal coefficients or constants are involved, you might want to use a calculator to perform some of the arithmetic.

1 1 0 2 12 2 3 1

1 3 9 1

1 3 9 1

7 01 3 1 3

. . . .

. .

. .

.. .

x x

x

x

x

+ ==

=

=

Divide both sides by 5.

Simplify.

Substitute x = 4.

true statement

Write the equation.

Combine like terms.

Use a calculator or pencil and paper to divide.

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Check:

1 1 0 2 12 2 3 1

1 1 0 2 127 7

. . . .

. .?

x x+ = −

( ) + ( ) =

.. .

. . .

. .

?

2 3 1

7 7 1 4 9 1

9 1 9 1

+ ==

c. 45

310

x= (This could be written

45

310

x = because 45

x is

the same as 45x

.)

Solution

45

310

45

310

11

43

238

5

5

x

x

x

x

=

⋅ = ⋅

⋅ =⋅

⋅⋅

=

54

54

Check:

45

310

45

310

310

310

38

x=

⋅ =

=

?

d. −x = 4

Solution

− =− =−

=

= −− −

x

x

x

x

4

1 4

1 4

41 1

Check:

− =

−( ) ==

x 4

4

4 4

4 ?

Substitute x = 7.

Simplify.

true statement

4. Solve the following equations.

a. 3x = 33

b. 2.5x + 0.4x = 15.2 − 3.6

c. 23

45

x=

d. –x = −15

a. x = 11 b. x = 4

c. x =65

d. x = 15

Write the equation.

Multiply both sides by 5

4.

Simplify.

Substitute x = .3

8

true statement

Write the equation.

−1 is the coefficient of x.

Divide by −1 so that the coefficient will become +1.

Substitute x = −4.

true statement Now work margin exercise 4.

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Solving Linear Equations: x + b = c and ax = c Section 5.3 586

Example 5

Application

The original price of a Blu-Ray player was reduced by $45.50. The sale price was $165.90. Solve the equation y − 45.50 = 165.90 to determine the original price of the Blu-Ray player.

Solution

y − 45.50 = 165.90y − 45.50 + 45.50 = 165.90 + 45.50 y = 211.40

The original price of the Blu-Ray player was $211.40.

5. The original price of a wool coat was reduced by $15.80. The reduced price was $84.79. Solve the equation y − 15.80 = 84.79 to determine the original price of the coat.

The original price of the wool coat was $100.59.

Use the addition principle by adding 45.50 to both sides.Simplify.

Now work margin exercise 5.

Solve the following equations.

1. −16 = x + 5 2. 6y − 1.5 = 7.5 3. 4x = −20

4. 3y5

33= 5. 1.7z + 2.4z = 8.2 6. −x = −8

7. 3x = 10 8. 5x − 4x + 1.6 = − 2.7 9. 45

3= x

Practice Problems

Practice Problem Answers 1. x = −21 2. y = 1.5 3. x = −5 4. y = 55 5. z = 2 6. x = 8

7. x =103

8. x = −4.3 9. 415

= x

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587 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

Exercises 5.3

Determine whether or not the given number is a solution to the given equation by substituting and then evaluating. See Example 1.

1. x + 4 = 2 given that x = −2 2. z + −( ) =12 6 given that z = 18

−2 is a solution 18 is a solution

3. x − 3 = −7 given that x = 4 4. x − 2 = −3 given that x = 1

4 is not a solution 1 is not a solution

5. −10 + x = −14 given that x = −4 6. −9 − x = −14 given that x = 5

−4 is a solution 5 is a solution

7. − + = −26 8x given that x = −18 8. 42 30+ = −z given that z = −72

−18 is a solution −72 is not a solution

9. x − − =3 25 given that x = −28 10. − + =2 13x given that x = −11

−28 is a solution −11 is a solution

Solve each of the following linear equations. (Remember that the objective is to isolate the variable on one side of the equation with a coefficient of +1.) See Examples 2 through 4.

11. x − =6 1 12. x − =10 9 13. y + =7 3 14. y + =12 5 15. x + = −15 4

x = 7 x = 19 y = −4 y = −7 x = −19

16. x + = −17 10 17. 22 15= −n 18. 36 20= −n 19. 6 12= +z 20. 18 1= +z

x = −27 n = 37 n = 56 z = −6 z = 17

21. x − = −20 15 22. x − = −10 11 23. y + = −3 4 2 5. . 24. y + = −1 6 3 7. . 25. x + =3 6 2 4. .

x = 5 x = −1 y = −5.9 y = −5.3 x = −1.2

26. x + =2 7 3 8. . 27. x + =1

2035

28. n − =27

314

29. 5 45x = 30. 9 108x =

x = 1.1 x = 1120

n = 12

x = 9 x = 12

31. 32 4= y 32. 51 17= y 33. 34

15x

= 34. 57

65x

= 35. 425

=y

y = 8 y = 3 x = 20 x = 91 y = 10

36. 4646

5=

y 37. 4 3 10 12x x− = − 38. 7 6 13 15x x− = + 39. 7 8 13 25x x− = −

y = 5 x = −2 x = 28 x = 12

40. 10 11 20 14n n− = − 41. 3 2 6 14n n− + = 42. 7 6 13 22n n− + = 43. 1 7 1 3 6 3. . .y y+ =

n = −6 n = 8 n = 9 y = 2.1

44. 2 5 7 5 4 2. . .y y+ = 45. 34

53

x = 46. 56

53

x = 47. 7 5 99 75. .x = −

y = 0.42 x = 209

x = 2 x = −13.3

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Solving Linear Equations: x + b = c and ax = c Section 5.3 588

48. − =14 0 7. x 49. 1 5 0 5 6 7 5 3. . . .y y− + = − 50. 2 6 1 6 5 1 2 9. . . .y y− − = −

x = −20 y = −12 y = 2.2

51. 10 912

910

x x− − = − 52. 6 534

112

x x− + = − 53. 1 4 0 4 2 7 1 3. . . .x x− + = −

x = −25

x = −56

x = −4

54. 3 5 2 5 6 3 1 0 2 5. . . . .y y− − = − − 55. 74

34

78

32

x x− + = 56.

52

32

45

75

110

n n− + = −

y = 2.8 x = 58

n = 12

57. 6 2 3 5 7 6. .= − + −n n 58. − = − −7 2 1 3 0 3 1 0. . . .n n 59. 1 7 5 1 1 7. . .x = − − 60. 3 2 2 8 9 2. . .x = −

n = 9.7 n = −6.2 x = −4 x = −2

Solve the following word problems. See Example 5.

61. World Languages: The Japanese writing system consists of three sets of characters, two with 81 characters (which all Japanese students must know), and a third, kanji, with over 50,000 characters (of which only some are used in everyday writing). If a Japanese student knows 2107 total characters, solve the equation x + =2 81 2107( ) to determine the number of kanji characters the student knows. 1945 kanji characters

62. Astronomy: The diameter of the Milky Way, the galaxy our solar system is in, is approximately 23,585 times the distance from the sun to the nearest star, Proxima Centauri. Considering that the Milky Way is roughly 100,000 light years across, solve the following equation to find the number of light years from the sun to this star. (Round your answer to the nearest hundredth.) 4.24 light years

63. Writing: An author is determined to have his first novel published by the publisher of George Orwell’s 1984, his favorite book. However, his contract with the publisher requires his novel to be at least 75,000 words, and he has only written 63,500. Solve the following equation to determine how many more words he must write. 11,500 words

63 500 75 000, ,+ =x

64. Sculpture: A sculptor has decided to begin a project to make scale models of famous landmarks out of stone. His first model will be of one of the moai, giant human figures carved from stone on Easter Island. If his

model is to be 1

12scale and the original moai

weighs 75 tons, solve the equation 12 75x = to determine how many tons his completed sculpture will weigh. 6.25 tons

23 585 100 000, ,x =

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589 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

Use a graphing calculator to help solve the following equations.

65. y + = −32 861 17 892. . y = −50.753 66. x − 41.625 = 59.354 x = 100.979

67. 17 61 16 61 27 059 9 845. . . .x x− + = x = −17.214 68. 14 83 8 65 13 83 17 437 1 0. . . . .y y− − = + y = 27.087

69. 2 637 648 702. .x = x = 246 70. − =0 3057 316 7052. .y y = −1036

71. − = + −x 145 6 17 89 10 32. . . x = −153.17 72. − = + −y 143 5 178 462 200. . y = −121.962

73. a. Is the expression 6 + 3 = 9 an equation? Explain.

b. Is 4 a solution to the equation 5 + x = 10? Explain.

Writing & Thinking

73. a. Yes. It is stating that 6 + 3 is equal to 9.

b. No. If we substitute 4 for x, we get the statement 9 = 10, which is not true.

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Solving Linear Equations: ax + b = c Section 5.4 590

5.4 Solving Linear Equations: ax + b = c

Objective A Solving Equations of the Form ax + b = c

Earlier we learned the equations to be solved were in one of two forms: x b c+ = or ax c= . In the first type, the coefficient of the variable x was always +1 and we used the addition principle to add or subtract constants to solve the equation. (We used the addition principle.) In the second type, we had to multiply both sides by the reciprocal of the coefficient of the variable (or divide both sides by the coefficient itself). Now we will apply both of these techniques in solving the same equation. That is, in the form ax b c+ = , the coefficient a may be a number other than 1. The general procedure for solving linear equations is now a combination of the procedures stated in Section 5.3.

1. Combine like terms on both sides of the equation.2. Use the addition principle of equality and add the opposite of the

constant b to both sides.3. Use the multiplication (or division) principle of equality to multiply

both sides by the reciprocal of the coefficient of the variable (or divide both sides by the coefficient itself). The coefficient of the variable will become +1.

4. Check your answer by substituting it for the variable in the original equation.

Procedure for Solving Linear Equations that Simplify to the Form ax + b = c

Example 1

Solving Linear Equations

a. 3 3 18x + = −

Solution

3 3 18

3 3 18

3

3 3

x

x

x

+ = −+ = −

= −− −

Write the equation.

Add to both sides.3−

221

3 21

73 3

Simplify.

Divide both sides by

Simplify.

3.x

x

=−

= −

Check:

3 3 18

3 3 18

21 3 18

18

7

x + = −

( ) + = −

− + = −− =

−?

?

Substitute =

Simplify.

7.x −

−−18 true statement

Solve equations of the form ax + b = c.

A

Objectives

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b. − = − −26 2 14 4y y

Solution

− = − −− = − −

26 2 14 4

26 2 14

2

y y

y

Write the equation.

Combine like terms.

66 2 14

12 2

12 2

14 14

2

+ +

− −

= − −− = −−

=−

y

y

y

Add to both sides.

Simplify.

14

226

Divide both sides by

Simplify.

2.−

= y

Check:

− = − −

− = ( ) − − ( )

− = − −

26 2 14 4

26 2 14 4

26 12 14 24

6 6

y y?

?

Substitute =

Si

6.y

mmplify.

true statement− = −26 26

− = − −

− = ( ) − − ( )

− = − −

26 2 14 4

26 2 14 4

26 12 14 24

6 6

y y?

?

Substitute =

Si

6.y

mmplify.

true statement− = −26 26

Examples 2a. and 2b. illustrate solving equations with decimal coefficients. You may choose to work with the decimal coefficients as they are. However, another approach, shown here, is to multiply both sides in such a way to give integer coefficients. Generally, integers are easier to work with than decimals.

Example 2

Solving Linear Equations with Decimal Coefficients

a. 16 53 18 2 7 43 0. . .− − =z Solution

16 53 18 2 7 43 0

16 53 18 2 7 43 0

1653 1820 7

100 100

. . .

. . .

− − =

− −( ) = ( )

− −

z

z

z 443 0

910 1820 0

910 1820 0

1820 910

910 910

=− =

− =− = −

− −z

z

z

Write the equation.

Multiply both sides by 100.(This results in integer coefficients.)

Simplify.

Combine like terms.

Add −910 to both sides.

Simplify.

Divide both sides by −1820.

Simplify.

−=

= =

− −1820 910

0 512

1820 1820z

z z. or

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Solving Linear Equations: ax + b = c Section 5.4 592

Check:

16 53 18 2 7 43 0

16 53 18 2 7 43 00 5

. . .

. . .?

?.

− − =

− ( ) − =

z

Substitute = .0.5z

116 53 9 10 7 43 0

0 0

. . .− − ==

Simplify.

true statement

b. 5 1 7 4 1 8 9 1. . . .x x+ − = −

Solution

5 1 7 4 1 8 9 1

5 1 7 4 1 8 9 1

51 74 18 91

10 10

. . . .

. . . .

x x

x x

x x

+ − = −+ −( ) = −( )

+ − = −333 74 91

33 74 91

33 165

33 165

5

74 74

33 33

x

x

x

x

x

+ = −+ = −

= −

=−

= −

− −

Write the equation.

Multiply both sides by 10.(This results in integer coefficients.)

Simplify.

Combine like terms.

Add −74 to both sides.

Simplify.

Divide both sides by the coefficient 33.

Simplify. Checking will show that −5 is a solution.

Examples 3a. and 3b. illustrate solving equations with coefficients that are fractions. You may choose to work with the coefficients as they are. However, another approach, shown here, is to multiply both sides by the LCM of the denominators which will give integer coefficients Generally, integers are easier to work with than fractions.

Example 3

Solving Linear Equations with Fractional Coefficients

a. 56

52

109

x − = −

Solution

56

52

109

56

52

109

56

52

18 18

18 18

x

x

x

− = −

= −

= −

− = −

18109

15 45 20x

Write the equation.

Multiply both sides by 18 (theLCM of the denominators).

Apply the distributive property.

Simplify.

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15 45 20

15 25

15 25

53

45 45

15 15

x

x

x

x

− = −=

=

=

+ + Add 45 to both sides.

Simplify.

Divide both sides by 15.

Simplify.

Check:

56

52

109

56

52

109

2518

4518

2

53

x − = −

− = −

− = −

?

?

Substitute =5

3.y

0018

2018

2018

Simplify.

true statement− = −

56

52

109

56

52

109

2518

4518

2

53

x − = −

− = −

− = −

?

?

Substitute =5

3.y

0018

2018

2018

Simplify.

true statement− = −

b. 12

34

72

23

0x x x+ + − =

Solution

Write the equation.

Multiply both sides by 12 (the LCM of the denominators).

Apply the distributive property.Simplify.

Combine like terms.

Add −42 to both sides.

Simplify.

Divide both sides by 7.

Simplify. Checking will show that −6 is the solution.

12

34

72

23

0

12

34

72

23

0

12

12 12

12 12

x x x

x x x

x

+ + − =

+ + −

= ( )

+334

72

23

0

6 9 42 8 0

7 42

12 12 12x x

x x x

x

+

= ( )+ + − =

+ = 00

7 42 0

7 42

7 42

6

42 42

7 7

x

x

x

x

+ == −

=−

= −

− −

Solve each of the following equations.

1. a. 2x + 5 = −21

b. −18 = 2y − 8 − 7y

2. a. 3.69 + 12.80 z − 6.25 = 0

b. 0 7 9 5 2 4 4 1. . . .x x+ − = −

3. a. 35

73

1715

x − =−

b. 13

38

92

812

0x x x+ + − =

1. a. x = −13 b. y = 2

2. a. z = 0.2 b. x = 8

3. a. x = 2 b. x = −108

Now work margin exercises 1 through 3.

notesAbout Checking: Checking can be quite time-consuming and need not be done for every problem. This is particularly important on exams. You should check only if you have time after the entire exam is completed.

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Solving Linear Equations: ax + b = c Section 5.4 594

Solve the following linear equations.

1. x x+ 14 8 7− = − 2. 2.4 .6 0.9= − −2 5 9y y.

3. nn

− − =23

12

16

4. 314

12 7

0x x

+ − =

Practice Problems

Practice Problem Answers 1. x = 3 2. y = −1 3. n = 2 4. x = −7

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595 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

Exercises 5.4

Solve each of the following linear equations. See Examples 1 through 3.

1. 3 11 2x + = 2. 3 10 5x + = − 3. 5 4 6x − = 4. 4 8 12y − = −

x = −3 x = −5 x = 2 y = −1

5. 6 10 22x + = 6. 3 7 19n + = 7. 9 5 13x − = 8. 2 4 12x − =

x = 2 n = 4 x = 2 x = 8

9. 1 3 4− =y 10. 5 2 9− =x 11. 14 9 5+ =t 12. 5 2 7+ = −x

y = −1 x = −2 t = −1 x = −6

13. − + =5 2 9 3 5x . . 14. 3 2 7 2 7x + = −. . 15. 10 3 4 18+ − =x 16. 5 5 6 9+ − =x

x = −0.12 x = −1.8 x = 4 x = 2

17. 15 7 7 8= + +x 18. 14 9 5 8= + +x 19. 5 3 2 2y y− + = 20. 6 8 7 7y y+ − = −

x = 0 x = 19

y = 0 y = 0

21. x x− + =4 25 31 22. 3 9 13 11y y+ − = 23. − = − +20 7 3 4y y 24. − = + +20 5 16y y

x = −2 y = 2 y = −6 y = −6

25. 4 10 35 1 2n n− + = − 26. − − + =5 3 2 34n n 27. 3 15 1n n− − = 28. 2 12 0n n+ + =

n = 6 n = −4 n = 8 n = −4

29. 5 4 0 2 0. .x x− = 30. 0 5 1 0 3= +. .x x 31. 12

772

x + = 32. 35

495

x + =

x = 0 x = 0 x = −7 x = −113

33. 12

83

56

− =x 34. 25

12

74

− =x 35. 32

13

113

= +x 36. 118

15

45

= +x

x = −18

x = −2710

x = −132

x = 238

37. 72

552

9− − =x 38. 83

273

6+ − =x 39. 58

14

12

310

x x− + =

x = −215

x = −47

x = −815

40. 12

34

53

56

x x+ − = 41. y2

15

3+ = 42. y3

23

7− =

x = 2 y = 285

y = 23

43. 78

34

58

= −x 44. 1

1045

310

= +x 45. y y7 28

12

34

+ + =

x = 2 x = −14

y = 75

46. 56

78

112

13

y y− − = 47. x x+ + = −1 2 6 9 3 0. . . 48. 3 0 75 1 72 3 23x x− − =. . .

y = −10 x = −4.5 x = 2.2

49. 10 0 5 32= − +x x. 50. 33 3 0 4= + −y y. 51. 2 5 0 5 3 5 2 5. . . .x x+ − =

x = −44 y = 50 x = 2

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Solving Linear Equations: ax + b = c Section 5.4 596

52. 4 7 0 5 0 3 0 1. . . .− − = −x x 53. 6 4 1 2 0 3 0 4. . . .+ + =x x 54. 5 2 1 3 1 5 0 4. . . .− − = −x x

x = 6 x = −4 x = 2

55. − = + −12 13 2 42 0 6 13 64. . . .y y 56. − = + −7 01 1 75 3 05 8 45. . . .x x 57. − + + =0 4 17 2 18 1. . .x x

y = 0.5 x = 0.3 x = 1.5

58. y y− + =0 75 13 76 14 66. . . 59. 0 17 3 15 02 0 456= − −. . .x x 60. 0 20 5 16 35 0 1245= − +. . .x x

y = 3.6 x = 0.2 x = −0.03

Solve the following word problems.

61. Music: The tickets for a concert featuring the new hit band, Flying Sailor, sold out in 2.5 hours. If there were 35,000 tickets sold, solve the equation 35 000 2 5 0, .− =x to find the number of tickets sold per hour. 14,000 tickets per hour

62. Parking Lots: A rectangular shaped parking lot is to have a perimeter of 450 yards. If the width must be 90 yards because of a building code, solve the equation 2 2 90 450l + ( ) = to determine the length of the parking lot. 135 yards

63. Temperature: Jeff, who lives in England, is reading a letter from his pen pal in the United States. His pen pal says that the temperature in his city was 97.7° Fahrenheit one day, so it was too hot to play soccer outside. Jeff doesn’t know how hot this is, because he is used to temperatures in Celsius. Help Jeff solve the equation below to determine the temperature in degrees Celsius. 36.5°C

1 8 32 97 7. .C + =

64. Height: The tallest man-made structure in the world is the Burj Khalifa in Dubai, which stands at 2717 feet tall. The tallest tree in the world is a Mendocino tree in California. If 7 of these trees were stacked on top of each other, they would still be 144.5 feet shorter than the Burj Khalifa. Solve the equation below to determine the height of the tree. 367.5 feet

7 144 5 2717x + =.

Use a graphing calculator to help solve the linear equations.

65. 0 15 5 23 17 815 15 003. . . .x x+ − = 66. 15 97 12 34 16 95 8 601. . . .y y− + = x = 6.1 y = −2.3

67. 13 45 20 17 36 24 696. . .x x− − = − 68. 26 75 30 23 28 4 4625. . .y y− + = x = 1.12 y = 5.79

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5.5 Solving Linear Equations: ax + b = cx + d

Objective A Solving Equations of the Form ax +b = cx +d

Now we are ready to solve linear equations of the most general form ax + b = cx + d where constants and variables may be on both sides. There may also be parentheses or other symbols of inclusion. Remember that the objective is to get the variable on one side of the equation by itself with a coefficient of +1.

1. Simplify each side of the equation by removing any grouping symbols and combining like terms on both sides of the equation.

2. Use the addition principle of equality and add the opposite of a constant term and/or variable term to both sides so that variables are on one side and constants are on the other side.

3. Use the multiplication (or division) principle of equality to multiply both sides by the reciprocal of the coefficient of the variable (or divide both sides by the coefficient itself). The coefficient of the variable will become +1.

4. Check your answer by substituting it for the variable in the original equation.

General Procedure for Solving Linear Equations that Simplify to the Form ax + b = cx + d

Example 1

Solving Equations with Variables on Both Sides

a. 5 3 2 18x x+ = −

Solution

5 3 2 18

5 3 2 183 3

x x

x x

+ = −+ = −− −

Write the equation.

Add to both sides.3−

55 2 21

5 2 21

3 21

2 2

x x

x x

x

x x

= −= −= −

− −Simplify.

Add to both sides

Si

2 .− x

mmplify.

Divide both sides by

Simplify.

3.33 3

21

7

x

x

=−

= −

Teaching Note:You might want to emphasize to your students that there is more than one possible first step. For example in Example 1a., the text shows adding −3 to both sides. You can show them that adding +18, −2x, or −5x as a first step will give the same solution.

Solve equations of the form ax + b = cx + d.

Understand the terms conditional equations, identities, and contradictions.

A

B

Objectives

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Solving Linear Equations: ax + b = cx + d Section 5.5 598

Check:

5 3 2 18

5 3 2 187 7

x x+ = −

− −( ) + = ( ) − ?

Substitute = 7.x −

?− + = − −− = −

35 3 14 18

32 32

Simplify.

true statementt

5 3 2 18

5 3 2 187 7

x x+ = −

− −( ) + = ( ) − ?

Substitute = 7.x −

?− + = − −− = −

35 3 14 18

32 32

Simplify.

true statementt

b. 4 1 2 13 5x x x+ − = − +

Solution

4 1 2 13 5

3

3 1 2 8

x x x

x x

+ − = − +

+ = −

Write the equation.

Combine like terms.

xx x

x x

x xx

+ = −

= −

= −

− −

− −

1 2 8

3 2 9

3 2 9

1 1

2 2

Add to both sides.

Simplify.

1−

xx

x

Add to both sides.

Simplify.

2− x

= −9Check:

4 1 2 13 5

4 1 29 9

x x x+ − = − +

− −( ) + − ( ) = ?

−−( ) − +

9 13 5

36

Substitute = 9.x −

?++ + = − − +

− = −1 9 18 13 5

26 26

Simplify.

true statement

Example 2

Solving Linear Equations Involving Decimals

6 2 5 7 3 6y y+ = −. .

Solution

6 2 5 7 3 6

6 2 5 7 3 63 6 3 6

y y

y y

+ = −+ = −+ +

. .

. .. .

Write the equation.

Add t3.6 oo both sides.

Simplify.

Add to both s6

6 6 1 7

6 6 1 76 6

y y

y yy y

+ =+ =− −

.

. − y iides.

Simplify.6 1. = y

Check:

6 2 5 7 3 6

6 2 5 7 3 66 1 6 1

y y+ = −

( ) + = ( ) −

. .

?. .. .

Subsstitute = .

Simp

6.1y

?. . . .36 6 2 5 42 7 3 6+ = − llify.

true statement39 1 39 1. .=

Teaching Note:Another good approach in Example 2 is to first multiply each term by 10 to clear the decimals. You might want to go further and give an example where they could multiply by 100 to clear decimals.

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Example 3

Solving Linear Equations with Fractional Coefficients

13

16

25

710

x x+ = −

Solution

13

16

25

710

13

16

25

710

30 30

x x

x x

+ = −

+

= −

Write the equation.

+

=

Multiply both sides by theLCM, .30

30 3013

16

x 330 3025

710

10 5

x

x

+

Apply the distributive property.

== −+ = −

= −− −

12 21

10 5 12 21

10 12

5 5

x

x x

x x

Simplify.

Add to both sides.5−

226

10 12 26

2 26

12 12

Simplify.

Add to both sides.

Si

12x x

x

x x− −= −− = −

− x

mmplify.

Divide both sides by

Simplify. Chec

2.−

=−

=− −2 26

13

x

x

2 2−

kking will show that

13 is the solution.

Example 4

Solving Equations with Parentheses

a. 2 7 4 1 26y y−( ) = +( ) −

Solution

2 7 4 1 26

2 14 4 4 26

y y

y y

−( ) = +( ) −− = + −

Write the equation.

Use the distribbutive property.

Combine like terms.2 14 4 22

2 14 4 2222

y y

y y

− = −− = −+ ++ 22 Add to both sides. Here we will

put the variable on the right22

ssideto get a positive coefficient of .

Simplify.

y

2 8 4

2 8 2

y y

y y

+ =+ =− 44

8 2

8 2

2

2 2

y

y

y

y−=

=

Add to both sides.

Simplify.

Divide both side

2 − y

ss by

Simplify. Checking will show

that 4 is the solut

2.

4 = y

iion.

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Solving Linear Equations: ax + b = cx + d Section 5.5 600

b. − +( ) − = − −( ) −2 5 13 2 6 3 2 41x x

Solution

− +( ) − = − −( ) −− − − = − + −2 5 13 2 6 3 2 41

10 26 2 18 12

x x

x x

Write the equation.

441

10 2

Use the distributive property.Be careful with the signs.

− −x 88 18 29

10 28 18 2918 18

= − −− − = − −+ +

x

x xx x

Combine like terms.

Add to18 x bboth sides.

Simplify.

Add to both sid28

8 28 29

8 28 2928 28

x

x

− = −− = −+ + ees.

Simplify.

Divide both sidesby

Simplify.

8.

8 1

8 1

18

8 8

x

x

x

= −

=−

= − Checking will show

that is the solution.−12

Solve each of the following equations.

1. a. 7x + 4 = 3x − 20

b. 6x + 3 − 2x = 2x − 15 + 4

2. 5y + 1.4 = 7y − 2.8.

3. 14

25

12

610

x x+ = −

4. a. 3(y − 5) = 8(y + 1) − 3

b. − 5 (3x + 7) − 11 = − 7 (x − 1) − 5

1. a. x = −6 b. x = −7

2. y = 2.1

3. x = 4

4. a. y = −4 b. x = −6

Now work margin exercises 1 through 4.

Objective B Conditional Equations, Identities, and Contradictions

When solving equations, there are times when we are concerned with the number of solutions that an equation has. If an equation has a finite number of solutions (the number of solutions is a countable number), the equation is said to be a conditional equation. As stated earlier, every linear equation has exactly one solution. Thus every linear equation is a conditional equation. However, in some cases, simplifying an equation will lead to a statement that is always true, such as 0 = 0. In these cases the original equation is called an identity and has an infinite number of solutions which can be written as all real numbers or R. If the equation simplifies to a statement that is never true, such as 0 = 2, then the original equation is called a contradiction and there is no solution. Table 1 summarizes these ideas.

Type of Equation Number of Solutions

conditional finite number of solutions

identity infinite number of solutions

contradiction no solution

Table 1

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601 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

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Example 5

Solutions of Equations

Determine whether each of the following equations is a conditional equation, an identity, or a contradiction.

a. 3x + 16 = −11

Solution

3 16 11

3 16 1116 16

x

x

+ = −+ = −− −

Write the equation.

Add to both sides16− ..

Simplify.

Divide both sides by .

Simplify.

3

3 27

3 27

93 3

x

x

x

= −

=−

= −

The equation has one solution and it is a conditional equation.

b. 3 25 3 6 10x x x−( ) + = +( )

Solution

3 25 3 6 10

3 75 3 6 60

x x x

x x x

−( ) + = +( )− + = +

Write the equation.

Use the disttributive property.

Combine like terms.6 75 6 60

6 75 66

x x

x xx

− = +− =− ++

− =−60

75 60

6x Add to both sides.

Simplify.

6− x

The last equation is never true. Therefore, the original equation is a contradiction and has no solution.

c. − −( ) + = −2 7 14x x x

Solution

− −( ) + = −− + + = −2 7 14

2 14 14

x x x

x x x

Write the equation.

Use the distributtive property

Combine like terms.

Add t

.

14 14

14 14

− = −− = −+ +

x x

x xx x x oo both sides.

Simplify.14 14=

The last equation is always true. Therefore, the original equation is an identity and has an infinite number of solutions. Every real number is a solution.

5. Determine whether each of the following equations is a conditional equation, an identity, or a contradiction.

a. 2x + 9 = −13

b. 4(x − 17) + 3x = 7(x + 7)

c. −3(x − 4) + x = 12 − 2x

a. conditional b. contradiction c. identity

Now work margin exercise 5.

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Solving Linear Equations: ax + b = cx + d Section 5.5 602

Solve the following linear equations.

1. x x x+ − = −14 6 2 7 2. 6 4 2 1 3 1 1 2. . . .x x+ = −

3. 23

12

16

xx− = + 4.

314

14

17

14

n n+ = −

5. 5 3 14 4 2− −( ) = − +( )y y

Determine whether each of the following equations is a conditional equation, an identity, or a contradiction.

6. 7 3 42 7 21x x−( ) + = + 7. − + = − +( ) +2 14 2 1 10x x

Practice Problems

Practice Problem Answers

1. x = 3 2. x = −1 3. x = −2 4. n = −7

5. y = −23

6. identity 7. contradiction

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603 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

Exercises 5.5

Solve each of the following linear equations. See Examples 2 through 4.

1. 3 2 8x x+ = − 2. 5 1 2 5x x+ = − 3. 4 3 6n n− = + 4. 6 3 7y y+ = −

x = −5 x = −2 n = 3 y = −2

5. 3 18 7 6y y+ = − 6. 2 5 8 10y y+ = + 7. 3 11 8 4x x+ = − 8. 9 3 5 9x x+ = −

y = 6 y = −56

x = 3 x = −3

9. 14 3n n= 10. 1 6 0 8. .x x= 11. 6 2 1 2 1y y− = −. . 12. 13 5 2 5x x+ = +

n = 0 x = 0 y = 0 x = 0

13. 2 1 3 3z z+( ) = + 14. 6 3 3 2x x− = +( ) 15. 16 23 3 16 2 2y y y y+ − = − +

z = −1 x = 3 y = 15

16. 5 2 4 3 1x x x x− + = + − 17. 0.25 + 3x + 6.5 = 0.75x 18. 0.9y + 3 = 0.4y + 1.5

x = 5 x = −3 y = −3

19. 6.5 + 1.2x = 0.5 − 0.3x 20. x − 0.1x + 0.8 = 0.2x + 0.1 21. 23

113

6x x+ = −

x = −4 x = −1 x = −21

22. 45

225

4n n+ = − 23. y y5

34 2

34

+ = + 24. 56

19

32

19

n n+ = +

n = −15 y = 0 n = 0

25. 38

12

18

12

y y−

= +

26. 12 2

113 2

1x x

+

= −

27. 23 3

34 2

x x x+ = − +

y = 1 x = −10 x = −32

28. 34

15

12

310

x x x+ = − 29. x x xx

+ − = −23

26

18

30. 312

25 10

720

x x xx

+ − = +

x = −23

x = 14

x = 7

60

31. 3 1 9 6 2 4+( ) = −( )x x 32. 4 5 8 3 10−( ) = +( )x x 33. 3 4 1 4 2 3 8x x−( ) = −( ) +

x = 3

17 x = −

157

x = −14

34. 7 2 1 5 6 13x x−( ) = +( ) − 35. 5 3 2 1 4 5 6− +( ) = −( ) +x x 36. − +( ) − = −( ) +2 5 4 6 2 2y y

x = 83

x = 85

y = −12

37. 8 4 2 3 5 3+ −( ) = − +( )x x 38. 8 3 5 9 9 2 14x x+( ) − = −( ) + 39. 4 7 0 3 0 5 0 1. . . .− = −x x

x = 23

x = −73

x = 6

40. 5 8 0 1 0 2 0 2. . . .− = −x x 41. 0 2 3 0 1 5. .x x+( ) = −( ) 42. 0 4 3 0 3 6. .x x+( ) = −( ) x = 20 x = −11 x = −30

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Solving Linear Equations: ax + b = cx + d Section 5.5 604

43. 12

4 813

4 7 3−( ) = +( ) −x x 44. 314

425

2 3+ −( ) = +( )x x

x = 12

x = 2419

45. 0 6 22 9 1 5 18 4. . . .x x− = − 46. 0 1 3 8 5 72 0 3. . . .y y+ = −

x = −5 y = 4.8

47. 0 12 0 25 5 895 4 3. . . .n n n+ − = 48. 0 15 32 21 0005 10 5. . .n n n+ − =

n = −1.5 n = 0.97

49. 0 7 14 1 0 3 32 9. . . .x x+( ) = +( ) 50. 0 8 6 21 0 2 24 84. . . .x x−( ) = −( ) x = 0 x = 0

Determine whether each of the following equations is a conditional equation, an identity, or a contradiction. See Example 5.

51. 2 3 1 5 3x −( ) + = conditional 52. − + = − −( )2 13 2 7x x contradiction

53. 5 13 2 7 3x x+ = − −( ) + conditional 54. 3 9 3 3 6x x x+ = − −( ) + identity

55. 7 1 3 3 4x x x−( ) = − −( ) + contradiction 56. 3 2 4 6 1x x x x−( ) + = −( ) + identity

57. 5 1 3 1 2 1x x x+( ) = +( ) + +( ) identity 58. 8 20 3 5 2 3 4x x x x− + = − −( ) + −( ) contradiction

59. 2 3 5 2 3x x x+ = −( ). conditional 60. 5 2 3 4 0 2 0 42. . . .x x x+ = −( ) conditional

Solve each of the following word problems.

61. Construction: A farmer is putting a shed on his property. He has two designs. One uses wood and would cost $2 per square foot plus an extra $8400 in materials. The other design is metal and would cost $4 per square foot plus an additional $8800. Both sheds are the same size,

and the wood shed costs 34

what the metal shed

costs. Solve the following equation to determine how many square feet the shed will be. 1800 square feet

2 8400

34

4 8800x x+ = +( )

62. Renting Buildings: Two rival shoe companies want to rent the same empty building for their office and shipping space. Schulster’s Shoes needs 1000 square feet for offices, 600 for shipping, and another 6 square feet for every packaged box of shoes. Shoes, Shoes, Shoes! needs 750 square feet for offices, 400 for shipping, and 9 square feet per packaged box of shoes. If only one company will fit exactly into the empty building and they both plan to have the same amount of inventory, solve the following equation to determine how many boxes of shoes both companies hope to have at any given time. 150 boxes

1000 600 6 750 400 9+ + = + +x x

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605 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

63. Ice Cream: An ice cream shop is having a special “Ice Cream Sunday” event in which they are giving away giant mixed sundaes of 3 scoops of vanilla ice cream and 2 scoops of chocolate. If they have 24 gallons of chocolate and 36 gallons of vanilla to start with, solve the given equation to determine how many sundaes they will have made when they run out of ice cream. (For this problem, we assume a gallon = 20 scoops.) 240 sundaes

361

203 24

120

2− ( ) = − ( )x x

64. Music: A guitarist and a drummer are getting ready for a gig. The length of the gig will depend on how much material they have prepared. For every hour of the show, the guitarist must practice for 5 days and the drummer for 3 days. Since the guitarist knows some of the songs, he saves 3 days of practice time. The drummer hurts his hand and loses 3 days of practice. If they plan to start and finish practicing at the same time, solve the following equation to determine how long the show will be. 3 hours

5 3 3 3x x− = +

Use a graphing calculator to help solve the linear equations.

65. 0.17x − 23.0138 = 1.35x + 36.234 66. 48.512 − 1.63x = 2.58x + 87.63553 x = −50.21 x = −9.293

67. 0 32 14 1 2 47 2 21795. . . .x x+( ) = + 68. 1 6 9 3 2 0 2 3 133 94. . . .+( ) = +( )x x x = 1.067 x = 4.58

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Applications: Number Problems and Consecutive Integers Section 5.6 606

5.6 Applications: Number Problems and Consecutive Integers

Objective A Number Problems

In earlier chapters, we discussed translating English phrases into algebraic expressions, and learned the Pólya four-step process as an approach to problem solving. Now we will use those skills to read number problems and translate the sentences and phrases in the problem into a related equation. The solution of this equation will be the solution to the problem.

Example 1

Number Problems

a. If a number is decreased by 36 and the result is 76 less than twice the number, what is the number?

Solution

Let n = the unknown number.

a number isdecreased by 36

the resultis

76 less thantw� ��� ��� ��� �� iice the number� ��� ���

n nn n

n

n n− = −

− = −− = −

− −

+

36 2 7636 2 76

36 76

36 76 == −=

+n

n

76

40

76

The number is 40.

b. Three times the sum of a number and 5 is equal to twice the number plus 5. Find the number.

Solution

Let x = the unknown number.

3 times the sum ofa number and 5

is equalto

twice the number

� ��� ��� ��� pplus 5� ��� ���

3 5 2 5

3 15 2 53 15 2 5

15 52 2

x x

x xx x

xx

x x

+( ) = ++ = +

+ = ++ =

+

− −

115 510

15 15− −== − −x The number is 10.

N u

Solve word problems involving translating number problems.

Solve word problems involving translating consecutive integers.

Learn how to solve word problems involving a variety of other applications.

A

B

C

Objectives

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607 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

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c. One integer is 4 more than three times a second integer. Their sum is 24. What are the two integers?

Solution

Let n = the second integer, then 3n + 4 = the first integer.

1 2 24

3 4 24

4 4

st integer nd integer( ) + ( ) =

+( ) + =+

n n

n

Their sum is 24.

==+ =

=

=

=+ =

− −24

4 4 24

4 20

4 20

5

3 4 19

4 4

4 4

n

n

n

n

n The two integers are 55 and 19.

1. a. If a number is decreased by 28 and the result is 92 less than twice the number, what is the number?

b. 6 times the sum of a number and 3 is equal to 3 times the number plus 3. Find the number.

c. One integer is 3 more than 4 times a second integer. Their sum is 28. What are the two integers?

a. The number is 64. b. The number is −5. c. The two integers are 5

and 23. Now work margin exercise 1.

Objective B Consecutive Integers

Remember that the set of integers consists of the whole numbers and their opposites.

Z = … …, , , , , , , , , ,− − − −{ }4 3 2 1 0 1 2 3 4

Even integers are integers that are divisible by 2. The even integers are

… …, , , , , , , , .− − −{ }6 4 2 0 2 4 6

Odd integers are integers that are not even. If an odd integer is divided by 2, the remainder will be 1. The odd integers are

… …, , , , , , , .− − −{ }5 3 1 1 3 5

notesIn this discussion we will be dealing only with integers. Therefore, if you get a result that has a fraction or decimal number (not an integer), you will know that an error has been made and you should correct some part of your work.

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Applications: Number Problems and Consecutive Integers Section 5.6 608

The following terms and the ways of representing the integers must be understood before attempting the problems.

Integers are consecutive if each is 1 more than the previous integer. Three consecutive integers can be represented as n, n + 1, and n + 2.

For example: 5, 6, 7

Consecutive Integers

Even integers are consecutive if each is 2 more than the previous even integer. Three consecutive even integers can be represented as 2n, 2n + 2, and 2n + 4, where n is an integer.

For example: 24, 26, 28

Consecutive Even Integers

Odd integers are consecutive if each is 2 more than the previous odd integer. Three consecutive odd integers can be represented as 2n + 1, 2n + 3, and 2n + 5, where n is an integer.

For example: 41, 43, 45.

Consecutive Odd Integers

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609 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

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Example 2

Consecutive Integers

Three consecutive odd integers are such that their sum is −3. What are the integers?

Solution

Let 2n + 1 = the first odd integer, 2n + 3 = the second odd integer, and 2n + 5 = the third odd integer.Set up and solve the related equation.

1st integer 2nd integer 3rd integer( ) + ( ) + ( ) = −

+ + +( ) +

3

2 1 2 3 2n n n ++( ) = −+ = −

+ = −= −

=−

= −+ = −+ = −

− −

5 3

6 9 3

6 9 3

6 12

6 12

2

2 1 3

2 3 1

9 9

6 6

n

n

n

n

n

n

n

22 5 1n + =

The three consecutive odd integers are −3, −1, and 1.

2. Three consecutive odd integers are such that their sum is −9. What are the integers?

−5, −3, and −1

Now work margin exercise 2.

Objective C Other Applications

As you learn more abstract mathematical ideas, you will find that you will use these ideas and the related processes to solve a variety of everyday problems as well as problems in specialized fields of study. Generally, you may not even be aware of the fact that you are using your mathematical knowledge. However, these skills and ideas will be part of your thinking and problem solving techniques for the rest of your life.

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Applications: Number Problems and Consecutive Integers Section 5.6 610

Example 3

Applications

a. Joe pays $800 per month to rent an apartment. If this is 25

of his monthly income, what is his monthly income?

Solution

Let x = Joe’s monthly income and 25

x = rent.

25

800

25

8001

2000

52

52

x

x

x

=

⋅ = ⋅

=Joe’s monthly income is $2000.

b. A student bought a calculator and a textbook for a total of $200.80 (including tax). If the textbook cost $20.50 more than the calculator, what was the cost of each item?

Solution

Let x = cost of the calculator, then x + 20 50. = cost of the textbook.The equation to be solved is:

x x+ + =20 50 200 80. .

cost of

text book

cost of

calculator

tota

� �� �� �

ll

spent��� ��

2 20 50 80

2 20 50 8020 50 20 5

x

x

+ =

+ =− −

. .

. .. .

200

200 00

2 2

2 30

2 30

15

20 50

x

x

x

x

180

180

90

=

=

=

+

.

.

.

.

cost of calculator

= 110 65. cost of textbook

The calculator costs $90.15 and the textbook costs $90.15 + $20.50 = $110.65, with tax included in each price.

3. a. Jim pays $1200 per month to send his son Taylor to basketball

camp. If this is 37

of

his monthly income, what is his monthly income?

b. Graeme bought bagpipes and a soccer net for a total of $507.30 (including tax). If the bagpipes cost $70.40 more than the soccer net, what was the cost of each item?

a. $2800 b. bagpipes: $288.85;

soccer net: $218.45

Now work margin exercise 3.

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611 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

Exercises 5.6

Read each problem carefully, translate the various phrases into algebraic expressions, set up an equation, and solve the equation. See Examples 1 and 2.

1. Five less than a number is equal to 13 decreased by the number. Find the number. x − 5 = 13 − x; 9

2. Three less than twice a number is equal to the number. What is the number? 2x − 3 = x; 3

3. Thirty-six is 4 more than twice a certain number. Find the number. 36 = 2x + 4; 16

4. Fifteen decreased by twice a number is 27. Find the number. 15 − 2x = 27; −6

5. Seven times a certain number is equal to the sum of twice the number and 35. What is the number? 7x = 2x + 35; 7

6. The difference between twice a number and 3 is equal to 6 decreased by the number. Find the number. 2x − 3 = 6 − x; 3

7. Fourteen more than three times a number is equal to 6 decreased by the number. Find the number. 3x + 14 = 6 − x; −2

8. Two added to the quotient of a number and 7 is equal to −3. What is the number? x7

2 3 35+ = − −;

9. The quotient of twice a number and 5 is equal to the number increased by 6. What is the number? 2

56 10

xx= + −;

10. Three times the sum of a number and 4 is equal to −9. Find the number. 3 4 9 7x +( ) = − −;

11. Four times the difference between a number and 5 is equal to the number increased by 4. What is the number? 4 5 4 8x x−( ) = + ;

12. When 17 is added to six times a number, the result is equal to 1 plus twice the number. What is the number? 6x + 17 = 1 + 2x; −4

13. If the sum of twice a number and 5 is divided by 11, the result is equal to the difference between 4 and the number. Find the number. 2 5

114 3

xx

+= − ;

14. If the difference between a number and 21 is divided by 2, the result is 4 times the number. What is the number? x

x−

= −21

24 3;

15. Twice a number increased by three times the number is equal to 4 times the sum of the number and 3. Find the number. 2 3 4 3 12x x x+ = +( );

16. Twice the difference between a number and 10 is equal to 6 times the number plus 16. What is the number? 2 10 6 16 9x x−( ) = + −;

17. The sum of two consecutive odd integers is 60. What are the integers? n n+ +( ) =2 60

29 31

;

,

18. The sum of two consecutive even integers is 78. What are the integers? n n+ +( ) =2 78

38 40

;

, 19. Find three consecutive integers whose sum is 69.

n n n+ +( ) + +( ) =1 2 69 22 23 24; , , 20. Find three consecutive integers whose sum is 93.

n n n+ +( ) + +( ) =1 2 93 30 31 32; , ,

21. The sum of four consecutive integers is 74. What are the integers?

n n n n+ +( ) + +( ) + +( ) =1 2 3 74 17 18 19 20; , , ,

22. Find four consecutive integers whose sum is 90. n n n n+ +( ) + +( ) + +( ) =1 2 3 90

21 22 23 24

;

, , ,

23. 171 minus the first of three consecutive integers is equal to the sum of the second and third. What are the integers?

171 1 2 56 57 58− = +( ) + +( )n n n ; , ,

24. If the first of three consecutive integers is subtracted from 120, the result is the sum of the second and third. What are the integers? 120 1 2 39 40 41− = +( ) + +( )n n n ; , ,

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Applications: Number Problems and Consecutive Integers Section 5.6 612

25. Four consecutive integers are such that if 3 times the first is subtracted from 208, the result is 50 less than the sum of the other three. What are the integers? 208 3 1 2 3 50

42 43 44 45

− = +( ) + +( ) + +( ) −n n n n ;

, , ,

26. Find two consecutive integers such that twice the first plus three times the second equals 83. 2 3 1 83 16 17n n+ +( ) = ; ,

27. Find three consecutive even integers such that the first plus twice the second is 54 less than four times the third. n n n+ +( ) = +( ) −2 2 4 4 54 42 44 46; , ,

28. Find three consecutive odd integers such that 4 times the first is 44 more than the sum of the second and third. 4 2 4 44 25 27 29n n n= +( ) + +( ) + ; , ,

29. Find three consecutive even integers such that if the first is subtracted from the sum of the second and third, the result is 66.

n n n+( ) + +( ) − =2 4 66 60 62 64; , ,

30. Find three consecutive even integers such that their sum is 168 more than the second. n n n n+ +( ) + +( ) = +( ) +2 4 2 168 82 84 86; , ,

31. Find three consecutive odd integers such that the sum of twice the first and three times the second is 7 more than twice the third. 2 3 2 2 4 7 3 5 7n n n+ +( ) = +( ) + ; , ,

32. Find three consecutive even integers such that the sum of three times the first and twice the third is twenty less than six times the second. 3 2 4 6 2 20 16 18 20n n n+ +( ) = +( ) − ; , ,

Solve the following word problems. See Example 3.

33. School Supplies: A mathematics student bought a graphing calculator and a textbook for a course in statistics. If the text costs $49.50 more than the calculator, and the total cost for both was $125.74, what was the cost of each item? c c+ +( ) =49 50 125 74. . ; calculator: $38.12, textbook: $87.62

34. Electronics: The total cost of a computer flash drive and a color printer was $225.50, including tax. If the cost of the flash drive was $170.70 less than the printer, what was the cost of each item?

c c−( ) + =170 70 225 50. . ; flash drive: $27.40; printer: $198.10

35. Real Estate: A real estate agent says that the current value of a home is $90,000 more than twice its value when it was new. If the current value is $310,000, what was the value of the home when it was new? The home is 25 years old. 2x + 90,000 = 310,000; $110,000

36. Guitars: Texas has a mean of 91,399 guitars sold each year, which is seven times the mean number of guitars sold each year in Wyoming. What is the mean number of electric guitars, sold each year in Wyoming? 7x = 91,399; 13,057 guitars

37. Classic Cars: A classic car is now selling for $1500 more than three times its original price. If the selling price is now $12,000, what was the car’s original price? 3x + 1500 = 12,000; $3500

38. Mail: On August 24, the Fernandez family received 19 pieces of mail, consisting of magazines, bills, letters, and ads. If they received the same number of magazines as letters, three more bills than letters, and five more ads than bills, how many magazines did they receive? x x x x+ + +( ) + + +( ) =3 3 5 19; 2 magazines

39. Class Size: A high school graduating class is made up of 542 students. If there are 56 more girls than boys, how many boys are in the class? x x+( ) + =56 542; 243 boys

40. Golfing: Lucinda bought two boxes of golf balls. She gave the pro-shop clerk a 50-dollar bill and received $10.50 in change. What was the cost of one box of golf balls? (Tax was included.) 50 − 2x = 10.50; $19.75

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613 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

41. Guitars: A guitar manufacturer spent $158 million on the production of acoustic and electric guitars last year. If the amount the company spent producing acoustic guitars was $68 million more than it spent on producing electric guitars, how much did the company spend producing electric guitars? x x+( ) + =68 158; $45 million on electric guitars

42. Telephone Calls: For a long-distance call, the telephone company charges 35¢ for each of the first three minutes and 15¢ for each additional minute. If the cost of a call was $12.30, how many minutes did the call last? 0 35 3 0 15 3 12 30. . . ;( ) + −( ) =x 78 minutes

43. Rental Cars: If the rental price on a car is a fixed price per day plus $0.28 per mile, what was the fixed price per day if the total paid was $140 and the car was driven for 250 miles in two days? 2 0 28 250 140x + ( ) =. ; $35

44. Rental Cars: The U-Drive Company charges $20 per day plus 22¢ per mile driven. For a one-day trip, Louis paid a rental fee of $66.20. How many miles did he drive? 20 + 0.22x = 66.20; 210 miles

45. Carpentry: A 29 foot board is cut into three pieces at a sawmill. The second piece is 2 feet longer than the first and the third piece is 4 feet longer than the second. What are the lengths of the three pieces? x + x + 2 + x + 6 = 29; 7 feet, 9 feet, 13 feet

46. Triangles: The three sides of a triangle are n, 4n − 4, 2n + 7 (as shown in the figure below). If the perimeter of the triangle is 59 cm, what is the length of each side? n + 4n − 4 + 2n + 7 = 59; 8 cm, 28 cm, 23 cm

n 2n + 7

4n − 4

47. Construction: Joe Johnson decided to buy a lot and build a house on the lot. He knew that the cost of constructing the house was going to be $25,000 more than the cost of the lot. He told a friend that the total cost was going to be $275,000. As a test to see if his friend remembered the algebra they had together in school, he challenged his friend to calculate what he paid for the lot and what he was going to pay for the house. What was the cost of the lot and the cost of the house? x + x + 25,000 = 275,000; lot: $125,000, house: $150,000

48. Triangles: The three sides of a triangle are x, 3x − 1, and 2x + 5 (as shown in the figure below). If the perimeter of the triangle is 64 inches, what is the length of each side? (Reminder: The perimeter of a triangle is the sum of the lengths of the sides.) x + 3x − 1 + 2x + 5 = 64; 10 inches, 29 inches, 25 inches

x2x + 5

3x − 1

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Applications: Number Problems and Consecutive Integers Section 5.6 614

Make up your own word problem that might use the given equation in its solution. Be creative! Then solve the equation and check to see that the answer is reasonable.

49. 5 8x x− = The difference between five times a number and the number is equal to 8; 2

50. 2 3 9x + = Twice a number increased by 3 equals 9; 3

51. n n+ +( ) =1 33 Find two consecutive integers whose sum is 33; 16, 17

52. n n n+ +( ) = +( )4 3 2 Find 3 consecutive even integers such that the sum of the first and the third is 3 times the second; −2, 0, 2

53. 3 1 53n n+( ) = + Find two consecutive integers such that 3 times the second is 53 more than the first; 25, 26

54. 2 2 1 4n n n+( ) − = + − Find a number such that twice the sum of the number and 2, decreased by 1 is equal to the sum of the

number and 4 minus the number; 12

Note: Answers for 49 − 54 will vary.

55. a. How would you represent four consecutive odd integers? n, n + 2, n + 4, n + 6

b. How would you represent four consecutive even integers? n, n + 2, n + 4, n + 6

c. Are these representations the same? Explain. Yes. Answers will vary.

Writing & Thinking

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615 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

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5.7 Applications: Percent Problems

Our daily lives are filled with decimal numbers and percents: stock market reports, batting averages, win-loss records, salary raises, measures of pollution, taxes, interest on savings, home loans, discounts on clothes and cars, and on and on. Since decimal numbers and percents play such a prominent role in everyday life, we need to understand how to operate with them and apply them correctly in a variety of practical situations.

Objective A A Brief Review of Decimals and Percentages

The word percent comes from the Latin per centum, meaning “per hundred.” So, percent means hundredths. Thus

72%, 72100

, and 0.72 all have the same meaning.

change a decimal number to percent:

0.036 = 3.6% Move the decimal point two places to the right and add the % sign.

change a percent to a decimal number:

56% = 0.56 Move the decimal point two places to the left and drop the % sign.

change a percent to a fraction (or mixed number):

Write the percent in hundredths form and reduce.

change a fraction (or mixed number) to a percent:

Write the fraction in decimal form and then change

the decimal to a percent.

Briefly review the concepts of decimal numbers and percentages.

Understand the concept of the basic percent equation.

Use the basic percent equation to solve application problems involving percentages.

A

B

C

Objectives

3030

1003

10% = =

14

0 25 25= =. %

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Applications: Percent Problems Section 5.7 616

Objective B The Basic Percent Equation

Consider the statement

“15% of 80 is 12.”

This statement has three numbers in it. In general, solving a percent problem involves knowing two of these numbers and trying to find the third. That is, there are three basic types of percent problems. We can break down the sentence in the following way.

Sentence 15% of 80 is 12

Equation 0.15 ⋅ 80 = 12

percent “of” “is” written changed changed as decimal to times to =

Basic Formula Rate ⋅ Base = Amount

The terms that we have just discussed are explained in detail in the following box.

R = RATE or percent (as a decimal or fraction)

B = BASE (number we are finding the percent of)

A = AMOUNT (a part of the base)

“of ” means to multiply.

“is” means equal (=).

The relationship among R, B, and A is given in the equation

R B A A R B⋅ ⋅= = (or ).

The Basic Formula R · B = A

Even though there are just three basic types of percent problems, many people have difficulty deciding whether to multiply or divide in a particular problem. Using the equation R B A⋅ = helps to avoid these difficulties. If the values of any two of the quantities in the formula are known, they can be substituted in the equation and then the missing number can be found by solving the equation. The process of solving the equation for the unknown quantity determines whether multiplication or division is needed.

The following examples illustrate how to substitute into the equation and how to solve the resulting equations.

Teaching Note:Some students might find the following diagram helpful as a mnemonic device:

A

R B

dividesign

times sign

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Example 1Percent of a Number

a. What is 72% of 800?

Solution

R = 0.72, B = 800, and A is unknown.

R B A

A

A

⋅ =

⋅ ==

0 72 800

576

. Here, simply multiply to find .A

So 576 is 72% of 800.

b. 57% of what number is 163.191?

Solution

R = 0.57,B is unknown, and A = 163.191.R B A

B

B

⋅ =⋅

=

=

0 57 163 191

0 57 163 1910 57 0 57

. .

. .. .

Now divide both sidess by 0.57 to find .B

B = 286 3.

R B A

B

B

⋅ =⋅

=

=

0 57 163 191

0 57 163 1910 57 0 57

. .

. .. .

Now divide both sidess by 0.57 to find .B

B = 286 3.

So 57% of 286.3 is 163.191.

c. What percent of 180 is 45?

Solution

R is unknown, B = 180, and A = 45.

R B A

R

R

⋅ =

⋅ =⋅

=

180 45

180 45180 180

Now divide both sides by 180 to find .R

RR

R

==

0 25

25

.

% Write the decimal in percent form.

So 25% of 180 is 45.

1. a. What is 30% of 660?

b. 63% of what number is 212.184?

c. What percent of 300 is 60?

a. 198 b. 336.8 c. 20%

Now work margin exercise 1.

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Applications: Percent Problems Section 5.7 618

Objective C Applications with Percentages

To sell goods that have been in stock for some time or simply to attract new customers, retailers and manufacturers sometimes offer a discount, a reduction in the selling price usually stated as a percent of the original price. The new, reduced price is called the sale price.

Example 2

Discounts

A bicycle was purchased at a discount of 25% of its original price of $1600. What was the sale price?

Solution

There are two ways to approach this problem. One way is to find the discount and then subtract this amount from $1600. Another way is to subtract 25% from 100% to get 75% and then find 75% of $1600. Here, we will calculate the discount and then subtract from the original price.

R B A⋅ =

⋅ =0 25 1600 400. discount

Sale Price =original price − discount = $1600 − $400 = $1200The sale price for the bicycle was $1200.

2. A ping pong table was purchased at a discount of 40% of its original price of $1040. What was the sale price?

$624

Now work margin exercise 2.

Sales tax is a tax charged on goods sold by retailers, and it is assessed by states and cities for income to operate various services. The rate of sales tax (a percent) varies by location.

Example 3

Sales Tax

Suppose 6% sales tax was added to the sale price $1200 of the bicycle in Example 2. What was the total paid for the bicycle?

Solution

The sales tax must be calculated on the sale price and then added to the sale price.

R B A⋅ =

⋅ =0 06 1200 72. sales tax

total paid = sale price + sales tax = $1200 + $72 = $1272The total paid for the bicycle, including sales tax, was $1272.

3. Suppose a 4% sales tax was added to the sale price of the ping pong table in margin exercise 2. What was the total paid for the ping pong table?

$648.96

Now work margin exercise 3.

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A commission is a fee paid to an agent or salesperson for a service. Commissions are usually a percent of a negotiated contract (as to a real estate agent) or a percent of sales.

Example 4Commission

A saleswoman earns a salary of $1200 a month plus a commission of 8% on whatever she sells after she has sold $8000 in furniture. What did she earn the month she sold $25,000 worth of furniture?

Solution

First subtract $8000 from $25,000 to find the amount on which the commission is based.

$25,000 − $8000 = $17,000

Now find the amount of the commission. R B A⋅ =

⋅0 08 17 000 1360. , = commission

Now add the commission to her salary to find what she earned that month.

$1200 + $1360 = $2560 earned for the month

4. A real estate agent earns a salary of $2300 a month plus a commission of 4% on the houses he sells after he has sold $1,000,000 of housing. What did he earn the month he sold $1,200,000 of housing?

$10,300

Now work margin exercise 4.

Problems involving percent come in a variety of forms. The next example illustrates another situation you may encounter in your own life.

Example 5

Commission

The Berrys sold their house. After paying the real estate agent a commission of 6% of the selling price and then paying $1486 in other costs and $90,000 on the mortgage, they received $49,514. What was the selling price of the house?

Solution

Use the relationship:

selling price − cost = profit.

S − C = P

Let S = selling price.

Teaching Note:Examples from real estate can be effective here. Some of your students may even be working in real estate and can relate their experiences in buying and selling real estate.

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Applications: Percent Problems Section 5.7 620

sellingprice

cost profit�

� �− =

− ( ) =S S0.06 + 1486 + 90,000 49 51, 4449 514

0 94S

SS

− − − ===

0.06 1486 90,000141,000150,000

S ,.

Check:

$150,000 $9000 $150,000selling price commission selling price

× 0.06 1486 100,486

$9

commission % other costs cost−0000 + 90,000 $49,514

$100,486

commission mortgage profit

coost

The selling price was $150,000.

5. The Rotbergs sold the delivery truck from their family cookie business on eBay. After paying eBay a commission of 5% of the selling price and then paying $300 in other costs, they received $7965. What was the selling price of the cookie truck?

6. Calculate the percent of profit for both a. and b. and tell which is the better investment.

a. $900 profit on an investment of $2700

b. $400 profit on an investment of $1100

5. $8700

6. a. has a percent of profit of 33.3% and b. has a 36.4% percent of profit. Therefore, b. is a better investment.

Percent of profit for an investment is the ratio (a fraction) that compares the money made to the money invested. If you make two investments of different amounts of money, then the amount of money you make on each investment is not a fair comparison. In comparing such investments, the investment with the greater percent of profit is considered the better investment. To find the percent of profit, form the fraction (or ratio) of profit divided by the investment and change the fraction to a percent.

Example 6

Percent of Profit

Calculate the percent of profit for both a. and b. and tell which is the better investment.

a. $300 profit on an investment of $2400 or b. $500 profit on an investment of $5000

Solution

Set up ratios and find the corresponding percents.

a.

b.

Clearly, $500 is more than $300, but 12.5% is greater than 10% and investment a. is the better investment.

$$

. . %300

2400300 1300 8

18

0 125 12 5profit

invested=

⋅⋅

= = = percent of pprofit

$$

. %500

5000500 1

500 101

100 1 10

profitinvested

=⋅

⋅= = = percent of proofit

Now work margin exercises 5 and 6.

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621 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

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1. Write 0.3 as a percent.

2. Write 6.4% as a decimal.

3. Write 1

5 as a percent.

4. Find 12% of 200.

5. Find the percent of profit if $480 is made on an investment of $1500.

Practice Problems

Practice Problem Answers1. 30% 2. 0.064 3. 20% 4. 24 5. 32%

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Applications: Percent Problems Section 5.7 622

Exercises 5.7

Write each of the following numbers in percent form.

1. 0.91 91% 2. 0.625 62.5% 3. 1.37 137% 4. 0.0075 0.75%

5. 58

62.5% 6. 87

100 87% 7.

34

75% 8. 112

150%

Write each of the following percents in decimal form.

9. 69% 0.69 10. 82% 0.82 11. 162% 1.62 12. 235% 2.35

13. 7.5% 0.075 14. 11.3% 0.113 15. 0.5% 0.005 16. 31.4% 0.314

Write each of the following percents in fraction form (reduced).

17. 35% 720

18. 72% 1825

19. 130% 1310 20. 40%

25

Solve the following word problems.

21. Budgeting: Heather’s monthly income is $2500. Using the given circle graph answer the following questions.

a. What percent of her income does Heather spend on rent each month? 32%

b. What percent of her income is spent on food and entertainment each month? 28%

c. What percent of her income does Heather save each month? 12%

$300

$300

$300$300

$400

$800RentUtilities

Food

Savings

Other$100

Entertainment

Car expenses

Monthly Expenses

22. Earthquakes: There were 4257 registered earthquakes in the U.S. in 2009. Using the given graph answer the following questions. (Round answers to the nearest tenth of a percent.)

a. What percent of U.S. earthquakes in 2009 were less than a magnitude of 3.0? 55.7%

b. What percent of earthquakes were of magnitude 3.0 – 3.9? 35.0%

c. What percent of earthquakes were 3.0 and above? 43.3%

3.0 – 3.9:1491

4.0 – 4.9: 294

2.0 – 2.9:2372

0 – 1.9: 42

5.0 and above: 58

Magnitude of U.S. Earthquakes in 2009Source: Data from the U.S. National

Earthquake Information Center

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623 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

23. Apple’s Revenue: In 2009, Apple’s revenue was $36 billion. (Round answers to the nearest tenth of a percent.)

a. What percent of Apple’s revenue came from computer sales? 38.9%

b. What percent of Apple’s revenue came from mobile phones? 19.4%

c. What percent of Apple’s revenue did not come from music related products? 66.7%

Computers14,000

Other1000

Apple’s Revenue in 2009(Data is in millions of dollars)Source: http://tech.fortune.cnn.com

Music Players8000

Mobile Phones

7000

Music Store4000

Software2000

24. Music Sales: In July, Daddy Fat’s Record Shop sold 3900 CDs.

a. What percent of their sales came from classical CDs? 12%

b. What percent of their sales came from R&B and rap CDs? 15%

c. What percent of their sales was neither rock nor country CDs? 36%

Daddy Fat's CD Sales for July

Rock R & B and Rap

Classical

Other

Country351

1404

1092

585

468

Find the missing rate, base, or amount. See Example 1.

25. 81% of 76 is __61.56__. 26. 102% of 87 is __ 88.74_. 27. ___ 40__% of 150 is 60.

28. ___150_% of 160 is 240. 29. 3% of __2180__ is 65.4. 30. 20% of ___225___ is 45.

31. What percent of 32 is 40? 125% 32. 1250 is 50% of what number? 2500

33. 100 is 125% of what number? 80 34. Find 24% of 244. 58.56

Solve the following word problems. See Example 2 through 5.

35. Percent of Profit What is the percent of profit if

a. $400 is made on an investment of $5000? 8%

b. $350 is made on an investment of $3500? 10%

c. Which was the better investment? b.

36. Percent of Profit What is the percent of profit if

a. $400 is made on an investment of $2000? 20%

b. $510 is made on an investment of $3000? 17%

c. Which was the better investment? a.

37. Sales Tax: In North Carolina sales tax on food is 4% and sales tax on all other products is 5.75%. At a superstore, Nevaeh buys $72 of food and $51 of other products. What will be the total cost with tax? $128.81

38. Salary: A calculator salesman’s monthly income is $1500 salary plus a commission of 7% of his sales over $5000. What was his income the month that he sold $11,470 in calculators? $1952.90

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Applications: Percent Problems Section 5.7 624

39. Salary Bonus: A computer programmer was told that he would be given a bonus of 5% of any money his programs could save the company. How much would he have to save the company to earn a bonus of $1000? $20,000

40. Property Tax: The property tax on a home was $2550. What was the tax rate if the home was valued at $170,000? 1.5%

41. Basketball: In one season a basketball player missed 14% of her free throws. How many free throws did she make if she attempted 150 free throws? 129 free throws

42. Test Taking: A student missed 6 problems on a statistics test and received a grade of 85%. If all the problems were of equal value, how many problems were on the test? 40 problems

43. Buying Textbooks: If sales tax is figured at 7.25%, how much tax will be added to the total purchase price of three textbooks, priced at $35.00, $55.00, and $70.00? What will be the total amount paid for the books? $11.60; $171.60

44. Buying a Car: A car dealer bought a used car for $2500. He marked up the price so that he would make a profit of 25% of his cost.

a. What was the selling price? $3125 b. If the customer paid 8% of the selling

price in taxes and fees, what was the total cost of the car to the customer? The car was 6 years old. $3375

45. Shopping: At a department store sale, sheets and pillowcases were discounted 25%. Sheets were originally marked $30.00 and pillowcases were originally marked $12.00.

a. What was the sale price of the sheets? $22.50

b. What was the sale price of the pillowcases? $9

46. Golfing: The Golf Pro Shop had a set of golf clubs that were marked on sale for $560. This was a discount of 20% off the original selling price.

a. What was the original selling price? $700 b. If the clubs cost the shop $420, what

was the percent of profit based on cost? 33.33%

47. Losing Weight: A man weighed 200 pounds. He lost 20 pounds in 3 months. Then he gained back 20 pounds 2 months later.

a. What percent of his weight did he lose in the first 3 months? 10%

b. What percent of his weight did he gain back? 11.11%

c. The loss and the gain are the same amount, but the two percents are different. Explain. The first percent is a percent of his original weight, while the second is a percent of his weight after he lost weight.

48. Technology: Data cartridges for computer backup were on sale for $27.00 (a package of two cartridges). This price was a discount of 10% off the original price.

a. Was the original price more than $27 or less than $27? more than $27

b. What was the original price? $30

49. Farming: A citrus farmer figures that his fruit costs 96¢ a pound to grow. If he lost 20% of the crop he produced due to a frost and he sold the remaining 80% at $1.80 a pound, how many pounds did he produce to make a profit of $30,000? 62,500 pounds

50. Buying a Motorcycle: The dealer’s discount to the buyer of a new motorcycle was $499.40. The manufacturer gave a rebate of $1000 on this model.

a. What was the original price if the dealer discount was 7% of the original price?

b. What would a customer pay for the motorcycle if taxes were 5% of the final selling price (after the rebate) and license fees were $642.60? a. $7134.29 b. $6559.23

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625 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

51. Farming: Farmer McGregor raises strawberries. They cost him $0.80 a basket to produce. He is able to sell only 85% of those he produces. If he sells his strawberries at $2.40 a basket, how many baskets must he produce to make a profit of $2480? 2000 baskets

52. A man and his wife wanted to sell their house and contacted a realtor. The realtor said that the price for the services (listing, advertising, selling, and paperwork) was 6% of the selling price. The realtor asked the couple how much cash they wanted after the realtor fee was deducted from the selling price. They said $141,000 and that therefore, the asking price should be 106% of $141,000 or a total of $149,460. The realtor said that this was the wrong figure and that the percent used should be 94% and not 106%.

a. Explain why 106% is the wrong percent and $149,460 is not the right selling price.

b. Explain why and how 94% is the correct percent and just what the selling price should be for the couple to receive $141,000 after the realtor’s fee is deducted.

a. Because the 6% commission is against the selling price, not the $141,000 the couple wanted.

b. The selling price is 100%. Therefore, $141,000 is 94% of the selling price (selling price (100%) − realtor fee (6%) = amount for couple (94%)). The selling price should be $150,000.

Writing & Thinking

With the class separated into teams of two to four students, each team is to analyze the following problem and decide how to answer the related questions. Then each team leader is to present the team’s answers and related ideas to the class for general discussion.

53. Business: Sam works at a picture framing store and gets a salary of $500 per month plus a commission of 3% on sales he makes over $2000. Maria works at the same store, but she has decided to work on a straight commission of 8% of her sales.

a. At what amount of sales will Sam and Maria make the same amount of money? $8800

b. Up to that point, who would be making more? Sam

c. After that point, who would be making more? Explain briefly. (If you were offered a job at that store, which method of payment would you choose?) Maria

Collaborative Learning

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Working with Formulas Section 5.8 626

5.8 Working with Formulas

Objective A Formulas

Formulas are general rules or principles stated mathematically. There are many formulas in such fields of study as business, economics, medicine, physics, and chemistry as well as mathematics. Some of these formulas and their meanings are shown here.

notesBe sure to use the letters just as they are given in the formulas. In mathematics, there is little or no flexibility between capital and small letters as they are used in formulas. In general, capital letters have special meanings that are different from corresponding small letters. For example, capital A may mean the area of a triangle and small a may mean the length of one side, two completely different ideas.

Listed here are a few formulas with a description of the meaning of each formula. There are of course many more formulas.

Formula Meaning

I = PrtThe simple interest I earned by investing money is equal to the product of the principal P times the rate of interest r times the time t in one year or part of a year.

C F= −( )59

32Temperature in degrees Celsius C equals

59

times the

difference between the Fahrenheit temperature F and 32.

d = rtThe distance traveled d equals the product of the rate of speed r and the time t.

P = 2l + 2wThe perimeter, P, of a rectangle is equal to twice the length l plus twice the width w.

L = 2πrhThe lateral surface area L (top and bottom not included) of a cylinder is equal to 2π times the radius r of the base times the height h.

F = maIn physics, the force F acting on an object is equal to its mass m times its acceleration a.

α + β + γ = 180°The sum of the angles (α, β, and γ ) of a triangle is 180°.Note: α, β, and γ are the Greek lowercase letters alpha, beta, and gamma, respectively.

P = a + b + cThe perimeter, P, of a triangle is equal to the sum of the lengths of the three sides a, b, and c.

Table 1

Understand the concept of a formula, and learn several common formulas.

Evaluate formulas for given values of the variables.

Solve formulas for specified variables in terms of the other variables.

A

B

C

Objectives

Teaching Note:You might want to discuss various situations involving interest. For simple interest with t > 1, the interest must be paid at the end of the year because the interest rate is an annual rate. For compound interest, interest is paid on interest and the formula involves exponents:

A Prn

nt

= +

1 .

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Objective B Evaluating Formulas

If you know values for all but one variable in a formula, you can substitute those values and find the value of the unknown variable using the techniques for solving equations discussed earlier. This section presents a variety of formulas from real-life situations, with complete descriptions of the meanings of these formulas. Working with these formulas will help you become familiar with a wide range of applications of algebra and provide practice in solving equations.

Example 1

Evaluating Formulas

a. A note is a loan for a period of 1 year or less, and the interest earned (or paid) is called simple interest. A note involves only one payment at the end of the term of the note and includes both principal and interest. The formula for calculating simple interest is:

I = Prt where I = interest (earned or paid) P = principal (the amount invested or borrowed) r = rate of interest (stated as an annual or yearly

rate) t = time (one year or part of a year)

Note: The rate of interest is usually given in percent form and converted to decimal or fraction form for calculations.

Maribel loaned $5000 to a friend for 3 months at an annual interest rate of 8%. How much will her friend pay her at the end of the 3 months?

Solution

Here, P = $5000, r = 8% = 0.08, and

t = 3 months = 3

1214

year year.=

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Working with Formulas Section 5.8 628

Find the interest by substituting in the formula I = Prt and evaluating.

I = ⋅ ⋅

= ⋅=

5000 0 081

45000 0 02100 00

0 02

.

.$ .

.

The interest is $100 and the amount to be paid at the end of 3 months is principal + interest = $5000 + $100 = $5100.

b. Given the formula C F= −( )59

32 , first find C if F = °212 (212

degrees Fahrenheit) and then find F if C = °20 .

Solution

F = 212°, so substitute 212 for F in the formula.

C = −( )

= ( )=

59

212 32

59

180

100

That is, 212°F is the same as 100°C. Water will boil at 212°F at sea level. This means that if the temperature is measured in degrees Celsius instead of degrees Fahrenheit, water will boil at 100°C at sea level.

C = 20°, so substitute 20 for C in the formula.

2059

32

20 3295

95

59

= −( )

⋅ = ⋅ −( )

F

F

Now solve for .

Multiply both sides

F

bby 9

5.

Simplify.

Add 32 to both sides.

36 3236 32

6832 32

= −= −=

+ +FFF SSimplify.

That is, a temperature of 20°C is the same as a comfortable spring day temperature of 68°F.

c. The lifting force F exerted on an airplane wing is found by multiplying some constant k by the area A of the wing’s surface and by the square of the plane’s velocity v. The formula is F = kAv 2. Find the force on a plane’s wing during take off if the area of the

wing is 120 ft2, k is 43

, and the plane is traveling 80 miles per hour during take off.

Solution

We know that k = 43

, A = 120, and v = 80.

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629 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

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Substitution gives

F

F

= ⋅ ⋅

= ⋅ ⋅

= ⋅=

43

120 80

43

120 6400

160 64001 024 000

2

, , lb The force is meeasured in pounds.

d. The perimeter of a triangle is 38 feet. One side is 5 feet long and a second side is 18 feet long. How long is the third side?

Solution

Using the formula P = a + b + c, substitute P = 38, a = 5, and b = 18. Then solve for the third side.

P = a + b + c 38 = 5 + 18 + c P = 38

c = ?

b = 18

a = 5 38 = 23 + c

38 − 23 = 23 + c − 23 15 = c

The third side is 15 feet long.

1. a. You get a loan for $2000 for 2 months at an annual interest rate of 6%. How much do you have to pay at the end of the loan period, 2 months?

b. Given C F= −59

32( ) ,

find F if C = 50°.

c. Given F kAv= 2 find the

force exerted if k = 54

,

A = 100 ft2, and v = 10

mph.

a. $2020 b. 122° c. 12,500 lb

Now work margin exercise 1.

Objective C Solving Formulas for Different Variables

We say that the formula d = rt is “solved for” d in terms of r and t. Similarly,

the formula A bh=12

is solved for A in terms of b and h, and the formula

P = R − C (profit is equal to revenue minus cost) is solved for P in terms of R and C. Many times we want to use a certain formula in another form. We want the formula “solved for” some variable other than the one given in terms of the remaining variables. Treat the variables just as you would constants in solving linear equations. Study the following examples carefully.

Example 2

Solving for Different Variables

a. Given d = rt, solve for t in terms of d and r. We want to represent the time in terms of distance and rate. We will use this concept later in word problems.

Solution

d rt

d rtr r

=

=

Treat and as if they were constants.

Divide both sides b

r d

yy

Simplify.

.r

dr

t=

d rt

d rtr r

=

=

Treat and as if they were constants.

Divide both sides b

r d

yy

Simplify.

.r

dr

t=

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Working with Formulas Section 5.8 630

b. Given VkP

= , solve for P in terms of V and k.

Solution

VkP

VkP

PV k

PV k

P P

V V

=

⋅ = ⋅

=

=

Multiply both sides by

Simplify.

Divide

.P

bboth sides by

Simplify.

.V

PkV

=

VkP

VkP

PV k

PV k

P P

V V

=

⋅ = ⋅

=

=

Multiply both sides by

Simplify.

Divide

.P

bboth sides by

Simplify.

.V

PkV

=

c. Given C F= −( )59

32 as in Example 1b., solve for F in terms of

C. This would give a formula for finding Fahrenheit temperature given a Celsius temperature value.

Solution

C F

C F

= −( )

⋅ = ⋅ −( )

59

32

59

3295

95

Treat as a constant.

Multiply both si

C

ddes by .

Simplify.

Add to both sides

9

5

32

95

32

95

3232 32

C F

C F

= −

= −+ + ..

Simplify.95

32C F+ =

Thus F C= +95

32 is solved for F, and C F= −( )59

32 is solved for C.

These are two forms of the same formula.

d. Given the equation 2x + 4y = 10, 1. solve first for x in terms of y, and then 2. solve for y in terms of x.

This equation is typical of the algebraic equations that we will discuss in later sections.

Teaching Note:You might ask the students if anyone observed that they could divide by 2 first. Explain that looking for a common factor (or divisor) can make the problem easier to solve later.

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Solution

1. Solving for x yields the following.

2 4 10

2 4 104 4

x y

x y y y

+ =+ =− −

Treat 4 as a constant.

Subtract 4 from bot

y

y hh sides.

(This is the same as adding 4

Simplify.

.)− y

2 10 4

2 102

x y

x

= −

=−−

= −

= −

4

102

42

5 2

2y

xy

x y

Divide both sides by 2.

Simplify.

2. Solving for y yields the following.

2 4 10

2 4 102 2

x y

x y x x

+ =+ =− −

Treat 2 as a constant.

Subtract from bo2

x

x tth sides.

Simplify.

Divide both sides by .4

4 10 2

4 10 2

14 4

y x

y x

y

= −

=−

=004

24

52 2

= −

x

yx

Simplify.

or we can write

or yx

y x=−

= − +5

212

52

. Both forms are correct.

e. Given 3x − y = 15, solve for y in terms of x.

Solution

Solving for y gives

3 15

3 15

15 3

3 3

1

x y

x y

y x

y

x x

− =− =

− = −−( )

− −

Subtract from both sides.3x

== −( )−1 15 3x Multiply both sides by (or divide both

sides by )

1

1

−− ..

Simplify using the distributive property.y x

y x

= − += −

15 3

3 15or

2. a. Given F = kAv2 solve for k in terms of F, A and v.

b. Given the equation 16y + 25z = 400,

1. solve for y in terms of z, and then

2. solve for z in terms of y.

c. Given 4x + 8y + 12z = 16, solve for x.

a. kF

Av= 2

b. yz z

zy y

=−

= −

=−

= −

400 2516

252516

400 1625

161625

c. x = 4 − 2y − 3z

Now work margin exercise 2.

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Working with Formulas Section 5.8 632

1. 2x − y = 5; solve for y. 2. 2x − y = 5; solve for x.

3. A bh=12

; solve for h. 4. L = 2πrh; solve for r.

5. P = 2l + 2w; solve for w.

Practice Problems

Practice Problem Answers

1. y = 2x − 5 2. xy

=+ 52

3. hAb

=2

4. rL

h=

2π 5. w

P l=

− 22

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633 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

Exercises 5.8

Refer to Example 1a for information concerning simple interest and the related formula I = Prt.

Simple Interest

1. You want to borrow $4000 at 12% for 3 months. How much interest would you pay? $120

2. For how many years must you leave $1000 in a savings account at 5.5% interest to earn $220 in interest? 4 years

3. What principal would you need to invest to earn $450 in simple interest in 6 months if the interest rate was 9%? $10,000

4. After 1 month, Gustav received $25 in simple interest on his savings account of $12,000. What was the interest rate? 2.5%

5. A savings account of $3500 is left for 9 months and draws simple interest at a rate of 7%.

a. How much interest is earned? $183.75 b. What is the balance in the account at the

end of the 9 months? $3683.75

6. Tim just deposited $2562.50 to pay off a 3 month loan of $2500.

a. How much of what he deposited was interest on the loan? $62.50

b. What rate of interest was he charged? 10%

In the following application problems, read the descriptive information carefully and then substitute the values given in the problem for the corresponding variables in the formulas. Evaluate the resulting expression for the unknown variable. See Example 1

VelocityIf an object is shot upward with an initial velocity v0 in feet per second, the velocity v in feet per second is given by the formula v = v0 − 32t, where t is time in seconds. (v0 is read “v sub zero.” The 0 is called a subscript.)

7. An object projected upward with an initial velocity of 106 feet per second has a velocity of 42 feet per second. How many seconds have passed? 2 seconds

8. Find the initial velocity of an object if the velocity after 4 seconds is 48 feet per second. 176 ft/sec

MedicineIn nursing, one procedure for determining the dosage for a child is

child’s dosageage of child in years

ageof child + 12 adult dosag = ⋅ ee.

9. If the adult dosage of a drug is 20 milliliters, how much should a 3-year-old child receive? 4 milliliters

10. If the adult dosage of a drug is 340 milligrams, how much should a 5-year-old child receive? 100 milligrams

InvestmentThe total amount of money in an account with P dollars invested in it is given by the formula A = P + Prt, where r is the rate expressed as a decimal and t is time (one year or part of a year).

11. If $1000 is invested at 6% interest, find the total amount in the account after 6 months. $1030

12. How long will it take an investment of $600, at an annual rate of 5%, to be worth $615? 0.5 year (6 months)

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Working with Formulas Section 5.8 634

CarpentryThe number N of rafters in a roof or studs in a wall can be found by the formula N

Ld

= + 1, where L is the

length of the roof or wall and d is the center-to-center distance from one rafter or stud to the next. Note

that L and d must be in the same units.

13. How many rafters will be needed to build a roof 26 ft long if they are placed 2 ft on center? 14 rafters

26 ft

14. A wall has studs placed 16 in. on center. If the wall is 20 ft long, how many studs are in the wall? 16 studs

20 ft

16 in.16 in.

15. How long is a wall if it requires 22 studs placed 16 in. on center? 336 in. or 28 ft

16. What should the center-to-center distance be if you are building a 33 ft long roof using 12 rafters? 3 ft

CostThe total cost C of producing x items can be found by the formula C = ax + k, where a is the cost per item and k is the fixed costs (rent, utilities, and so on).

17. Find the total cost of producing 30 items if each costs $15 and the fixed costs are $580. $1030

18. The total cost to produce 80 dolls is $1097.50. If each doll costs $9.50 to produce, find the fixed costs. $337.50

19. It costs a company $3.60 to produce a calculator. Last week the total costs were $1308. If the fixed costs are $480 weekly, how many calculators were produced last week? 230 calculators

20. Each week an electronics company builds 60 mp3 players for a total cost of $5340. If the fixed costs for a week are $750, what is the cost to produce each mp3 player? $76.50

ProfitThe profit P is given by the formula P = R − C, where R is the revenue and C is the cost.

21. Find the revenue (income) of a company that shows a profit of $3.2 million and costs of $1.8 million. $5 million

22. Find the revenue of a company that shows a profit of $3.2 million and costs of $5.7 million. $8.9 million

DepreciationMany items decrease in value as time passes. This decrease in value is called depreciation. One type of depreciation is called linear depreciation. The value V of an item after t years is given by V = C − Crt, where C is the original cost and r is the annual rate of depreciation expressed as a decimal.

23. If you buy a car for $6000 and it depreciates linearly at a rate of 10% per year, what will be its value after 6 years? $2400

24. A contractor buys a 4 year-old piece of heavy equipment valued at $20,000. If the original cost of this equipment was $25,000, find the annual rate of depreciation. 5%

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635 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

Distance, Rate, and TimeThe distance traveled d is given by the formula d = rt, where r is the rate of speed and t is the time it takes.

25. How long will a truck driver take to travel 350 miles if his mean speed is 50 mph? 7 hours

26. What is the mean rate of speed of a biker who bikes 21.92 miles in 68.5 min? 0.32 miles per minute

27. What is Jonathan’s mean rate of speed if he hikes 10.4 miles in 6.4 hours? 1.625 miles per hour

28. How long will it take a train traveling at 40 mph to go 140 miles? 3.5 hours

Solve each formula for the indicated variable.

29. P = a + b + c; solve for b. 30. P = 3s; solve for s. 31. F = ma; solve for m.

b = P − a − c sP

=3

mFa

=

32. C = πd; solve for d. 33. A = lw; solve for w. 34. P = R − C; solve for C.

dC

wAl

= C = R − P

35. R = np; solve for n. 36. v = k + gt; solve for k. 37. I = A − P; solve for P.

nRp

= k = v − gt P = A − I

38. L = 2πrh; solve for h. 39. Am n

=+2

; solve for m. 40. P = a + 2b; solve for a.

hL

r=

2π m = 2A − n a = P − 2b

41. I = Prt; solve for t. 42. REI

= ; solve for E. 43. P = a + 2b; solve for b.

tI

Pr= E = RI b

P a=

−2

44. c 2 = a 2 + b2; solve for b2. 45. α + β + γ = 180°; solve for β. 46. y = mx + b; solve for x.

b c a2 2 2= − β = 180° − α − γ xy b

m=

47. V = lwh; solve for h. 48. v = −gt + v00; solve for t. 49. A bh=

12

; solve for b.

hVlw

= tv v

g=

−0 bAh

=2

50. REI

= ; solve for I. 51. A = πr 2; solve for π. 52. ARL

=2

; solve for L.

IER

= π =A

r 2 L

RA

=2

53. Kmv

g=

2

2; solve for g. 54. x + 4y = 4; solve for y. 55. 2x + 3y = 6; solve for y.

gmv

K=

2

2 y

x=

−44

yx

=−6 23

56. 3x − y = 14; solve for y. 57. 5x + 2y = 11; solve for x. 58. −2x + 2y = 5; solve for x.

y = 3x − 14 xy

=−11 25

xy

=−2 5

2

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Working with Formulas Section 5.8 636

59. A h b c= +( )12

; solve for b. 60. A P rt= +( )1 ; solve for r. 61. Rx

=−( )3 12

8; solve for x.

bA hc

h

bAh

c

=−

= −

2

2

or rA P

Pt=

− xR

xR

=+

= +

8 363

83

12

or

62. − − = − +( )2 5 3x x y ; solve for x.

63. 3y − 2 = x + 4y + 10; solve for y.

64. V r h=13

2π ; solve for h.

x = 5 − 3y y = −x − 12 hV

r=

32π

Make up a formula for each of the following situations.

65. Concert Tickets: Each ticket for a concert costs $t per person and parking costs $9.00. What is the total cost per car C if there are n people in a car? C = nt + 9

66. Car Rentals: ABC Car Rental charges $25 per day plus $0.12 per mile. What would you pay per day for renting a car from ABC if you were to drive the car x miles in one day? C = 0.12x + 25

67. Computers: Top-of-the-Line computer company knows that the cost (labor and materials) of producing a computer is $325 per computer per week and the fixed overhead costs (lighting, rent, etc.) are $5400 per week. What are the company’s weekly costs of producing n computers per week? C = 325n + 5400

68. Computers: If the Top-of-the-Line computer company (see Exercise 67) sells its computers for $683 each, what is its profit per week if it sells the same number n that it produces? (Remember that profit is equal to revenue minus costs, or P = R − C.) P = 358n − 5400

69. The formula zx x

s=

− is used extensively in

statistics. In this formula, x represents one of a set of numbers, x represents the mean of those numbers in the set, and s represents a value called the standard deviation of the numbers. (The standard deviation is a positive number and is a measure of how “spread out” the numbers are.) The values for z are called z-scores, and they measure the number of standard deviation units a number x is from the mean x.

a. If x = 70, what will be the z-score for x = 70? Does this z-score depend on the value of s? Explain.

b. If x = 70, for what values of x will the corresponding z-scores be negative?

c. Calculate your z-score on each of the last two test scores in this class. (Your instructor will give you the mean and standard deviation for each test.) What do these scores tell you about your performance on the two exams? a. 0; No, because the numerator will be zero and thus the whole fraction will be equal to zero for all values of s. b. x < 70 c. Answers will vary.

70. Suppose that, for a particular set of exam scores, x = 72 and s = 6. Find the z-score that corresponds to a score of

a. 78 b. 66 c. 81 d. 60 a. 1 b. −1 c. 1.5 d. -2

x x0

0.050.100.150.200.250.300.350.400.45

Writing & Thinking

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637 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

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5.9 Applications: Distance-Rate-Time, Interest, Mean, Cost

Objective A Distance-Rate-Time

Word problems (or applications) are designed to teach you to read carefully, to organize, and to think clearly. Whether or not a particular problem is easy for you depends a great deal on your personal experiences and general reasoning abilities. The problems generally do not give specific directions to add, subtract, multiply, or divide. You must decide what relationships are indicated through careful analysis of the problem.

Problems involving distance usually make use of the relationship indicated by the formula d = rt, where d = distance, r = rate, and t = time. A chart or table showing the known and unknown values is quite helpful and is illustrated in the next example.

Example 1

Distance-Rate-Time

A motorist has a mean of 45 mph for the first part of a trip and 54 mph for the last part of the trip. If the total trip of 303 miles took 6 hours, what was the time for each part?

Solution

Analysis of strategy:

What is being asked for?

Total time minus time for 1st part of trip gives time for 2nd part of trip.

Solve distance-rate-time problems using linear equations.

Solve simple interest problems using linear equations.

Solve mean problems using linear equations.

Solve cost problems using linear equations.

A

B

C

D

Objectives

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Applications: Distance-Rate-Time, Interest, Mean, Cost Section 5.9 638

Let t = time for 1st part of trip and 6 − t = time for 2nd part of trip.

rate ⋅ time = distance

1st Part 45 t 45t

2nd Part 54 6 − t 54 6 −( )t

t hr 6 − t hr

303 miles

54 mph45 mph

+ =

1st part

distance���

2nd part

distance���

total

distance���

45t + 54 6 −( )t = 303 Solve the equation.

45t + 324 − 54t = 303 324 − 9t = 303 − 9t = −21

t = −−219

= 73

hr 1st part of the trip

6 − t = 673

113

− = hr 2nd part of the trip

Check:

4573

15 7 105 1

54113

18 11 198 2

⋅ = ⋅ =

⋅ = ⋅ =

miles ( st part

miles ( nd p

)

aart)

105 + 198 = 303 miles in total

The first part took 73

213

hr or hr.

The second part took 113

323

hr or hr.

1. What is the time for each part of a trip if the speed of the first part is 60 mph and the speed of the second part is 45 mph and the total distance of 200 miles is covered in 4 hours?

The first part of the trip

took 43

or 113

hours and

the second part of the trip

took 83

or 223

hours.

Form an equation relating

the given information.

Now work margin exercise 1.

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Objective B Interest

To solve problems related to interest on money invested for one year, you need to know the basic relationship among the principal P (amount invested), the annual rate of interest r, and the amount of interest I (money earned). This relationship is described in the formula P r I⋅ = . (This is the formula for simple interest, I = Prt, with t = 1.) We use this relationship in Example 2.

Example 2

Interest

Kara has had $40,000 invested for one year, some in a savings account which paid 7% and the rest in a high-risk stock which yielded 12% for the year. If her interest income last year was $3550, how much did she have in the savings account and how much did she invest in the stock?

Solution

Let x = amount invested at 7% and 40,000 − x = amount invested at 12%.

principal ⋅ rate = interest

Savings Account x 0.07 0 07. x( )Stock 40,000 − x 0.12 0 12 40 000. , −( )x

Check:

25 000 0 07 1750 15 000 0 12 1800, . , .( ) = ( ) =and and $1750 + $1800 = $3550.

Kara had $25,000 in the savings account at 7% interest and invested $15,000 in the stock at 12% interest.

2. Mark has had $15,000 invested for one year, some in a low-risk stock yielding 4% and the rest in a high-risk stock yielding 15%. If his investment has earned $1975 in interest, how much is in the low-risk stock and how much is in the high-risk stock?

low-risk stock: $2500; high-risk stock: $12,500 Total amount invested

minus amount invested at 7% represents amount invested at 12%.

Multiply both sides of the equation by 100 to eliminate the decimal.

amount invested at 7%

amount invested at 12%

interest at 7% interest at 12% total interest� �� �� � �� �� � ��+ = ���

0.07 0.12 3550

7 12 40,000 355,0007 48

x

x xx

( ) −( )

−( )

+ =

+ =+

40 000, x

00,000 12 355,0005 125,000

25,00040,000 15,000

−− −

xxxx

====

Now work margin exercise 2.

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Applications: Distance-Rate-Time, Interest, Mean, Cost Section 5.9 640

Objective C Mean

The mean of a set of numbers was defined and discussed earlier in the book. In this section we use the concept of mean to find unknown numbers and to read and understand graphs in more detail.

Example 3

Mean

Suppose that you have scores of 85, 92, 82 and 88 on four exams in your English class. What score will you need on the fifth exam to have a mean of 90?

Solution

Let x = your score on the fifth exam.

The sum of all the scores, including the unknown fifth exam, divided by 5 must equal 90.

85 92 82 885

90

3475

90

5347

55 90

347 450

103

+ + + +=

+=

⋅+

= ⋅

+ ==

x

x

x

x

x

Assuming that each exam is worth 100 points, you cannot attain an mean of 90 on the five exams.

3. Suppose on four of your five biology tests you scored 90, 95, 97, and 90 respectively. You can exempt the final if your mean on the 5 tests is greater than or equal to an 85. What do you need to make on your last test to have a mean of 85?

You need to make at least a 53 on the last test to exempt the final.

Now work margin exercise 3.

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Example 4

Bar Graphs

The bar graph shows the enrollment at the main campuses at six Big Ten universities. Use the graph to find the following (note that the units on the graph are in thousands):

Enrollment at Main Campuses of

Big Ten Universities in 2009

0102030405060

Indiana

Iowa

Michig

an State

Northweste

rn

Ohio Sta

te

Penn State

Nu

mb

er o

f Stu

den

ts E

nro

lled

(in

100

0’s)

42

30

47

15

5544

Big Ten Universities

a. Find the mean enrollment over the six schools. (Round to the nearest thousand.)

Solution

Find the sum: 42 30 47 15 55 44 233+ + + + + = .Divide by 6: 233 6 39÷ ≈ .

The mean enrollment is approximately 39,000 students.

b. Which university has the lowest enrollment?

Solution

Northwestern has the lowest enrollment: 15,000 students.

c. Find the difference in enrollment between Ohio State and Penn State.

Solution

The difference is 55,000 − 44,000 = 11,000 students.

4. Using the graph in Example 4, find the difference in enrollment between the largest university and the smallest university.

40,000 students

Now work margin exercise 4.

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Applications: Distance-Rate-Time, Interest, Mean, Cost Section 5.9 642

Objective D Cost

Example 5

A jeweler paid $350 for a ring. He wants to price the ring for sale so that he can give a 30% discount on the marked selling price and still make a profit of 20% on his cost. What should be the marked selling price of the ring?

Solution

Again, we make use of the relationship:

S − C = P (selling price − cost = profit).

Let x = marked selling price then x − 0.30x = actual selling price, and 350 = cost.

Check:

Step 1: Step 2:

$600 $600

0.30

marked selling price marked selling price

discount × %% discount

discount actual selling price

− $180

$180 $420

Step 3:

$420 actual selling price

$350 cost

$70 profit

As a double check

,, $350

0.20

$70

cost

profit %

profit.

×

The jeweler should set the selling price at $600.

5. A used car salesman paid $1500 for a car. He plans to sell the car at a 25% discount on marked selling price, but still wants to make a profit of 15% on his cost. What should be the maked selling price of the car?

$2300

actual

selling pricecost profit

� �� ��� �− =

− − = ( )x x0 30 350 0 20 350. .00 70 350 70

0 70 420600

..

xx

x

− ===

The profit is 20% of what he paid originally.

Now work margin exercise 5.

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643 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

Exercises 5.9

Solve each of the following application problems.

1. Hiking: What is Nathan’s mean rate of speed if he hikes 12.6 miles in 7.5 hours? 1.68 mph

2. Biking: What is Amy’s mean rate of speed if she bikes 39.6 miles in 2.2 hours? 18 mph

3. Traveling by Car: Jamie plans to take the scenic route from Los Angeles to San Francisco. Her GPS tells her it is a 420 mile trip. If she figures her mean speed will be 48 mph, how long will the trip take her? 8.75 hours

4. Traveling by Car: Scott's mean speed on his drive from Memphis, TN is 60 mph. If the total trip is 285 miles, how long should he expect the drive to take? 4.75 hours

5. Biking: Jane rides her bike to Lake Junaluska. Going to the lake, her mean speed is 12 mph. On the return trip, her mean speed is 10 mph. If the round trip takes a total of 5.5 hours, how long does the return trip take?

rate ⋅ time = distance

Going 12 5.5 − t

Returning 10 t

3 hours

6. Plane Speeds: Two planes which are 2475 miles apart fly toward each other. Their speeds differ by 75 mph. If they pass each other in 3 hours, what is the speed of each?

rate ⋅ time = distance

1st plane r 3

2nd plane r + 75 3

375 mph; 450 mph

7. Traveling by Car: Marcus drives from Chicago to Detroit in 6 hours. On the return trip, his speed is increased by 10 mph and the trip takes 5 hours. Find his rate on the return trip. How far apart are the towns?

rate ⋅ time = distance

Going r 6

Returning r + 10 5

60 mph; 300 miles

8. Hiking: Tim and Barb have 8 hours to spend on a mountain hike. They can walk up the trail at a mean rate of 2 mph and can walk down at a mean of 3 mph. How long should they plan to hike uphill before turning around?

rate ⋅ time = distance

Uphill 2 x

Downhill 3 8 − x

4.8 hours

9. Traveling by Car: The Reeds are moving

across Texas. Mr. Reed leaves 312

hours

before Mrs. Reed. If his mean speed is 40 mph and her mean speed is 60 mph, how long will Mrs. Reed have to drive before she overtakes Mr. Reed? 7 hours

10. Traveling by Car: After traveling for

40 minutes, Mr. Koole had to slow to 23

his

original speed for the rest of the trip due to heavy traffic. The total trip of 84 miles took 2 hours. Find his original speed. 54 mph

11. Train Speeds: A train leaves Cincinnati at 2:00 P.M. A second train leaves the same station in the same direction at 4:00 P.M. The second train travels 24 mph faster than the first. If the second train overtakes the first at 7:00 P.M., what is the speed of each of the two trains? 36 mph; 60 mph

12. Running: Maria runs through the countryside at a rate of 10 mph. She returns along the same route at 6 mph. If the total trip took 1 hour 36 minutes, how far did she run in total? 12 miles

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Applications: Distance-Rate-Time, Interest, Mean, Cost Section 5.9 644

13. Traveling by Car: The distance from Atlanta, Georgia to Washington, DC is 620 miles. Driving in the middle of the night it takes about 9 hours to get to Washington. Due to higher traffic volume, it takes 2 more hours to travel there during the day. What is the mean rate of the driver during the day and during the night? (Round to the nearest whole number.) day: 56 mph; night: 69 mph

14. Traveling by Car: Mr. Kent drove to a conference. The first half of the trip took 3 hours due to traffic. Traffic let up for the second half of the trip and he was able to increase his speed by 20 mph to make sure he got there on time. Find his rates of speed if he traveled 2 hours at the second rate. 1st half : 40 mph; 2nd half: 60 mph

15. Exercising: Jayden walked to his friend’s house at a rate of 4 mph to borrow his friend’s bicycle. Coming back home, he rode the bicycle at a mean rate of 12 mph. The total time for the round trip was 1 hour 30 minutes. How far away does Jayden’s friend live? 4.5 miles

16. Exercising: Once a week Felicia walks/runs for a total of 6 miles. Felicia spends twice as much time walking as she does running. If she walks at a rate of 4 mph and runs three times faster than she walks, what is the time for each part? walks for 0.6 hours or 36 minutes; runs for 0.3 hours or 18 minutes

17. Investing: Amanda invests $25,000, part at 5% and the rest at 6%. The annual return on the 5% investment exceeds the annual return on the 6% investment by $40. How much did she invest at each rate? $14,000 at 5%; $11,000 at 6%

18. Investing: Mr. Hill invests $10,000, part at 5.5% and part at 6%. The annual interest from the 5.5% investment exceeds the annual interest from the 6% investment by $251. How much did he invest at each rate? $7400 at 5.5%; $2600 at 6%

19. Investing: The annual interest earned on a $6000 investment was $120 less than the interest earned on $10,000 invested at 1% less interest per year. What was the rate of interest on each amount? 4.5% on $10,000; 5.5% on $6000

20. Investing: Two investments totaling $16,000 produce an annual income of $1140. One investment yields 6% a year, while the other yields 8% per year. How much is invested at each rate? $7000 at 6%; $9000 at 8%

21. Investing: The annual interest on a $4000 investment exceeds the interest earned on a $3000 investment by $80. The $4000 is invested at a 0.5% higher rate of interest than the $3000. What is the interest rate of each investment? 6.5% on $4000; 6% on $3000

22. Investing: D’Andra makes two investments that total $12,000. One investment yields 8% per year and the other 10% per year. The total interest for one year is $1090. Find the amount invested at each rate. $5500 at 8%; $6500 at 10%

23. Investing: A company is planning to invest $42,000 into two simple interest accounts. The annual interest rate on one of the accounts is 4.5% while the rate on the other is 6%. How much should the company invest in each account so that the two accounts will produce an equal annual interest income? $24,000 at 4.5%; $18,000 at 6%

24. Shopping: A particular style of shoe costs the dealer $81 per pair. At what price should the dealer mark them so he can sell them at a 10% discount off the selling price and still make a 25% profit? $112.50

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645 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

25. Investing: Sebastian would like both of his investments, a total of $12,000, to bring him the same annual interest income. One of his investments is at 5.5% annual interest rate and the other at 7%. Find the amount of money that Sebastian should invest in each account. $6720 at 5.5%; $5280 at 7%

26. Used Car Sales: A car dealer paid $2850 for a used car. For his upcoming Labor Day sale, he wants to offer a 5% discount off the posted selling price, but would still like to make a 40% profit. What price should he advertise for that car? $4200

27. Investing: Gabriella got some money from her grandparents as a graduation present. She decided to invest all of it. Part of the money was invested at a 2.5% interest rate, and the rest at a 4% interest rate. She invested $200 more in the 4% account than the 2.5% account. If her annual interest income was $47, how much did she invest at each rate? $600 at 2.5%; $800 at 4%

28. Investing: Jordan is earning 1.5% interest from money invested in a savings account and 4% interest on a mutual bond fund. If the total of his investments is $18,000 and the annual interest from the savings account is less than the annual interest from the bond by $60, how much has Jordan invested at each rate? $12,000 at 1.5%; $6000 at 4%

29. Investing: A small company invested $20,500 such that a part of the money is in an account with a 4% interest rate and the rest at a 5% rate. The annual interest from the 5% account is $35 more than the interest earned from the 4% account. Find the amount of money the company invested at each rate. $11,000 at 4%; $9500 at 5%

30. Shopping: During the month of January, a department store would like to have a sale of 40% off of women’s long boots. The store purchased the boots for $33 per pair. How much should the selling price be, if the manager of the store wants to make a 10% profit per pair? $60.50

31. Shopping: Carla plans on buying a new pair of sandals for the summer. They are on sale for 20% off of the original price. What was the original price of the sandals, if she pays $34.83, with 7.5% sales tax? $40.50

32. Investing: Last year an individual invested some money at a 5% interest rate and $2200 less than that amount at a 6% interest rate. If his interest income was $880, how much did he invest at each rate? $9200 at 5%; $7000 at 6%

33. Shopping: A store purchased a certain style of leather jacket at $70 per jacket. If the store wants to sell the jackets at a 20% discount and still make a profit of 30%, what should be the marked selling price for each jacket? $113.75

34. Electronics: After receiving a 10% off coupon in the mail, Mark decided that it was time to buy a Blu-Ray player. Using the coupon and paying 8% sales tax, the final price came to $213.84. What was the listed price of the Blu-Ray player? $220

35. Exam Scores: Marissa has five exam scores of 75, 82, 90, 85, and 77 in her chemistry class. What score does she need on the final exam to have a mean grade of 80 (and thus earn a grade of B)? (All exams have a maximum of 100 points.) 71

36. Exam Scores: Gerald had scores of 80, 92, 89, and 95 on four exams in his algebra class. What score will he need on his fifth exam to have an overall mean grade of 90? (All exams have a maximum of 100 points.) 94

37. Biking: While riding her bike to the park and back home five times, Stacey timed herself at 60 min, 62 min, 55 min (the wind was helping), 58 min, and 63 min. She had set a goal of having a mean time of 60 minutes for her rides. How many minutes will she need on her sixth ride to attain her goal? 62 min

38. Cell Phones: For every 4 week period, Lauren wants to make a mean of 6 phone calls per week from her prepaid cell phone. The first week she made 9 phone calls, the second week 6 phone calls, and the third week 5 phone calls. How many phone calls does Lauren need to make in the fourth week to make sure she stays on track with her goal? 4 phone calls

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Applications: Distance-Rate-Time, Interest, Mean, Cost Section 5.9 646

39. TV Watching: While growing up, Jason was allowed to watch TV a mean of 3 hours a day over a one week period. One particular week he watched 1 hour, 2 hours, 1 hour, 3 hours, 3 hours, and 5 hours. How many hours could Jason watch the seventh and last day of the week and still obey his parents? 6 hours

40. Budgeting: A college student realized that he was spending too much money on video games. For the remaining 5 months of the year his goal is to spend a mean of $50 a month towards his hobby. How much can he spend in December, taking into consideration that the other 4 months he spent $70, $25, $105, $30 respectively? $20

41. Exam Scores: Wade has scores of 59, 68, 76, 84 and 69 on the first five tests in his social studies class. He knows that the final exam counts as two tests. What score will he need on the final to have a mean of 70? (All tests and exams have a maximum of 100 points.) 67

42. Exam Scores: A statistics student has grades of 86, 91, 95, and 76 on four hour-long exams. What score must he receive on the final exam to have a mean grade of 90 if:

a. the final is equivalent to a single hour-long exam (100 points maximum)? not possible

b. the final is equivalent to two hour-long exams (200 points maximum)? 192

43. Temperature: Given the monthly temperatures over a year for Christchurch, New Zealand: Source: http://www.weather.com

63 6360

55

49

44 4446

50

5457

61

40

45

50

55

60

65

F)

Monthly Temperatures in Christchurch, New Zealand

Months of the Year

Tem

per

atu

re i

n D

egre

es (

˚

a. Find the mean temperature for the year. (Round to the nearest tenth.) 53.8°F

b. Find the minimum mean temperature for the year. 44°F

c. Find the difference in mean temperature between the months of June and December. −17°F

44. Mean Rainfall: Given the means of monthly rainfall over a year for Visakhapatnam, India: Source: http://www.weather.com

0.5 0.5 0.71.5

1.9

4.1

5.2

6.7 6.37.1

3.1

0.2

0.01.02.03.04.05.06.07.08.0

Mean Monthly Rainfall in Visakhapatnam, India

Months of the Year

Rai

nfa

ll in

Inch

es

a. Find the mean rainfall for the year. (Round to the nearest tenth.) 3.2 inches

b. Find the maximum mean rainfall for the year. 7.1 inches

c. Find the difference in mean rainfall between the months of October and December. 6.9 inches

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647 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

45. Company Profits: Given the yearly profits of the world’s largest companies in 2009: Source:

http://money.cnn.com/magazines/fortune/global500/2009/

full_list/

Companies

Pro

fits i

n B

illio

ns

of

Do

llars

0

10

20

30

40

50

Royal Dutch Shell

Exxon Mobil

WalmartStores

BP Chevron Total

26

45

13

2124

16

Profits of World's Largest Companies in 2009

a. Find the mean profits of the companies in the year 2009. (Round to the nearest tenth of a billion.) $24.2 billion

b. Find the yearly profits of Royal Dutch Shell in the year 2009. $26 billion

c. What was the difference in profits between Exxon Mobil and BP? $24 billion

46. Airport Usage: Given the number of passengers at the following airports: Source:

Airports Council International – North America

88

66

64

61

57

0 20 40 60 80 100

Atlanta, GA

Chicago, IL

London

Tokyo

Los Angeles, CA

Number of Passengers at World Airports in 2009

Number of Passengers (Approximated in Millions)

Airp

ort

s

a. Find the mean number of passengers.

b. Find the difference in passengers between Atlanta, GA and Tokyo. 27 million

c. What was the total number of passengers to go through London? 66 million

a. 67.2 million

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Key Terms and Ideas Chapter 5 648

Chapter 5: Index of Key Terms and Ideas

Algebraic Vocabulary page 565Constant: A single numberTerm: Any constant or variable or the indicated product and/or quotient of constants and variablesNumerical coefficient: The number written next to a variableAlgebraic expression: A combination of variables and numbers using any of the operations of addition, subtraction, multiplication, or division, as well as exponents

Like Terms pages 565 – 566Like terms (or similar terms) are terms that are constants or terms that contain the same variables that are raised to the same exponents.

Combining Like Terms pages 566 – 568 To combine like terms, add (or subtract) the coefficients and keep the common variable expression.

Evaluating Algebraic Expressions pages 568 – 5691. Combine like terms, if possible.2. Substitute the values given for any variables.3. Follow the rules for order of operations.

Section 5.1: Simplifying and Evaluating Algebraic Expressions

Translating English Phrases and Algebraic Expressions pages 573 – 574

Addition Subtraction Multiplication DivisionExponentPowers)

add subtract (from) multiply divide square ofsum difference product quotient cube ofplus minus times

more than less than twice

increased by decreased byof (with fractions

and percent)total less

Section 5.2: Translating English Phrases and Algebraic Expressions

Solution page 580 A solution to an equation is a number that gives a true statement when substituted for the variable in the equation.

Linear Equations in x page 580If a, b, and c are constants and a ≠ 0, then a linear equation in x (or first-degree equation in x) is an equation that can be written in the form ax + b = c.

Section 5.3: Solving Linear Equations: x + b = c and ax = c

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Addition Principle of Equality pages 581 – 582If the same algebraic expression is added to both sides of an equation, the new equation has the same solutions as the original equation. Symbolically, if A, B, and C are algebraic expressions, then the equationsA = B and A + C = B + C have the same solutions.

Equivalent Equations pages 581 – 582 Equations with the same solutions are said to be equivalent equations.

Procedure for Solving Linear Equations that Simplify to the Form x + b = c pages 582 –5831. Combine any like terms on each side of the equation.2. Use the addition principle of equality and add the opposite of the constant

b to both sides. The objective is to isolate the variable on one side of the equation (either the left side or the right side) with a coefficient of +1.

3. Check your answer by substituting it for the variable in the original equation.

Multiplication (or Division) Principle of Equality page 584If both sides of an equation are multiplied by (or divided by) the same nonzero constant, the new equation has the same solutions as the original equation. Symbolically, if A and B are algebraic expressions and C is any

nonzero constant, then the following equations have the same solutions.

A B AC BC

AC

BC

= = ≠

= ≠

and where C 0

and where C 0

Procedure for Solving Linear Equations that Simplify to the Form ax = c page 584 – 5861. Combine any like terms on each side of the equation.

2. Use the multiplication (or division) principle of equality and multiply both sides of the equation by the reciprocal of the coefficient of the variable. (Note: This is the same as dividing both sides of the equation by the coefficient.) Thus the coefficient of the variable will become +1.

3. Check your answer by substituting it for the variable in the original equation.

Section 5.3: Solving Linear Equations: x + b = c and ax = c (cont.)

Procedure for Solving Linear Equations that Simplify to the Form ax + b = c pages 591 – 5921. Combine like terms on both sides of the equation.2. Use the addition principle of equality and add the opposite of the constant

b to both sides.3. Use the multiplication (or division) principle of equality to multiply both

sides by the reciprocal of the coefficient of the variable (or divide both sides by the coefficient itself). The coefficient of the variable will become +1.

4. Check your answer by substituting it for the variable in the original equation.

Section 5.4: Solving Linear Equations: ax + b = c

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Key Terms and Ideas Chapter 5 650

Procedure for Solving Linear Equations that Simplify to the Form ax + b = cx + d pages 598 – 5991. Simplify each side of the equation by removing any grouping symbols and

combining like terms on both sides of the equation. 2. Use the addition principle of equality and add the opposite of a constant

term and/or variable term to both sides so that variables are on one side and constants are on the other side.

3. Use the multiplication (or division) principle of equality to multiply both sides by the reciprocal of the coefficient of the variable (or divide both sides by the coefficient itself). The coefficient of the variable will become +1.

4. Check your answer by substituting it for the variable in the original equation.

Types of Equations and Their Solutions pages 601– 602Type of Equation Number of Solutions Conditional Finite number of solutions Identity Infinite number of solutions Contradiction No solutions

Section 5.5: Solving Linear Equations: ax + b = cx + d

Consecutive Integers pages 608 – 609 Integers are consecutive if each is 1 more than the previous integer. Three consecutive integers can be represented as n, n + 1, and n + 2.

Consecutive Even Integers page 609Even integers are consecutive if each is 2 more than the previous even integer. Three consecutive even integers can be represented as 2n, 2n + 2, and 2n + 4, where n is an integer.

Consecutive Odd Integers page 609Odd integers are consecutive if each is 2 more than the previous odd integer. Three consecutive odd integers can be represented as 2n + 1, 2n + 3, and 2n + 5, where n is an integer.

Section 5.6: Applications: Number Problems and Consecutive Integers

The Basic Percent Formula page 617R ⋅ B = A (or A = R ⋅ B)R = RATE or percent (as a decimal or fraction)B = BASE (number we are finding the percent of)A = AMOUNT (a part of the base)“of” means to multiply.“is” means equal =( ).

Discount page 619A reduction in the selling price stated as a percent of the original price.

Sales Tax paeg 619A tax charged on goods sold by retailers (the rate, a percent, varies by location).

Section 5.7: Applications: Percent Problems

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Commission page 620 – 621A fee paid to an agent or salesperson for a service.

Percent of Profit page 621A fraction (or ratio) of profit divided by the investment, changed to a percent.

Section 5.7: Applications: Percent Problems (cont.)

Lists of Common Formulas page 627Formula MeaningI = Prt The simple interest I earned by investing money is equal

to the product of the principal P times the rate of interest r times the time t in one year or part of a year.

C F= −( )59

32 Temperature in degrees Celsius C equals 59

times the

difference between the Fahrenheit temperature F and 32.d = rt The distance traveled d equals the product of the rate of

speed r and the time t.P = 2l + 2w The perimeter, P, of a rectangle is equal to twice the

length l plus twice the width w.L = 2πrh The lateral surface area, L, (top and bottom not included)

of a cylinder is equal to 2π times the radius r of the base times the height h.

F = ma In physics, the force, F, acting on an object is equal to its mass m times its acceleration a.

α + β + γ = 180° The sum of the angles (α, β, and γ ) of a triangle is 180°.P = a + b + c The perimeter, P, of a triangle is equal to the sum of the

lengths of the three sides a, b, and c.

Evaluating Formulas pages 628 – 630If you know values for all but one variable in a formula, you can substitute those values and find the value of the unknown variable using the techniques for solving equations.

Solving Formulas for Different Variables pages 630 – 632We want the formula “solved for” some variable other than the one given in terms of the remaining variables. Treat the variables just as you would constants in solving linear equations.

Section 5.8: Working with Formulas

Mean Problems page 641Use the concept of mean to find unknown numbers and to read and understand graphs.

Cost Problems page 643S − C = P (selling price − cost = profit).

Section 5.9: Applications: Distance-Rate-Time, Interest, Mean, Cost

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Chapter Test Chapter 5 652

Chapter 5: Test

For the following expressions, a. simplify, and b. evaluate the simplified expression for x = −2 and y = 3.

1. 7 8 32 2x x x+ − a. 5 72x x+ b. 6 2. 5 6 2y y y− − + a. 6y − 6 b. 12

3. 2 5 3 4x x−( ) + +( ) a. 5x + 2 b. −8 4. 3 62 2y y y y+ − + a. 2 72y y+ b. 39

Translate each problem as directed.

5. Translate each algebraic expression into an equivalent English phrase. (There may be more than one correct translation.)

a. 5 18n + c. 42 7− y

the product of 5 and a number increased by 18 42 less the product of 7 and a number

b. 3 6x +( ) d. 6 11x −

3 multiplied by the sum of a number and 6 the product of 6 and a number decreased by 11

6. Write the algebraic expression described by each English phrase.

a. the product of a number and 6 decreased by 3 c. 4 less than twice the sum of a number and 15

6x − 3 2 15 4x +( ) −

b. twice the sum of a number and 5 d. the quotient of a number and 10 increased by 2 5x +( ) twice the number

x

x10

2+

Solve each of the following linear equations.

7. 0 2 8= +x x = −4 8. y + = −13 16 y = −29 9. 53

1 4x + = − x = −3

10. 3 6 8 15+ − =x x x = −6 11. 4 5 2 5x x x x− − = + − x = 5 12. 8 3 4 2 6+( ) = − −( )y y y = 0

13. 32

12

334

a a+ = − + a = −98

14. 632

913

56

+ = + −x x x x =32

15. 0 7 2 0 4 8. .x x+ = + x = 20 16. 4 2 1 3 2 4 5x x−( ) + = −( ) − x = −2

Determine whether each of the following equations is a conditional equation, an identity, or a contradiction.

17. 9 4 3 6 2x x x x+ − +( ) = −( ) identity 18. 14

4 112

2 1x x+( ) = − −( ) conditional

Set up an equation for each word problem and solve the equation.

19. One number is 5 more than twice another number. Their sum is −22. Find the numbers.

2 5 22 9 13y y+( ) + = − − −; ,

20. Find two consecutive integers such that twice the first added to three times the second is equal to 83. 2 3 1 83 16 17n n+ +( ) = ; ,

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653 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

21. Find three consecutive odd integers such that three times the second is equal to 27 more than the sum of the first and the third.

3 2 4 27 25 27 29n n n+( ) = + +( ) + ; , ,

22. Basketball: Brandon scored 21 points in his basketball game last night. If he made two free throws (1 point each) and one 3-point shot, how many 2-point shots did he make?

2 1 1 3 2 21 8( ) + ( ) + =x ; 2-point shots

Find the missing rate, base, or amount.

23. 62% of 180 is _____. 24. 48 is _____% of 150.

111.6 32%

Solve each of the following application problems.

25. Investing: Which is the better investment, an investment of $6000 that earns a profit of $360 or an investment of $10,000 that earns a profit of $500? Explain your answer in terms of percent.

The $6000 investment is the better investment since it has a higher percent of profit, 6%, than the $10,000 investment, 5%.

27. Triangles: The triangle shown here indicates that the sides can be represented as x, x + 1, and 2x − 1. What is the length of each side if the perimeter is 12 meters? 3 meters, 4 meters, and 5 meters

�2x − 1x + 1

x

26. Investing: Julie made an investment of $2000 and lost 25% in one year. In the next year she made 25%. How much money did she have at the end of the two years?

$1875

28. Shopping: A man’s suit was on sale for $547.50. If this price was a discount of 25% from the original price, what was the original price?

$730

Solve each formula for the indicated variable.

29. N mrt p= + ; solve for m. mN p

rt=

− 30. 5 3 7 0x y+ − = ; solve for y. yx

=−7 53

Solve each of the following application problems.

31. Text Messaging: The cost to send/receive t text messages in a given month on AT&T’s Messaging 1500 plan is C t= −( ) +0 05 1500 15 00. . when t ≥ 1500. How much did Jena spend the month she sent/received 1780 text messages. $29.00

32. Traveling by Bus: A bus leaves Kansas City headed for Phoenix traveling at a rate of 48 mph. Thirty minutes later, a second bus follows, traveling at 54 mph. How long will it take the second bus to overtake the first? 4 hours

33. Investing: George and Marie have investments in two accounts totaling $25,000. One investment yields 5% a year and the other yields 3.5%. If their annual income from the 5% account yields $145 more each year than the other account, what amount do they have in each account? $12,000 at 5%, $13,000 at 3.5%

34. Shopping: The manager of a jewelry store purchased a selection of watches from the manufacturer for $190 each. Next weekend he plans to have a 20% off sale. What price should he mark on each watch so that he will still make a profit of 40% on his cost?

$332.50

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Chapter Test Chapter 5 654

35. Test Scores: In his calculus class a student has the following test scores: 77, 82, and 73. If his last test will have a maximum of 100 points, what score does he need to finish the class with an mean of 80? 88

36. Payment Plans: When purchasing an item on an installment plan, the total cost C equals the down payment d plus the product of the monthly payment p and the number of months t. (C = d + pt). Use this information to answer the following questions.

a. A refrigerator costs $857.60 if purchased on the installment plan. If the monthly payments are $42.50 and the down payment is $92.60, how long will it take to pay for the refrigerator? 18 months

b. A used car will cost $3250 if purchased on an installment plan. If the monthly payments are $115 for 24 months, what will be the down payment? $490

37. Cell Phone Usage: Given the amount of U.S. cellular telephone subscribers: Source: International

Telecommunications Union

a. Find the mean number of subscribers from 2002-2005, inclusive. 175.05 million

b. Find the difference in the number of subscribers from 1999 and 2008. 184.5 million

c. How many subscribers were there in 2001? 128.5 million

US Cellular Telephone Subscribers

86.0

109.5

128.5

141.8

160.6

184.8

213.0

241.8

255.4

0 50 100 150 200 250 300

2008

1999

2000

2001

2002

2003

2004

20052006

2007

Yea

r

270.5

Number of Subscribers (in Millions)

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655 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

Cumulative Review: Chapters 1 - 5

Find the prime factorizaton.

1. 380 2 2 5 19⋅ ⋅ ⋅

Simplify the expressions using the rules for order of operations. Reduce all answers to lowest terms.

2. 12

38

34

56

+ ÷ − 16

3. 13

25

58

32

12

+ ⋅ ÷ − 0

4. 23

518

38

2

÷ − 4940

5. 9

10215

12

23

+

÷ −

315

For the following expressions, a. simplify, and b. evaluate the simplified expression for x = −2 and y = 3.

6. − +( ) +4 3 2y y a. −2y − 12 b. −18 7. 3 4 5x x+( ) − + a. 4x + 7 b. −1

8. 2 4 52 2x x x x+( ) − −( ) a. x x2 13+ b. −22 9. 3 5

42 7

x xx

−( )− − a. x − 7 b. −9

Solve the following problems.

10. 6 3 5 4x x x+ − + 13 5x − 11. a a a7 14 21

+ − a6

12. Translate each English phrase into an algebraic expression:

a. twice a number decreased by 3 2x − 3

b. the quotient of a number and 6 increased

by 3 times the number xx

63+

c. thirteen less than twice the sum of a number and 10 2 10 13x +( ) −

13. Translate each algebraic expression into an English phrase:

a. 6 + 4n the sum of 6 and 4 times a number

b. 18 5n −( ) 18 times the difference of a number and 5

Solve each of the following linear equations.

14. x + = −4 2 5 6. . 15. − + =5 13x 16. 17 51y = − 17. − = −16 96y x = −9.8 x = 18 y = −3 y = 6

18. 9x − 8 = 4x − 13 19. 6 2 5− −( ) =x 20. 10 4 5 3 4+ = − +( )x x x

x = −1 x = 3 x = −11

21. 4.4 + 0.6x = 1.2 − 0.2x 22. − +( ) − = − −( ) −2 5 12 5 6 3 2 10x x

x = −4 x =318

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Cumulative Review Chapters 1-5 656

23. 34

12

58

110

y y− = + 24. 12

15

215

1+ = +y y 25. − + =34

38

16

y y y

y =245

y =152

y = 0

Determine whether each of the following equations is a conditional equation, an identity, or a contradiction.

26. 0 2 25 1 6 0 1. . .x x+ = − + 27. 3 5 4 2 5x x x−( ) + = + −( ) 28. 4 3 6 12x x x− +( ) = −

conditional contradiction contradiction

29. − + +( ) = + +( )x x x x7 3 3 3 7 30. 8 4 3 2 6 2 16− −( ) + = +( ) −x x x

identity identity

Solve each of the following problems.

31. The two triangles shown here are similar triangles. Find the values of x and y. x = 2.4; y = 1.2

25

4A B

C

D E

F3

x

y

32. If the difference between 12

1118

and is

divided by the sum of 23

715

and , what is the

quotient? −551

33. From the sum of −2 and −11, subtract the product of −4 and 2. −5

34. Entertaining: Lucia is making punch for a party. She is making 5 gallons. The recipe

calls for 23

cup of sugar per gallon of punch.

How many cups of sugar does she need? 103

313

cups (or cups)

35. College Admission: The percentages of applicants accepted into the eight Ivy League colleges are as follows: 8%, 19%, 10%, 7%, 18%, 11%, 13%, and 10%. What is the mean percentage of applicants accepted to an Ivy League college? Source: College Board 12%

Set up an equation for each problem and solve for the unknown quantity.

36. If twice a certain number is increased by 3, the result is 8 less than three times the number. Find the number. 11

37. Computers: Computers are on sale at a 30% discount. If you know that you will pay 6.5% in sales taxes, what will you pay for a computer that is priced at $680? $506.94

38. Find three consecutive even integers such that the sum of the first and twice the second is equal to 14 more than twice the third. 18, 20, 22

39. Commission: Barry makes $500 a month plus commission working at a clothing store on the weekends. If he receives a 3% commission on all clothes he sells, what was his income the month he sold $1860 worth of clothes? $555.80

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657 Chapter 5 Algebraic Expressions, Linear Equations, and Applications

40. Find four consecutive integers such that the sum of the first, second, and fourth integers is 2 less than the third. −2, −1, 0, 1

41. Investing: Which is the better investment: a. a profit of $800 on an investment of $10,000 or b. a profit of $1200 on an investment of $15,000? Explain in terms of percents. These are equally good investments since the percent of profit, 8%, is the same for each investment.

42. Electronics: The total cost of a portable hard drive and a spare laptop battery was $130.75. If the hard drive cost $38.45 more than the battery, what was the cost of each item? hard drive: $84.60; battery: $46.15

Solve for the indicated variable.

43. h vt t v= − 16 2; solve for . 44. A P Prt r= + ; solve for . 45. 5 3 14x y y− = ; solve for .

vh t

t=

+ 16 2

rA P

Pt=

− y

x= −

−14 53

46. V r h h=13

2π ; solve for . 47. C d d= π ; solve for . 48. 3 5 10x y y+ = ; solve for .

hV

r=

32π

dC

yx

=−10 35

Solve the following word problems.

49. Traveling by Train: Diego took the train into New York City to see a concert. On the way to the concert the train had a mean of 90 miles per hour. On the way home from the concert the train made a lot more stops and had a mean of only 50 miles per hour. If Diego spent a total of 2.8 hours on the train, how long did the return trip take?

1.8 hours

50. Investing: Two investments totaling $22,000 produce an annual income of $742.50. One investment yields 4.5% a year, while the other yields 3% per year. How much is invested at each rate?

$16,500 at 3%; $5500 at 4.5%

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Cumulative Review Chapters 1-5 658

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