Chapter 1 · Chapter 1 8. Words: Your recorded balance − Forgotten check = Statement balance...

Copyright © Big Ideas Learning, LLC Algebra 1 1All rights reserved. Worked-Out Solutions

Chapter 1 Maintaining Mathematical Profi ciency (p. 1)

1. −5 + (−2) = −7 2. 0 + (−13) = −13

3. −6 + 14 = 8

4. 19 − (−13) = 19 + 13 = 32

5. −1 − 6 = −1 + (−6) = −7

6. −5 − (−7) = −5 + 7 = 2

7. 17 + 5 = 22 8. 8 + (−3) = 5

9. 11 − 15 = 11 + (−15) = −4

10. −3(8) = −24 11. −7 ⋅ (−9) = 63

12. 4 ⋅ (−7) = −28 13. −24 ÷ (−6) = 4

14. −16 ÷ 2 = −8 15. 12 ÷ (−3) = −4

16. 6 ⋅ 8 = 48 17. 36 ÷ 6 = 6

18. −3(−4) = 12

19. a. Sample answer: To add integers with the same sign, add

their absolute values, and the sum has the same sign

as both addends. To add integers with different signs,

subtract their absolute values, and the difference has the

same sign as the addend with the greatest absolute value;

−6 + 2 = −4

b. Sample answer: To subtract integers, change the

subtraction sign to an addition sign, and change the

integer following the sign to its opposite. Then, follow the

rules for adding integers; 5 − (−3) = 5 + 3 = 8

c. Sample answer: To multiply integers, multiply their

absolute values. If the integers have the same sign, then

the product is positive. If the integers have different signs,

then the product is negative; (−6)(−4) = 24

d. Sample answer: To divide integers, divide their absolute

values. If the integers have the same sign, then the

quotient is positive. If the integers have different signs,

then the quotient is negative; −15 ÷ 3 = −5

Chapter 1 Mathematical Practices (p. 2)

1. Population change = 310 million − 280 million

= 30 million people

Time change = 2010 − 2000

= 10 years

Rate of change = 30 million people

—— 10 years

= 3 million people per year

2. Gas mileage = 240 mi

— 8 gal

= 30 mi/gal

3. 18 in. = 1.5 ft

Volume = ℓwh

= (5 ft) × (3 ft) × (1.5 ft)

= 22.5 ft3

Amount of water = 3 —

4 (22.5 ft3)

= 16.875 ft3

Drain time = 16.875 ft3

— 1 ft3/min

= 16.875 min

It takes about 17 minutes for the water to drain.

1.1 Explorations (p. 3)

1.

Quadri-lateral

m∠ A (deg-rees)

m∠ B(deg-rees)

m∠ C(deg-rees)

m∠ D(deg-rees)

m∠ A + m∠ B+ m∠ C + m∠ D

a. 110 90 92 68 360°

b. 65 147 58 90 360°

c. 91 79 75 115 360°

Sample answer: Because a protractor is used, the measurements

are precise.

2. Conjecture: The sum of the angle measures of a quadrilateral

is 360°.

Sample answer:

A B

D C

100° + 40° + 140° + 80° = 360°

A

DC

B

90° + 60° + 115° + 95° = 360°

AB

D

C

150° + 30° + 75° + 105° = 360°

Divide the quadrilateral into two triangles.

The sum of the angle measures of a triangle is 180°,

so the sum of the angle measures of a quadrilateral is

2(180°) = 360°.

Chapter 1

2 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.

Chapter 1

3. a. x + 80 + 85 + 100 = 360

x + 265 = 360

− 265 − 265

x = 95

So, x = 95.

b. x + 78 + 60 + 72 = 360

x + 210 = 360

− 210 − 210

x = 150

So, x = 150.

c. x + 90 + 30 + 90 = 360

x + 210 = 360

− 210 − 210

x = 150

So, x = 150.

4. Sample answer: If you notice a pattern, you can use inductive

reasoning to write a rule. Then you can test your rule using

several examples. You can use the rule to write an equation

that can be used to solve a problem.

5. Sample answer: The corners can be arranged so the angles

complete a full circle, which is 360°.

360°

1.1 Monitoring Progress (pp. 4 –7)

1. n + 3 = −7 Write the equation.

− 3 − 3 Subtract 3 from each side.

n = −10 Simplify.

Check: n + 3 = −7

−10 + 3 =?

−7

−7 = −7 ✓

The solution is n = −10.

2. g − 1 —

3 = − 2 —

3 Write the equation.

+ 1 — 3 + 1 —

3 Add 1 —

3 to each side.

g = − 1 — 3 Simplify.

Check: g − 1 —

3 = − 2 —

3

− 1 — 3 − 1 —

3 =

? − 2 —

3

− 2 — 3 = − 2 —

3 ✓

The solution is g = − 1 — 3 .

3. −6.5 = p + 3.9 Write the equation.

− 3.9 − 3.9 Subtract 3.9 from each side.

−10.4 = p Simplify.

Check: −6.5 = p + 3.9

−6.5 =?

−10.4 + 3.9

−6.5 = −6.5 ✓

The solution is p = −10.4.

4. y —

3 = −6 Write the equation.

3 ⋅ ( y — 3 ) = 3 ⋅ (−6) Multiply each side by 3.

y = −18 Simplify.

Check: y — 3 = −6

−18

— 3 =

? −6

−6 = −6 ✓

The solution is y = −18.

5. 9π = πx Write the equation.

9π

— π

= πx

— π

Divide each side by π.

9 = x Simplify.

Check: 9π = πx

9π =?

π(9)

9π = 9π ✓

The solution is x = 9.

6. 0.05 w = 1.4 Write the equation.

0.05 w

— 0.05

= 1.4

— 0.05

Divide each side by 0.05.

w = 28 Simplify.

Check: 0.05 w = 1.4

0.05(28) =?

1.4

1.4 = 1.4 ✓

The solution is w = 28.

7. Let t be the time it would take to run 400 meters. Use the

Distance Formula.

d = r ⋅ t

400 = 10.35 ⋅ t

400

— 10.35

= 10.35t —

10.35

38.65 ≈ t

Usain Bolt would run 400 meters in about 38.65 seconds.


Chapter 1

8. Words: Your recorded

balance −Forgotten

check =Statement

balance

Variable: Let c be the amount of the forgotten check.

Equation: 68 − c = 26

68 − c = 26

− 68 − 68

−c = − 42

c = 42

The check you forgot to record was for $42.

1.1 Exercises (pp. 8–10)

Vocabulary and Core Concept Check

1. Addition, +, and subtraction, −, are inverses of each other.

Multiplication, ×, and division, ÷, are inverses of each other.

2. −2x = 10 −5x = 25

−2x — −2

= 10

— −2

−5x

— −5

= 25

— −5

x = −5 x = −5

yes; The equations are equivalent because they have the same

solution, x = −5.

3. Division Property of Equality; In order to write an equivalent

equation that has x by itself on one side, you must undo

multiplying by 14. So, you would divide each side by 14.

4. The equation x − 6 = 5 does not belong. Sample answer: You would use the Addition Property of Equality to solve

it, whereas you would use the Multiplication Property of

Equality to solve the other three equations.

Monitoring Progress and Modeling with Mathematics

5. x + 5 = 8 Write the equation.


x = 3 Simplify.

Check: x + 5 = 8

3 + 5 =?

8

8 = 8 ✓


6. m + 9 = 2 Write the equation.


m = −7 Simplify.

Check: m + 9 = 2

−7 + 9 =?

2

2 = 2 ✓

The solution is m = −7.

7. y − 4 = 3 Write the equation.

+ 4 + 4 Add 4 to each side.

y = 7 Simplify.

Check: y − 4 = 3

7 − 4 =?

3

3 = 3 ✓

The solution is y = 7.

8. s − 2 = 1 Write the equation.


s = 3 Simplify.

Check: s − 2 = 1

3 − 2 =?

1

1 = 1 ✓

The solution is s = 3.

9. w + 3 = −4 Write the equation.


w = −7 Simplify.

Check: w + 3 = −4

−7 + 3 =?

−4

−4 = −4 ✓

The solution is w = −7.

10. n − 6 = −7 Write the equation.


n = −1 Simplify.

Check: n − 6 = −7

−1 − 6 =?

−7

−7 = −7 ✓


11. −14 = p − 11 Write the equation.


−3 = p Simplify.

Check: −14 = p − 11

−14 =?

−3 − 11

−14 = −14 ✓

The solution is p = −3.

12. 0 = 4 + q Write the equation.


−4 = q Simplify.

Check: 0 = 4 + q 0 =

? 4 + (−4)

0 = 0 ✓

The solution is q = −4.


Chapter 1

13. r + (−8) = 10 Write the equation.

− (−8) − (−8) Subtract −8 from each side.

r = 10 + 8 Rewrite subtraction.

r = 18 Simplify.

Check: r + (−8) = 10

18 + (−8) =?

10

10 = 10 ✓

The solution is r = 18.

14. t − (−5) = 9 Write the equation.

t + 5 = 9 Rewrite subtraction.


t = 4 Simplify.

Check: t − (−5) = 9 4 − (−5) =

? 9

4 + 5 =?

9

9 = 9 ✓

The solution is t = 4.

15. Words: Discounted

ticket price =Original

price − 12.95

Variable: Let p be the original price.

Equation: 44 = p − 12.95

44 = p − 12.95

+ 12.95 + 12.95

56.95 = p

The original price of an amusement park ticket is $56.95.

16. Words: Your fi nal

score + 12 =Friend’s

fi nal score

Variable: Let x be your fi nal score.

Equation: x + 12 = 195

x + 12 = 195

−12 −12

x = 183

Your fi nal score is 183 points.

17. x + 100 + 120 + 100 = 360

x + 320 = 360

− 320 − 320

x = 40

So, x = 40.

18. x + 48 + 77 + 150 = 360

x + 275 = 360

− 275 − 275

x = 85

So, x = 85.

19. x + 122 + 92 + 76 = 360

x + 290 = 360

− 290 − 290

x = 70

So, x = 70.

20. x + 60 + 115 + 85 = 360

x + 260 = 360

− 260 − 260

x = 100

So, x = 100.

21. 5g = 20 Write the equation.

5g — 5 = 20

— 5 Divide each side by 5.

g = 4 Simplify.

Check: 5g = 20

5(4) =?

20

20 = 20 ✓

The solution is g = 4.

22. 4q = 52 Write the equation.

4q — 4 = 52


q = 13 Simplify.

Check: 4q = 52

4(13) =?

52

52 = 52 ✓

The solution is q = 13.

23. p ÷ 5 = 3 Write the equation.

5 ⋅ (p ÷ 5) = 5 ⋅ (3) Multiply each side by 5.

p = 15 Simplify.

Check: p ÷ 5 = 3

15 ÷ 5 =?

3

3 = 3 ✓

The solution is p = 15.

24. y ÷ 7 = 1 Write the equation.

7 ⋅ ( y ÷ 7) = 7 ⋅ (1) Multiply each side by 7.

y = 7 Simplify.

Check: y ÷ 7 = 1

7 ÷ 7 =?

1

1 = 1 ✓



Chapter 1

25. −8r = 64 Write the equation.

−8r — −8

= 64

— −8

Divide each side by −8.

r = −8 Simplify.

Check: −8r = 64

−8(−8) =?

64

64 = 64 ✓

The solution is r = −8.

26. x ÷ (−2) = 8 Write the equation.

−2 ⋅ [x ÷ (−2)] = −2 ⋅ 8 Multiply each side by −2.

x = −16 Simplify.

Check: x ÷ (−2) = 8

−16 ÷ (−2) =?

8

8 = 8 ✓

The solution is x = −16.

27. x — 6 = 8 Write the equation.

6 ⋅ ( x — 6 ) = 6 ⋅ (8) Multiply each side by 6.

x = 48 Simplify.

Check: x — 6 = 8

48

— 6 =

? 8

8 = 8 ✓


28. w —

−3 = 6 Write the equation.

−3 ⋅ ( w — −3

) = −3 ⋅ (6) Multiply each side by −3.

w = −18 Simplify.

Check: w — −3

= 6

−18

— −3

=?

6

6 = 6 ✓


29. −54 = 9s Write the equation.

−54

— 9 =

9s —

9 Divide each side by 9.

−6 = s Simplify.

Check: −54 = 9s

−54 =?

9(−6)

−54 = −54 ✓

The solution is s = −6.

30. −7 = t —

7 Write the equation.

7 ⋅ (−7) = 7 ⋅ ( t — 7 ) Multiply each side by 7.

−49 = t Simplify.

Check: −7 = t —

7

−7 =?

−49

— 7

−7 = −7 ✓

The solution is t = −49.

31. 3 — 2 + t =

1 —

2

− 3 — 2 − 3 —

2

t = − 2 — 2 , or −1

Check: 3 — 2 + t = 1 —

2

3 —

2 + ( − 2 —

2 ) =?

1 —

2

1 —

2 =

1 —

2 ✓


32. b − 3 —

16 =

5 —

16

+ 3 — 16

+ 3 — 16

b = 8 —

16 , or

1 —

2

Check: b − 3 —

16 =

5 —

16

8 —

16 −

3 —

16 =

?

5 —

16

5 —

16 =

5 —

16 ✓

The solution is b = 1 —

2 .

33. 3 —

7 m = 6

7 —

3 ⋅

3 —

7 m =

7 —

3 ⋅ 6

m = 14

Check: 3 — 7 m = 6

3 —

7 (14) =

? 6

6 = 6 ✓

The solution is m = 14.


Chapter 1

34. − 2 — 5 y = 4

− 5 — 2 ⋅ ( − 2 —

5 y ) = − 5 —

2 ⋅ 4

y = −10

Check: − 2 — 5 y = 4

− 2 — 5 (−10) =

? 4

4 = 4 ✓


35. 5.2 = a − 0.4

+ 0.4 + 0.4

5.6 = a

Check: 5.2 = a − 0.4

5.2 =?

5.6 − 0.4

5.2 = 5.2 ✓

The solution is a = 5.6.

36. f + 3π = 7π

− 3π − 3π

f = 4π

Check: f + 3π = 7π

4π + 3π =?

7π

7π = 7π ✓

The solution is f = 4π.

37. −108π = 6π j

−108π —

6π =

6π j —

6π

−18 = j

Check: −108π = 6π j

−108π =?

6π (−18)

−108π = −108π ✓

The solution is j = −18.

38. x ÷ (−2) = 1.4

−2 ⋅ [x ÷ (−2)] = −2 ⋅ 1.4

x = −2.8

Check: x ÷ (−2) = 1.4

−2.8 ÷ (−2) =?

1.4

1.4 = 1.4 ✓

The solution is x = −2.8.

39. A positive 0.8 should have been added to each side.

−0.8 + r = 12.6

+ 0.8 + 0.8

r = 13.4

The solution is r = 13.4.

40. Each side should have been multiplied by −3.

− m — 3 = − 4

−3 ⋅ ( − m

— 3 ) = −3⋅ (−4)

m = 12


41. C; Because each carton contains 18 eggs, the total number

of eggs is equal to the product of 18 and the number of

cartons, x.

18x = 162

18x

— 18

= 162 —

18

x = 9

The baker orders 9 cartons of eggs.

42. Words: Temperature

at 5 p.m. −Change in

temperature =Temperature

at 10 p.m.

Variable: Let T be the change in temperature.

Equation: 20 − T = −5

20 − T = −5

− 20 − 20

−T = −25

T = 25

The temperature fell 25°F from 5 p.m. to 10 p.m.

43. Words: Length = 1.9 ⋅ Width

Variable: Let w be the width.

Equation: 9.5 = 1.9 ⋅ w

9.5 = 1.9w

9.5 =

1.9w 1.9 1.9

5 = w

The American fl ag is 5 feet wide.


Chapter 1

44. Words: Current

balance = $308 +Balance 4

years ago

Variable: Let b be the balance 4 years ago.

Equation: 4708 = 308 + b

4708 = 308 + b

− 308 − 308

4400 = b

The balance 4 years ago was $4400.

45. Multiplication Property of Equality

4 ⋅ [ x − 1 —

2 =

x —

4 + 3 ] ⇒ 4x − 2 = x + 12

46. a. Words: Total

area = 4 ⋅ Area of

rectangle +Area of

square

Variable: Let A be the area of one rectangle.

Equation: 81 = 4 ⋅ A + 0.5A

81 = 4A + 0.5A

81 = 4.5A

81

— 4.5

= 4.5A

— 4.5

18 = A

Each rectangular mat is 18 square feet.

b. Guess: length = 8 ft, width = 4 ft

Check: A =ℓw

18 =?

8(4)

18 ≠ 32

Revise: length = 6 ft, width = 3 ft

Check: A =ℓw

18 =?

6(3)

18 = 18 ✓

Each rectangular mat is 6 feet by 3 feet.

47. a. Words: Total

spent =Number

of CDs ⋅ Amount you

spend per CD

Variable: Let p be the amount you spend on each CD.

Equation: 30.40 = 4 ⋅ p

30.40 = 4p

30.40

— 4 =

4p —

4

7.6 = p

You spend $7.60 on each CD.

b. Words: Amount you

spend per CD = 80% ⋅ Original

price

Variable: Let p be the original price.

Equation: 7.6 = 0.80 ⋅ p

7.6 = 0.8 p

7.6

— 0.8

= 0.8 p

— 0.8

9.5 = p

Each CD costs $9.50.

Because 3 CDs at the original price cost 3(9.5) = $28.50,

$25 is not enough to buy them all.

48. Equation Value of x Reason

x – c = 0 increases Because x = c, as c increases,

so does x.

cx = 1 decreases Because x = 1 —

c , as c increases,

x decreases.

cx = c stays the

same

Because x = 1, as c increases,

x stays the same.

x —

c = 1

increases Because x = c, as c increases,

so does x.

x − c = 0 cx = 1 cx = c x —

c = 1

+c +c cx

— c =

1 —

c

cx —

c =

c —

c c⋅

x —

c = c ⋅ 1

x = c x = 1 —

c x =1 x = c


Chapter 1

49. a. 5x = 10 − 5

5x = 5

5x

— 5 =

5 —

5

x = 1

When a = 5 and b = 10, x is a positive integer.

b. −2x = 9 − 5

−2x = 4

−2x —

−2 =

4 —

−2

x = − 2

When a = −2 and b = 9, x is a negative integer.

50. a. The entire circle represents 100%.

b. The percent of the partitions of a circle graph should sum

to 100 percent.

You can solve the equation to fi nd x, the percent of cats.

7 + 9 + 5 + 48 + x = 100

69 + x = 100

− 69 − 69

x = 31

At a local pet store, 31% of the animals sold are cats.

51. Let g be the number of girls and b be the number of boys in

the marching band.

1 —

6 g = 6

2 —

7 b = 10

6 ⋅ 1 —

6 g = 6 ⋅ 6

7 —

2 ⋅

2 —

7 b =

7 —

2 ⋅ 10

g = 36 b = 35

36 girls + 35 boys = 71 students

The marching band has 71 students.

52. Sample answer: A game has 30 pieces, each player should

get the same number of pieces, and all 30 pieces should be

used. If you and 4 of your friends are playing, how many

pieces should each player get? Let x be how many pieces

each player should get.

5x = 30

5x

— 5 =

30 —

5

x = 6

You and your 4 friends should each get 6 pieces.

53. V = Bh

84π = B(7)

84π —

7 =

7B —

7

12π = B

The area of the base of the cylinder is 12π square inches.

54. V = Bh

1323 = 147h

1323

— 147

= 147h

— 147

9 = h

The height of the rectangular prism is 9 centimeters.

55. V = 1 —

3 Bh

15π = 1 —

3 B(5)

3 —

5 (15π) =

3 —

5 ( 5 —

3 B )

9π = B

The area of the base of the cone is 9π square meters.

56. V = 1 — 3 Bh

35 = 1 — 3 ⋅ 30 ⋅ h

35 = 10h

35

— 10

= 10h —

10

3.5 = h The height of the square pyramid is 3.5 feet.

5 7. a. Batting

average = Number of hits

—— Number of at-bats

.296 = h —

446

446 ⋅ (.296) = 446 ⋅ h —

446

132 ≈ h Player A had 132 hits in the 2011 regular season.

b. no; If player B had fewer hits, then he must have had

fewer at-bats in order to have a greater batting average.

Maintaining Mathematical Profi ciency

58. 8(y + 3) = 8 ⋅ y + 8 ⋅ 3

= 8y + 24

59. 5 — 6 ( x + 1 —

2 + 4 ) = 5 —

6 ⋅ x +

5 —

6 ⋅

1 —

2 + 5 —

6 ⋅ 4

= 5 — 6 x +

5 —

12 +

20 —

6

= 5 — 6 x +

5 —

12 + 40

— 12

= 5 — 6 x +

45 —

12

= 5 — 6 x + 15

— 4


Chapter 1

60. 5(m + 3 + n) = 5 ⋅ m + 5 ⋅ 3 + 5 ⋅ n

= 5m + 15 + 5n

= 5m + 5n + 15

61. 4(2p + 4q + 6) = 4 ⋅ 2p + 4 ⋅ 4q + 4 ⋅ 6

= 8p + 16q + 24

62. 5 L

— min

⋅ 60 min

— 1 h

= 300 L —

h

63. 68 mi —

h ⋅

1 h —

60 min ⋅

1 min —

60 sec = 68 mi

— 3600 sec

≈ 0.02 mi

— sec

64. 7 gal

— min

⋅ 1 min

— 60 sec

⋅ 4 qt

— 1 gal

= 28 qt

— 60 sec

≈ 0.47qt

— sec

65. 8 km —

min ⋅

60 min —

1 h ⋅

1 mi —

1.61 km = 480 mi

— 1.61 h

≈ 298 .14 mi

— h

1.2 Explorations (p.11)

1. a. (30 + x) + 9x + 30 = 180 Write the equation.

30 + x + 9x + 30 = 180 Associative Property of

Addition

x + 9x + 30 + 30 = 180 Commutative Property of

Addition

10x + 60 = 180 Combine like terms.


10x = 120 Simplify.

10x

— 10

= 120 —


x = 12 Simplify.

So, x = 12 and the measures of the angles of the triangle

are 30°, 9x° = (9 ⋅ 12)° = 108°, and (30 + x)° = (30 + 12)° = 42°.

b. (x + 10) + (x + 20) + 50 = 180 Write the equation.

x + 10 + x + 20 + 50 = 180 Associative Property

of Addition

x + x + 10 + 20 + 50 = 180 Commutative Property

of Addition

2x + 80 = 180 Combine like terms.

− 80 − 80 Subtract 80 from each

side.

2x = 100 Simplify.

2x —

2 = 100


x = 50 Simplify,

So, x = 50 and the measures of the angles of the

triangle are 50°, (x + 20)° = (50 + 20)° = 70°, and

(x + 10)° = (50 + 10)° = 60°.

c.

50 + (2x + 30) + (2x + 20) + x = 360 Write the

equation.

50 + 2x + 30 + 2x + 20 + x = 360 Associative

Property of

Addition

2x + 2x + x + 50 + 30 + 20 = 360 Commutative

Property of

Addition

5x + 100 = 360 Combine like

terms.

−100 −100 Subtract 100

from each side.

5x = 260 Simplify.

5x —

5 =

260 —

5 Divide each

side by 5.

x = 52 Simplify.


quadrilateral are 50°, (2x + 30)° = (2 ⋅ 52 + 30)° = 134°, (2x + 20)° = (2 ⋅ 52 + 20)° = 124°, and

x° = 52°.

d.

(x − 17) + (x + 35) + (x + 42) + x = 360 Write the

equation.

x − 17 + x + 35 + x + 42 + x = 360 Associative

Property of

Addition

x + x + x + x − 17 + 35 + 42 = 360 Commutative

Property of

Addition


terms.

− 60 − 60 Subtract 60

from each side.

4x = 300 Simplify.

4x

— 4 = 300

— 4 Divide each

side by 4.

x = 75 Simplify.


quadrilateral are (x − 17)° = (75 − 17)° = 58°,

(x + 35)° = (75 + 35)° = 110°, (x + 42)° = (75 + 42)° = 117°, and x° = 75°.


Chapter 1

e. (5x + 2) + (3x + 5) + (8x + 8) Write the

+ (5x + 10) + (4x + 15) = 540 equation.

5x + 2 + 3x + 5 + 8x Associative

+ 8 + 5x + 10 + 4x + 15 = 540 Property of

Addition

5x + 3x + 8x + 5x + 4x Commutative

+ 2 + 5 + 8 + 10 + 15 = 540 Property of

Addition


terms.

− 40 − 40 Subtract 40

from each side.

25x = 500 Simplify.

25x

— 25

= 500 —

25 Divide each side

by 25.

x = 20 Simplify.

So, x = 20 and the measures of the angles of the pentagon

are (5x + 2)° = (5 ⋅ 20 + 2)° = 102°,

(3x + 5)° = (3 ⋅ 20 + 5)° = 65°,

(8x + 8)° = (8 ⋅ 20 + 8)° = 168°,

(5x + 10)° = (5 ⋅ 20 + 10)° = 110°, and

(4x + 15)° = (4 ⋅ 20 + 15)° = 95°.

f.

2(3x + 16) + (2x + 8) + (4x − 18) Write the

+ (3x − 7) + (2x + 25) = 720 equation.

6x + 32 + (2x + 8) + (4x − 18) Distributive

+ (3x − 7) + (2x + 25) = 720 Property

6x + 32 + 2x + 8 + 4x − 18 Associative

+ 3x − 7 + 2x + 25 = 720 Property of

Addition

6x + 2x + 4x + 3x + 2x + 32 Commutative

+ 8 − 18 − 7 + 25 = 720 Property of

Addition


terms.

− 40 − 40 Subtract 40

from each side.

17x = 680 Simplify.

17x

— 17

= 680

— 17

Divide each side

by 17.

x = 40 Simplify.

So, x = 40 and the measures of the angles of the hexagon

are (3x + 16)° = (3 ⋅ 40 + 16)° = 136°, 136°,

(2x + 8)° = (2 ⋅ 40 + 8)° = 88°,

(4x − 18)° = (4 ⋅ 40 − 18)° = 142°,

(3x − 7)° = (3 ⋅ 40 − 7)° = 113°, and

(2x + 25)° = (2 ⋅ 40 + 25)° = 105°.

Sample answer: You can check the measures of the angles

using a protractor. You can also check to make sure the

angle measures add up to the sum given by the formula.

2. a. Answer should include, but is not limited to: Check that

polygons have 3 or more straight sides of varying lengths

and that all are closed fi gures. Suggest that students draw

large polygons, because it will be easier to measure the

angles.

b. Answer should include, but is not limited to: The sum of

the angle measures of each polygon should satisfy the

formula S = 180(n − 2). Some might be a little off due

to rounding. Have students round to the nearest whole

number of degrees.

c. Answer should include, but is not limited to: The same

value for x should satisfy each expression for the angle

measures of the polygon.

d. and e. Answer should include, but is not limited to: Partners should confi rm that the calculated measures

of the angles are the same as (or at least close to) the

measures obtained with a protractor. In addition, the sum

of the calculated measures of the polygon should satisfy

the formula S = 180(n − 2).

3. Sample answer: If you notice a pattern, you can use inductive

reasoning to write a rule. Then you can test your rule using

several examples. You can use the rule to write an equation

that can be used to solve a problem.

4. Connecting a vertex with each of the other vertices in a

polygon creates n − 2 triangles, each of which has a total

angle measure of 180°.

5. S = 180(n − 2)

1080 = 180(n − 2)

1080

— 180

= 180(n − 2) —

180

6 = n − 2 + 2 + 2 8 = n 8 sides; Use the formula S = 180(n − 2) and inverse

operations to work backward and fi nd that a polygon, whose

angle measures sum to 1080°, is an octagon.

1.2 Monitoring Progress (pp. 12–15)

1. −2n + 3 = 9 −3 −3

− 2n = 6

−2n — −2 = 6

— −2

n = −3

Check: −2n + 3 = 9 −2(−3) + 3 =

? 9

6 + 3 =?

9 9 = 9 ✓



Chapter 1

2. −21 = 1 — 2 c − 11

+11 +11

−10 = 1 — 2 c

2 ⋅ (−10) = 2 ⋅ 1 —

2 c

−20 = c Check: −21 = 1 —

2 c − 11

−21 =?

1 —

2 (−20) − 11

−21 =?

−10 − 11

−21 = −21 ✓

The solution is c = −20.

3. −2x − 10x + 12 = 18

−12x + 12 = 18

− 12 − 12

−12x = 6

−12x — −12 = 6

— −12

x = − 1 — 2

Check: −2x − 10x + 12 = 18

−2 ( − 1 —

2 ) − 10 ( −

1 —

2 ) + 12 =

? 18

1 + 5 + 12 =?

18

18 = 18 ✓

The solution is x = − 1 — 2 .

4. 3(x + 1) + 6 = −9

3(x) + 3(1) + 6 = −9

3x + 3 + 6 = −9

3x + 9 = −9

− 9 − 9 3x = −18

3x — 3 = −18

— 3

x = −6

Check: 3(x + 1) + 6 = −9

3[(−6) + 1] + 6 =?

−9

3(−5) + 6 =?

−9

−15 + 6 =?

−9

−9 = −9 ✓


5. 15 = 5 + 4(2d − 3)

15 = 5 + 4(2d) − 4(3)

15 = 5 + 8d − 12

15 = 8d − 7 + 7 + 7 22 = 8d

22

— 8 = 8d

— 8

2.75 = d Check: 15 = 5 + 4(2d − 3)

15 =?

5 + 4[2(2.75) − 3]

15 =?

5 + 4(5.5 − 3)

15 =?

5 + 4(2.5)

15 =?

5 + 10

15 = 15 ✓

The solution is d = 2.75.

6. 13 = −2(y − 4) + 3y

13 = −2(y) − 2( −4) + 3y

13 = −2y + 8 + 3y

13 = y + 8 − 8 − 8 5 = y Check: 13 = −2(y − 4) + 3y

13 =?

−2(5 − 4) + 3(5)

13 =?

−2(1) + 15

13 =?

−2 + 15

13 = 13 ✓


7. 2x(5 − 3) − 3x = 5 2x(2) − 3x = 5 4x − 3x = 5 x = 5 Check: 2x(5 − 3) − 3x = 5 2(5)(5 − 3) − 3(5) =

? 5

2(5)(2) − 15 =?

5

20 − 15 =?

5

5 = 5 ✓



Chapter 1

8. −4(2m + 5) −3m = 35

−4(2m) + (−4)(5) − 3m = 35

−8m + (−20) − 3m = 35

−11m − 20 = 35

+ 20 + 20

−11m = 55

−11m —

−11 = 55

— −11

m = −5

Check: −4(2m + 5) − 3m = 35

−4[2(−5) + 5] − 3(−5) =?

35

−4(−10 + 5) + 15 =?

35

−4(−5) + 15 =?

35

20 + 15 =?

35

35 = 35 ✓


9. 5(3 − x) + 2(3 − x) = 14

5(3) − 5(x) + 2(3) − 2(x) = 14

15 − 5x + 6 − 2x = 14

−7x + 21 = 14

− 21 − 21

−7x = −7

−7x —

−7 = −7

— −7

x = 1 Check: 5(3 − x) + 2(3 − x) = 14

5(3 − 1) + 2(3 − 1) =?

14

5(2) + 2(2) =?

14

10 + 4 =?

14

14 = 14 ✓


10. d = 1 — 2 n + 26

50 = 1 — 2 n + 26

− 26 − 26

24 = 1 — 2 n

2 ⋅ 24 = 2 ⋅ 1 —

2 n

48 = n A fi re hose needs 48 pounds per square inch of water

pressure to reach a fi re 50 feet away.

11. Words: Perimeter = 2 ⋅ Width + 2 ⋅ Three times

the width

Variable: Let w be the width of the rectangular pen.

Equation: 96 = 2 ⋅ w + 2 ⋅ (3w)

96 = 2w + 2(3w)

96 = 2w + 6w

96 = 8w

96

— 8 = 8w

— 8

12 = w ℓ = 3w = 3(12) = 36

The pen should be 36 feet by 12 feet.



1. To solve the equation 2x + 3x = 20, fi rst combine 2x and 3x

because they are like terms.

2. Sample answer: One way to solve the equation

2(4x − 11) = 10 is to fi rst use the Distributive Property to

eliminate the parentheses. Then undo subtraction to isolate

the x-term. Finally, undo the multiplication to solve for

x. Another method is to fi rst undo multiplication and use

the Division Property of Equality to divide each side by 2.

Then undo subtraction to isolate the x-term. Finally, undo

multiplication again to solve for x.


3. 3w + 7 = 19

− 7 − 7 3w = 12

3w — 3 = 12

— 3

w = 4

Check: 3w + 7 = 19

3(4) + 7 =?

19

12 + 7 =?

19

19 = 19 ✓

The solution is w = 4.


Chapter 1

4. 2g − 13 = 3

+ 13 + 13

2g = 16

2g — 2 =

16 —

2

g = 8

Check: 2g − 13 = 3

2(8) − 13 =?

3

16 − 13 =?

3

3 = 3 ✓


5. 11 = 12 − q

− 12 − 12

−1 = −q

−1 —

−1 =

−q —

−1

1 = q

Check: 11 = 12 − q

11 =?

12 − 1

11 = 11 ✓

The solution is q = 1.

6. 10 = 7 − m

− 7 − 7 3 = −m

3 —

−1 =

−m —

−1

−3 = m

Check: 10 = 7 − m

10 =?

7 − (−3)

10 =?

7 + 3 10 = 10 ✓


7. 5 = z —

−4 − 3

+ 3 + 3

8 = z —

−4

−4 ⋅ 8 = −4 ⋅ ( z —

−4 )

−32 = z

Check: 5 = z —

−4 − 3

5 =?

−32

— −4

− 3

5 =?

8 − 3

5 = 5 ✓

The solution is z = −32.

8. a —

3 + 4 = 6

− 4 − 4

a —

3 = 2

3 ⋅ a —

3 = 3 ⋅ 2

a = 6

Check: a —

3 + 4 = 6

6 —

3 + 4 =

? 6

2 + 4 =?

6

6 = 6 ✓

The solution is a = 6.

9. h + 6

— 5 = 2

5 ⋅ h + 6

— 5 = 5 ⋅ 2

h + 6 = 10

− 6 − 6 h = 4

Check: h + 6

— 5 = 2

4 + 6

— 5 =

? 2

10

— 5 =

? 2

2 = 2 ✓

The solution is h = 4.


Chapter 1

10. d − 8

— −2

= 12

−2 ⋅ d − 8

— −2

= −2 ⋅ 12

d − 8 = −24

+ 8 + 8 d = −16

Check: d − 8 —

−2 = 12

−16 − 8

— −2

=?

12

−24

— −2

=?

12

12 = 12 ✓

The solution is d = −16.

11. 8y + 3y = 44

11y = 44

11y — 11

= 44

— 11

y = 4

Check: 8y + 3y = 44

8(4) + 3(4) =?

44

32 + 12 =?

44

44 = 44 ✓


12. 36 = 13n − 4n

36 = 9n

36

— 9 =

9n —

9

4 = n

Check: 36 = 13n − 4n

36 =?

13(4) − 4(4)

36 =?

52 − 16

36 = 36 ✓

The solution is n = 4.

13. 12v + 10v + 14 = 80

22v + 14 = 80

− 14 − 14

22v = 66

22v — 22

= 66

— 22

v = 3

Check: 12v + 10v + 14 = 80

12(3) + 10(3) + 14 =?

80

36 + 30 + 14 =?

80

80 = 80 ✓

The solution is v = 3.

14. 6c − 8 − 2c = −16

4c − 8 = −16

+ 8 + 8 4c = −8

4c — 4 =

−8 —

4

c = −2

Check: 6c − 8 − 2c = −16

6(−2) − 8 − 2(−2) =?

−16

−12 − 8 +4 =?

−16

−20 + 4 =?

−16

−16 = −16 ✓

The solution is c = −2.

15. a = 3400t + 600

21,000 = 3400t + 600

− 600 − 600

20,400 = 3400t

20,400

— 3400

= 3400t

— 3400

6 = t

The plane is at an altitude of 21,000 feet 6 minutes after

liftoff.


Chapter 1

16. Words: Repair

bill =Parts

cost +Labor cost

per hour ⋅ Hours

of labor

Variable: Let t be the number of hours of labor spent

repairing the car.

Equation: 553 = 265 + 48 ⋅ t

553 = 265 + 48t

− 265 − 265

288 = 48t

288

— 48

= 48t

— 48

6 = t

The repair bill includes charges for 6 hours of labor.

17. 4(z + 5) = 32

4(z) + 4(5) = 32

4z + 20 = 32

− 20 − 20

4z = 12

4z — 4 =

12 —

4

z = 3

Check: 4(z + 5) = 32

4(3 + 5) =?

32

4(8) =?

32

32 = 32 ✓

The solution is z = 3.

18. −2(4g − 3) = 30

−2(4g) − 2(−3) = 30

−8g + 6 = 30

− 6 − 6 −8g = 24

−8g — −8

= 24 —

−8

g = −3

Check: −2(4g − 3) = 30

−2[4(−3) − 3] =?

30

−2(−12 − 3) =?

30

−2(−15) =?

30

30 = 30 ✓

The solution is g = −3.

19. 6 + 5(m + 1) = 26

6 + 5(m) + 5(1) = 26

6 + 5m + 5 = 26

5m + 11 = 26

− 11 − 11

5m = 15

5m — 5 =

15 —

5

m = 3

Check: 6 + 5(m +1) = 26

6 + 5(3 + 1) =?

26

6 + 5(4) =?

26

6 + 20 =?

26

26 = 26 ✓


20. 5h + 2(11 − h) = −5

5h + 2(11) − 2(h) = −5

5h + 22 − 2h = −5

3h + 22 = −5

− 22 − 22

3h = −27

3h — 3 =

−27 —

3

h = −9

Check: 5h + 2(11 − h) = −5

5(−9) + 2[11 − (−9)] =?

−5

−45 + 2(11 + 9) =?

−5

−45 + 2(20) =?

−5

−45 + 40 =?

−5

−5 = −5 ✓

The solution is h = −9.


Chapter 1

21. 27 = 3c − 3(6 − 2c)

27 = 3c − 3(6) − 3(−2c)

27 = 3c − 18 + 6c

27 = 9c − 18

+ 18 + 18

45 = 9c

45

— 9 =

9c —

9

5 = c

Check: 27 = 3c − 3(6 − 2c)

27 =?

3(5) − 3[6 − 2(5)]

27 =?

15 − 3(6 − 10)

27 =?

15 − 3(−4)

27 =?

15 + 12

27 = 27 ✓

The solution is c = 5.

22. −3 = 12y − 5(2y − 7)

−3 = 12y − 5(2y) − 5(−7)

−3 = 12y − 10y + 35

−3 = 2y + 35

− 35 − 35

−38 = 2y

−38

— 2 =

2y —

2

−19 = y

Check: −3 = 12y − 5(2y − 7)

−3 =?

12(−19) − 5[2(−19) − 7]

−3 =?

−228 − 5(−38 −7)

−3 =?

−228 − 5(−45)

−3 =?

−228 + 225

−3 = −3 ✓


23. −3(3 + x) + 4(x − 6) = −4

−3(3) + (−3)(x) + 4(x) − 4(6) = −4

−9 − 3x + 4x − 24 = −4

x − 33 = −4

+ 33 + 33

x = 29

Check: −3(3 + x) + 4(x − 6) = −4

−3(3 + 29) + 4(29 − 6) =?

−4

−3(32) + 4(23) =?

−4

−96 + 92 =?

−4

−4 = −4 ✓


24. 5(r + 9) − 2(1 − r) = 1

5(r) + 5(9) − 2(1) − 2(−r) = 1

5r + 45 − 2 + 2r = 1

7r + 43 = 1

− 43 − 43

7r = −42

7r — 7 =

−42 —

7

r = −6

Check: 5(r + 9) − 2(1 − r) = 1

5(−6 + 9) − 2[1 − (−6)] =?

1

5(3) − 2(1 + 6) =?

1

15 − 2(7) =?

1

15 − 14 =?

1

1 = 1 ✓


25. 45 + 2k + k = 180

3k + 45 = 180

− 45 − 45

3k = 135

3k — 3 = 135

— 3

k = 45

So, k = 45 and the measures of the angles of the triangle are

45°, 2k° = 2 ⋅ 45 = 90°, and k° = 45°.


Chapter 1

26. a + 2a + a + 2a = 360

6a = 360

6a — 6 =

360 —

6

a = 60

So, a = 60 and the measures of the angles of the

quadrilateral are a° = 60°, 2a° = 2 ⋅ 60 = 120°, a° = 60°, and 2a° = 2 ⋅ 60 = 120°.

27. (2b − 90) + 3 — 2 b + b + (b + 45) + 90 = 540

11

— 2 b + 45 = 540

− 45 − 45

11

— 2 b = 495

2 —

11 ⋅ 11

— 2 b =

2 —

11 ⋅ 495

b = 90

So, b = 90 and the measures of the angles of the pentagon

are (2b − 90)° = 2 ⋅ 90 − 90 = 90°, 3 —

2 b° =

3 —

2 ⋅ 90 =

135°, b° = 90°, (b + 45)° = 90 + 45 = 135°, and 90°.

28. x + 120 + 100 + 120 + (x + 10) + 120 = 720

2x + 470 = 720

−470 −470

2x = 250

2x — 2 = 250

— 2

x = 125

So, x = 125 and the measures of the angles of the hexagon

are x°= 125°, 120°, 100°, 120°, (x + 10)° = 125 + 10 =

135°, and 120°.

29. 2n + 13 = 75

− 13 − 13

2n = 62

2n — 2 =

62 —

2

n = 31

The number is 31.

30. 3n − 4 = −19

+ 4 + 4 3n = −15

3n — 3 =

−15 —

3

n = −5

The number is −5.

31. 8 + n —

3 = −2

− 8 − 8

n —

3 = −10

3 ⋅ n —

3 = 3 ⋅ (−10)

n = −30

The number is −30.

32. 2n + 1 —

2 n = 10

( 4 — 2 +

1 —

2 ) n = 10

5 —

2 n = 10

2 —

5 ⋅

5 —

2 n =

2 —

5 ⋅ 10

n = 4

The number is 4.

33. 6(n + 15) = −42

6(n) + 6(15) = −42

6n + 90 = −42

− 90 − 90

6n = −132

6n — 6 =

−132 —

6

n = −22

The number is −22.

34. 4(n − 7) = 12

4(n) − 4(7) = 12

4n − 28 = 12

+ 28 + 28

4n = 40

4n — 4 =

40 —

4

n = 10

The number is 10.


Chapter 1

35. Words:

Total

earnings =

Gas

station

hours

worked

⋅Gas

station

hourly

wage

+

Land-

scaper

hourly

wage

⋅Land-

scaper

hours

worked

Variable: Let t be the hours you must work as a landscaper.

Equation: 400 = 30(8.75) + 11 ⋅ t

400 = 30(8.75) + 11t

400 = 262.5 + 11t

− 262.5 − 262.5

137.5 = 11t

137.5

— 11

= 11t

— 11

12.5 = t

Check: dollars =?

hours ⋅ dollars

— hour

+ dollars

— hour

⋅ hours

dollars =?

dollars + dollars

dollars = dollars ✓

You must work 12.5 hours as a landscaper to earn $400 per

week.

36. Words: Area of

swimming

pool

surface

=

Length

of deep

end ⋅Width

of

deep

end

+

Length

of

shallow

end

⋅Width

of

shallow

end

Variable: Let d be the length of the deep end.

Equation: 210 = d ⋅ 10 + 9 ⋅ 10

210 = 10d + 90

− 90 − 90

120 = 10d

120

— 10

= 10d

— 10

12 = d

Check: square feet =?

feet ⋅ feet + feet ⋅ feet

square feet =?

square feet + square feet

square feet = square feet ✓

The deep end is 12 feet long.

37. Words: Total

cost = Cost of

salad + 2 ⋅

Cost of

one taco +

Sales tax as

a decimal

⋅ (

Cost of

salad + 2 ⋅ Cost of

one taco )

+ Tip

Variable: Let t be the cost of one taco.

Equation: 13.80 = 2.5 + 2 ⋅ t + 0.08 ⋅ (2.5 + 2 ⋅ t) + 3 13.8 = 2.5 + 2t + 0.08(2.5 + 2t) + 3

13.8 = 2.5 + 2t + 0.2 + 0.16t + 3

13.8 = 2.16t + 5.7

− 5.7 − 5.7

8.1 = 2.16t

8.1

— 2.16

= 2.16t —

2.16

3.75 = t

Check: dollars =?

dollars + dollars + %(dollars + dollars)

+ dollars

dollars =?

dollars + dollars + dollars + dollars

+ dollars

dollars = dollars ✓

The cost of one taco is $3.75.

38. − 1 — 2 (5x − 8) − 1 = 6 Write the equation.

− 1 — 2 (5x − 8) = 7 Add 1 to each side.

5x − 8 = −14 Multiply each side by −2.

5x = −6 Add 8 to each side.

x = − 6 — 5 Divide each side by 5.

39. 2(x + 3) + x = −9 Write the equation.

2(x) + 2(3) + x = −9 Distributive Property

2x + 6 + x = −9 Simplify.

3x + 6 = −9 Combine like terms.

3x = −15 Subtract 6 from each side.

x = −5 Divide each side by 3.

40. The negative sign was not distributed correctly to each term

inside the parentheses.

−2(7 − y) + 4 = −4

−2(7) − 2(−y) + 4 = −4

−14 + 2y + 4 = −4

−10 + 2y = −4

+ 10 + 10

2y = 6

2y — 2 =

6 —

2

y = 3 The solution is y = 3.


Chapter 1

41. In order to undo multiplying by 1 —

4 , you should divide each

side by 1 —

4 , or multiply each side by 4.

1 —

4 (x − 2) + 4 = 12

− 4 − 4

1 —

4 (x − 2) = 8

4 ⋅ 1 — 4 (x − 2) = 4 ⋅ 8

x − 2 = 32

+ 2 + 2 x = 34


42. P = 2ℓ+ 2w

228 = 2(2w + 6) + 2w

228 = 2(2w) + 2(6) + 2w

228 = 4w + 12 + 2w

228 = 6w + 12

− 12 − 12

216 = 6w

216

— 6 =

6w —

6

36 = w

2w + 6 = 2 ⋅ 36 + 6 = 78

The court is 78 feet by 36 feet.

43. P = 2ℓ+ 2w

190 = 2 ( 11 —

8 y ) + 2(y)

190 = 11

— 4 y + 2y

190 = ( 11 —

4 +

8 —

4 ) y

190 = 19

— 4 y

4 —

19 ⋅ 190 =

4 —

19 ⋅ 19

— 4 y

40 = y

11

— 8 y =

11 —

8 ⋅ 40 = 55

The Norwegian fl ag is 55 inches by 40 inches.

44. P = s + (s + 6) + (s + 6) + s + 2s

102 = 6s + 12

− 12 − 12

90 = 6s

90

— 6 =

6s —

6

15 = s

s + 6 = 15 + 6 = 21

2s = 2 ⋅ 15 = 30

The school crossing sign has two sides that are each

15 inches, two sides that are each 21 inches, and one side

that is 30 inches.

45. a. 2(4 − 8x) + 6 = −1

2(4) − 2(8x) + 6 = −1

8 − 16x + 6 = −1

14 − 16x = −1

− 14 − 14

−16x = −15

−16x

— −16

= −15

— −16

x = 15

— 16

b. 2(4 − 8x) + 6 = −1

− 6 − 6 2(4 − 8x) = −7

2(4 − 8x)

— 2 =

−7 —

2

4 − 8x = − 7 — 2

− 4 − 4

−8x = − 15 —

2

− 1 — 8 ⋅ −8x = − 1 —

8 ⋅ ( − 15

— 2 )

x = 15

— 16

The solution is x = 15

— 16

.

Sample answer: Method 1 is preferred because it requires

fewer operations with fractions and was therefore less

complicated, making mistakes less likely.


Chapter 1

46. Words:

Total

cost = ( Ticket

price +

Convenience

charge ) ⋅ Number

of tickets + Processing

charge

Variable: Let t be the number of tickets purchased.

Equation: 220.70 = (32.50 + 3.30) ⋅ t + 5.90

220.7 = (32.5 + 3.3)t + 5.9

220.7 = 35.8t + 5.9

− 5.9 − 5.9

214.8 = 35.8t

214.8

— 35.8

= 35.8t

— 35.8

6 = t

For an order that costs $220.70, 6 tickets are purchased.

47. Words

Total

value

=

Value

per

dime⋅

Number

of

dimes+

Value

per

quarter⋅

Number

of

quarters

Variable: Let d be the number of dimes.

Equation: 2.80 = 0.10 ⋅ d + 0.25 ⋅ (d + 8)

2.8 = 0.1d + 0.25(d + 8)

2.8 = 0.1d + 0.25d + 2

2.8 = 0.35d + 2

− 2 − 2

0.8 = 0.35d

0.8

— 0.35

= 0.35d

— 0.35

2.29 ≈ d

no; Sample answer: Because it is not possible to have a

decimal number of dimes, it is not possible for the number

of quarters to be 8 more than the number of dimes when the

total of the dimes and quarters is $2.80.

48. Sample answer:

Component

Student’s score Weight Score × weight

Class

Participation92% 0.20 92% × 0.20 = 18.4%

Homework 95% 0.20 95% × 0.20 = 19%

Midterm

Exam88% 0.25 88% × 0.25 = 22%

Final Exam f 0.35 f × 0.35 = 0.35f

Total 1 18.4 + 19 + 22 + 0.35f

18.4 + 19 + 22 + 0.35f = 90

59.4 + 0.35f = 90

− 59.4 − 59.4

0.35f = 30.6

0.35f — 0.35

= 30.6

— 0.35

f ≈ 87.4

The student must earn at least an 88% on the fi nal exam in

order to earn an A in the class.

49. Let n be an integer. Then 2n is an even integer. The next even

integer is 2 more than 2n: 2n + 2.

The third even integer is 2 more than 2n + 2: 2n + 2 + 2 = 2n + 4.

So, the total is 2n + (2n + 2) + (2n + 4).

2n + (2n + 2) + (2n + 4) = 54

2n + 2n + 2 + 2n + 4 = 54

6n + 6 = 54

− 6 − 6 6n = 48

6n — 6 =

48 —

6

n = 8

2n = 2 ⋅ 8 = 16

2n + 2 = 2 ⋅ 8 + 2 = 18

2n + 4 = 2 ⋅ 8 + 4 = 20

So, the consecutive even integers are 16, 18, and 20.


Chapter 1

50. a. greater than; Only one of the numbers is greater than 20,

and two are less than 20. In order for the average to be 20,

the fourth meeting must have had more than 20 students

in attendance.

b. Sample answer: about 25 students

c. Sample answer: You can write and solve an equation

to fi nd how many students were in attendance at the

fourth meeting. If you let x be the number of students in

attendance at the fourth meeting, the equation is

18 + 21 + 17 + x

—— 4 = 20.

18 + 21 + 17 + x ——

4 = 20

4 ⋅ 18 + 21 + 17 + x ——

4 = 4 ⋅ 20

18 + 21 + 17 + x = 80

56 + x = 80

− 56 − 56

x = 24

There were 24 students at the fourth meeting. This is close

to the estimate in part (b).

51. bx = −7

bx

— b =

−7 —

b

x = − 7 — b

52. x + a = 3

— 4

− a − a

x = 3 —

4 − a

53. ax − b = 12.5

+ b + b ax = 12.5 + b

ax

— a =

12.5 + b —

a

x = 12.5 + b

— a

54. ax + b = c

− b − b ax = c − b

ax

— a =

c − b —

a

x = c − b

— a

55. 2bx − bx = −8

x(2b − b) = −8

x(b) = −8

x(b)

— b =

−8 —

b

x = − 8 — b

56. cx − 4b = 5b

+ 4b + 4b

cx = 9b

cx

— c =

9b —

c

x = 9b

— c


57. 4m + 5 − 3m = 4m − 3m + 5

= m + 5

58. 9 − 8b + 6b = 9 + (−8b) + 6b

= 9 + (−2b)

= 9 − 2b

59. 6t + 3(1 − 2t) − 5 = 6t + 3(1) − 3(2t) − 5

= 6t + 3 − 6t − 5

= 6t − 6t + 3 − 5

= 0 − 2

= −2

60. a. x − 8 = −9 b. x − 8 = −9

−1 − 8 =?

−9 2 − 8 =?

−9

−9 = −9 ✓ −6 ≠ −9

x = −1 is a solution. x = 2 is not a solution.

61. a. x + 1.5 = 3.5 b. x + 1.5 = 3.5

−1 + 1.5 =?

3.5 2 + 1.5 =?

3.5

0.5 ≠ 3.5 3.5 = 3.5 ✓

x = −1 is not a solution. x = 2 is a solution.

62. a. 2x − 1 = 3 b. 2x − 1 = 3

2(−1) − 1 =?

3 2(2) − 1 =?

3

−2 − 1 =?

3 4 − 1 =?

3

−3 ≠ 3 3 = 3 ✓



Chapter 1

63. a. 3x + 4 = 1 b. 3x + 4 = 1

3(−1) + 4 =?

1 3(2) + 4 =?

1

−3 + 4 =?

1 6 + 4 =?

1

1 = 1 ✓ 10 ≠ 1


64. a. x + 4 = 3x b. x + 4 = 3x

−1 + 4 =?

3(−1) 2 + 4 =?

3(2)

3 ≠ −3 6 = 6 ✓


65. a. −2(x − 1) = 1 − 3x b. −2(x − 1) = 1 − 3x

−2(−1 − 1) =?

1 − 3(−1) −2(2 − 1) =?

1 − 3(2)

−2(−2) =?

1 + 3 −2(1) =?

1 − 6

4 = 4 ✓ −2 ≠ −5



1. x + 5 + 2 + x + 2 + 5 = 3 + 3 — 2 x + 5 + 4

+ (5 − 3) + ( 3 — 2 x − 4 )

2x + 14 = 3x + 10

− 2x − 2x

14 = x + 10

− 10 − 10

4 = x The solution is x = 4.

Sample answer: Add the side lengths of each polygon to get

the perimeters, set them equal to each other, and solve for x.

The perimeter of the fi rst polygon is

4 + 5 + 2 + 4 + 2 + 5 = 22.

The perimeter of the second polygon is also

3 + 3 — 2 (4) + 5 + 4 + 2 + ( 3 —

2 ⋅ 4 − 4 )

= 3 + 6 + 5 + 4 + 2 + 2 = 22.

2. a. 5 + 5 + 4 + x + 4 = 1 — 2 ⋅ x ⋅ 3 + 4x

x + 18 = 3 — 2 x + 4x

x + 18 = 11 —

2 x

− x − x

18 = 9 — 2 x

2 —

9 ⋅ 18 = 2 —

9 ⋅

9 —

2 x


Sample answer: Add the side lengths to get the perimeter.

Add the area of the triangle to the area of the rectangle to

get the total area. Then set the perimeter equal to the area

and solve for x.

The perimeter is 5 + 5 + 4 + 4 + 4 = 22 feet.

The area is 1 —

2 ⋅ 4 ⋅ 3 + 4 ⋅ 4 = 6 + 16 = 22 square feet.

b. 6 + x + 6 + x + 1 + 1 = 6x − 1(2)

2x + 14 = 6x − 2 − 2x − 2x

14 = 4x − 2 + 2 + 2 16 = 4x

16

— 4 = 4x

— 4


Sample answer: Add the side lengths to get the perimeter.

Subtract the area of the small rectangle from the area

of the large rectangle to get the total area. Then set the

perimeter equal to the area and solve for x.

The perimeter of the fi gure is

6 + 4 + 6 + 4 + 1 + 1 = 22 feet.

The area of the fi gure is

6(4) − 1(2) = 24 − 2 = 22 square feet.

c. 1 — 2 ⋅ 2 ⋅ π ⋅ 2 + x + 4 + x = 1 —

2 ⋅ π ⋅ 2

2 + x ⋅ 4

2x + 2π + 4 = 2π + 4x

− 2x − 2x

2π + 4 = 2π + 2x

− 2π − 2π

4 = 2x

4 —

2 = 2x

— 2


Sample answer: Add the circumference of the semicircle

to the remaining three side lengths to fi nd the perimeter.

Add the area of the semicircle to the area of the rectangle

to fi nd the total area. Then set the perimeter equal to the

area and solve for x.

The perimeter of the fi gure is

1 —

2 ⋅ 2 ⋅ π ⋅ 2 + 2 + 4 + 2 = 2π + 8 feet.

The area of the fi gure is

1 —

2 ⋅ π ⋅ 22 + 2 ⋅ 4 = 2π + 8 square feet.


Chapter 1

3. To solve an equation that has variables on both sides, collect

the variable terms on one side of the equation and the

constant terms on the other side of the equation, then solve.

4. Sample answer: Some sample equations are 3x = 4x + 4

{x = −4}, 5x − 7 = 7x − 1{x = −3}, and 2(3x − 4) + 3 = 4x + 5{x = 5}. Note that students may write equations with

no solution or infi nitely many solutions.


1. −2x = 3x + 10

− 3x − 3x

−5x = 10

−5x — −5

= 10 — −5

x = −2

Check: −2x = 3x + 10

−2(−2) =?

3(−2) + 10

4 =?

−6 + 10

4 = 4 ✓


2. 1 —

2 (6h − 4) = −5h + 1

1 —

2 (6h) − 1 —

2 (4) = −5h + 1

3h − 2 = −5h + 1 − 3h − 3h

−2 = −8h + 1

− 1 − 1 −3 = −8h

−3

— −8

= −8h — −8

3 —

8 = h

Check: 1 — 2 (6h − 4) = −5h + 1

1 —

2 [ 6 ( 3 —

8 ) − 4 ] =? −5 ( 3 —

8 ) + 1

1 —

2 ( 9 —

4 − 4 ) =? − 15

— 8 + 1

1 —

2 ( 9 —

4 − 16

— 4 ) =? − 15

— 8 + 8 —

8

1 —

2 ( − 7 —

4 ) =? − 7 —

8

− 7 — 8 = − 7 —

8 ✓

The solution is h = 3 — 8 .

3. − 3 — 4 (8n + 12) = 3(n − 3)

− 3 — 4 (8n) − 3 —

4 (12) = 3(n) − 3(3)

− 6n − 9 = 3n − 9 + 6n + 6n

−9 = 9n − 9 + 9 + 9 0 = 9n

0 —

9 = 9n

— 9

0 = n Check: − 3 —

4 (8n + 12) = 3(n − 3)

− 3 — 4 (8(0) + 12) =

? 3(0 − 3)

− 3 — 4 (0 + 12) =

? 3(−3)

− 3 — 4 (12) =

? −9

−9 = −9 ✓


4. 4(1 − p) = −4p + 4

4(1) − 4(p) = −4p + 4

4 − 4p = −4p + 4

+ 4p + 4p

4 = 4

The statement 4 = 4 is always true. So, the equation is an

identity and has infi nitely many solutions.

5. 6m − m = 5 —

6 (6m − 10)

5m = 5 —

6 (6m) −

5 —

6 (10)

5m = 5m − 25

— 3

− 5m − 5m

0 = − 25 —

3

The statement 0 = − 25 —

3 is never true. So, the equation has

no solution.

6. 10k + 7 = −3 − 10k

+ 10k + 10k

20k + 7 = −3

− 7 − 7 20k = −10

20k — 20

= −10

— 20

k = − 1 — 2

The solution is k = − 1 — 2 .


Chapter 1

7. 3(2a − 2) = 2(3a − 3)

3(2a) − 3(2) = 2(3a) − 2(3)

6a − 6 = 6a − 6

− 6a − 6a

−6 = −6

The statement −6 = −6 is always true. So, the equation is

an identity and has infi nitely many solutions.

8. Words: Distance upstream = Distance downstream

Variable: Let x be the speed (in miles per hour) of the boat

upstream.

Equation: x mi

— 1 h

⋅ 3.5 h = (x + 2) mi

— 1 h

⋅ 2.5 h

3.5x = 2.5(x + 2)

3.5x = 2.5x + 5

− 2.5x − 2.5x

1x = 5

x = 5

3.5x = 3.5(5) = 17.5

The boat travels 17.5 miles upstream.



1. −2(4 − x) = 2x + 8

−2(4) − 2(−x) = 2x + 8

−8 + 2x = 2x + 8

− 2x − 2x

−8 = 8

no; The equation gives a statement that is never true, so the

equation has no solution and is not an identity.

2. 3(3x − 8) = 4x + 6

3(3x) − 3(8) = 4x + 6

9x − 24 = 4x + 6

− 4x − 4x

5x − 24 = 6

+ 24 + 24

5x = 30

5x — 5 =

30 —

5

x = 6

Sample answer: To solve 3(3x − 8) = 4x + 6, the fi rst step

is to use the Distributive Property on the left side of the

equal sign to remove the parentheses and then simplify so

that the equation becomes 9x − 24 = 4x + 6. In order to

eliminate the x-term from the right side, subtract 4x from

each side. Then, in order to isolate the remaining x-term,

undo subtraction by adding 24 to each side. Finally, you can

solve for x by undoing multiplication, so divide each side

by 5. The solution is x = 6.


3. 15 − 2x = 3x

+ 2x + 2x

15 = 5x

15

— 5 =

5x —

5

3 = x

Check: 15 − 2x = 3x

15 − 2(3) =?

3(3)

15 − 6 =?

9

9 = 9 ✓


4. 26 − 4s = 9s

+ 4s + 4s

26 = 13s

26

— 13

= 13s

— 13

2 = s

Check: 26 − 4s = 9s

26 − 4(2) =?

9(2)

26 − 8 =?

18

18 = 18 ✓


5. 5p − 9 = 2p +12

− 2p − 2p

3p − 9 = 12

+ 9 + 9 3p = 21

3p — 3 =

21 —

3

p = 7

Check: 5p −9 = 2p + 12

5(7) − 9 =?

2(7) + 12

35 − 9 =?

14 + 12

26 = 26 ✓



Chapter 1

6. 8g + 10 = 35 + 3g

− 3g − 3g

5g + 10 = 35

− 10 – 10

5g = 25

5g — 5 =

25 —

5

g = 5

Check: 8g + 10 = 35 + 3g

8(5) + 10 =?

35 + 3(5)

40 + 10 =?

35 + 15

50 = 50 ✓


7. 5t + 16 = 6 − 5t

+ 5t + 5t

10t + 16 = 6

− 16 − 16

10t = −10

10t — 10

= −10

— 10

t = −1

Check: 5t + 16 = 6 − 5t

5(−1) + 16 =?

6 − 5(−1)

−5 + 16 =?

6 + 5

11 = 11 ✓


8. −3r + 10 = 15r − 8

+ 3r + 3r

10 = 18r − 8

+ 8 + 8 18 = 18r

18

— 18

= 18r

— 18

1 = r

Check: −3r + 10 = 15r − 8

−3(1) + 10 =?

15(1) − 8

−3 + 10 =?

15 − 8

7 = 7 ✓

The solution is r = 1.

9. 7 + 3x − 12x = 3x + 1

7 − 9x = 3x + 1

+ 9x + 9x

7 = 12x + 1

− 1 − 1 6 = 12x

6 —

12 =

12x —

12

1 —

2 = x

Check: 7 + 3x − 12x = 3x + 1

7 + 3 ( 1 — 2 ) − 12 ( 1 —

2 ) =? 3 ( 1 —

2 ) + 1

7 + 3 —

2 − 6 =

? 3 —

2 + 1

14

— 2 +

3 —

2 −

12 —

2 =

? 3 —

2 +

2 —

2

17

— 2 −

12 —

2 =

? 5 —

2

5 —

2 =

5 —

2 ✓

The solution is x = 1 —

2 .

10. w − 2 + 2w = 6 + 5w

3w − 2 = 6 + 5w

− 3w − 3w

−2 = 6 + 2w

− 6 − 6 −8 = 2w

−8

— 2 =

2w —

2

−4 = w

Check: w − 2 + 2w = 6 + 5w

−4 − 2 + 2(−4) =?

6 + 5(−4)

−4 − 2 − 8 =?

6 − 20

−6 − 8 =?

6 − 20

−14 = −14 ✓



Chapter 1

11. 10(g + 5) = 2(g + 9)

10(g) + 10(5) = 2(g) + 2(9)

10g + 50 = 2g + 18

− 2g − 2g

8g + 50 = 18

− 50 = − 50

8g = −32

8g

— 8 =

−32 —

8

g = −4

Check: 10(g + 5) = 2(g + 9)

10(−4 + 5) =?

2(−4 + 9)

10(1) =?

2(5)

10 = 10 ✓

The solution is g = −4.

12. −9(t − 2) = 4(t − 15)

−9(t) − 9(−2) = 4(t) − 4(15)

−9t + 18 = 4t − 60

+ 9t + 9t

18 = 13t − 60

+ 60 + 60

78 = 13t

78

— 13

= 13t

— 13

6 = t

Check: −9(t − 2) = 4(t − 15)

−9(6 − 2) =?

4(6 − 15)

−9(4) =?

4(−9)

−36 = −36 ✓


13. 2 —

3 (3x + 9) = −2(2x + 6)

2 —

3 (3x) + 2 —

3 (9) = −2(2x) − 2(6)

2x + 6 = −4x − 12

+ 4x + 4x

6x + 6 = −12

− 6 − 6 6x = −18

6x — 6 = −18

— 6

x = −3

Check: 2 — 3 (3x + 9) = −2(2x + 6)

2 —

3 [ 3(−3) + 9 ] =

? −2 [ 2(−3) + 6 ]

2 —

3 (−9 + 9) =

? −2(−6 + 6)

2 —

3 (0) =

? −2(0)

0 = 0 ✓


14. 2(2t + 4) = 3 —

4 (24 − 8t)

2(2t) + 2(4) = 3 —

4 (24) −

3 —

4 (8t)

4t + 8 = 18 − 6t

+ 6t + 6t

10t +8 = 18

− 8 − 8 10t = 10

10t — 10

= 10

— 10

t = 1

Check: 2(2t + 4) = 3 — 4 (24 − 8t)

2 [ 2(1) + 4 ] =?

3 —

4 [ 24 − 8(1) ]

2(2 + 4) =?

3 —

4 (24 − 8)

2(6) =?

3 —

4 (16)

12 = 12 ✓



Chapter 1

15. 10(2y + 2) − y = 2(8y − 8)

10(2y) + 10(2) − y = 2(8y) − 2(8)

20y + 20 − y = 16y − 16

19y + 20 = 16y − 16

− 16y − 16y

3y + 20 = −16

− 20 − 20

3y = −36

3y — 3 =

−36 —

3

y = −12

Check: 10(2y + 2) − y = 2(8y − 8)

10 [ 2(−12) + 2 ] − (−12) =?

2 [ 8(−12) − 8 ]

10(−24 + 2) + 12 =?

2(−96 − 8)

10(−22) + 12 =?

2(−104)

−220 + 12 =?

−208

−208 = −208 ✓


16. 2(4x + 2) = 4x − 12(x − 1)

2(4x) + 2(2) = 4x −12(x) − 12(−1)

8x + 4 = 4x −12x + 12

8x + 4 = −8x +12

+ 8x + 8x

16x + 4 = 12

− 4 − 4 16x = 8

16x — 16

= 8 —

16

x = 1 —

2

Check: 2(4x + 2) = 4x −12(x − 1)

2 [ 4 ( 1 — 2 ) + 2 ] =? 4 ( 1 —

2 ) − 12 ( 1 —

2 − 1 )

2(2 + 2) =?

2 − 12 ( − 1 — 2 )

2(4) =?

2 + 6

8 = 8 ✓


2 .

17. 50h = 190 − 45h

+ 45h + 45h

95h = 190

95h — 95

= 190

— 95

h = 2 You and your friend will meet after you have been driving

toward each other for 2 hours.

18. 1.5r +15 = 2.25r

− 1.5r − 1.5r

15 = 0.75r

15

— 0.75

= 0.75r

— 0.75

20 = r

You must rent 20 movies to spend the same amount at each

store.

19. 3t + 4 = 12 + 3t

− 3t − 3t

4 = 12

The statement 4 = 12 is never true. So, the equation has no

solution.

20. 6d + 8 = 14 + 3d

− 3d − 3d

3d + 8 = 14

− 8 − 8 3d = 6

3d — 3 =

6 —

3

d = 2

The equation has one solution: d = 2.

21. 2(h + 1) = 5h − 7

2(h) + 2(1) = 5h − 7

2h + 2 = 5h − 7

− 2h − 2h

2 = 3h − 7

+ 7 + 7 9 = 3h

9 —

3 =

3h —

3

3 = h

The equation has one solution: h = 3.


Chapter 1

22. 12y + 6 = 6(2y + 1)

12y + 6 = 6(2y) + 6(1)

12y + 6 = 12y + 6

− 12y − 12y

6 = 6



23. 3(4g + 6) = 2(6g + 9)

3(4g) + 3(6) = 2(6g) + 2(9)

12g + 18 = 12g + 18

− 12g − 12g

18 = 18



24. 5(1 + 2m) = 1 —

2 (8 + 20m)

5(1) + 5(2m) = 1 —

2 (8) +

1 —

2 (20m)

5 + 10m = 4 + 10m

− 10m − 10m

5 = 4


solution.

25. In order to undo subtraction, 3c should have been added to

each side.

5c − 6 = 4 − 3c

+ 3c + 3c

8c − 6 = 4

+ 6 + 6 8c = 10

8c — 8 =

10 —

8

c = 5 —

4

The solution is c = 5 —

4 .

26. Because the statement 0 = 0 is always true, the equation has

infi nitely many solutions. It is better to subtract a variable

term from each side before subtracting a constant from

each side.

6(2y + 6) = 4(9 + 3y)

6(2y) + 6(6) = 4(9) + 4(3y)

12y + 36 = 36 + 12y

− 12y − 12y

36 = 36

Because the statement 36 = 36 is always true, the equation

has infi nitely many solutions.

27. Words: Total cost of

Company A’s

Internet service

=

Total cost of

Company B’s

Internet service

Variable: Let m be the number of months you have Internet

service.

Equation: 60.00 + 42.95m = 25.00 + 49.95m

60 + 42.95m = 25 + 49.95m

−42.95m −42.95m

60 = 25 + 7m

− 25 − 25

35 = 7m

35

— 7 = 7m

— 7

5 = m

After 5 months, you would pay the same total amount for

each Internet service.

28. Words: 4% of total

protein needed

daily

+ 48 = Total protein

needed daily

Variable: Let p be the total amount (in grams) of protein you

need daily.

Equation: 0.04 ⋅ p + 48 = p

0.04p + 48 = p

− 0.04p − 0.04p

48 = 0.96p

48

— 0.96

= 0.96p —

0.96

50 = p

You need 50 grams of protein daily.

29. 8(x + 6) − 10 + r = 3(x + 12) + 5x

8(x) + 8(6) − 10 + r = 3(x) + 3(12) + 5x

8x + 48 − 10 + r = 3x + 36 + 5x

8x + 38 + r = 8x + 36

− 8x − 8x

38 + r = 36

− 38 − 38

r = −2

So, r = −2.


Chapter 1

30. 4(x − 3) − r + 2x = 5(3x − 7) −9x

4(x) − 4(3) − r + 2x = 5(3x) − 5(7) −9x

4x − 12 − r + 2x = 15x − 35 − 9x

6x − 12 − r = 6x − 35

− 6x − 6x

−12 − r = −35

+ 12 + 12

−r = −23

−r

— −1

= −23

— −1

r = 23

So, r = 23.

31. 2πr2 + 2πrh = πr2h

2π(2.5)2 + 2π(2.5)(x) = π(2.5)2(x)

12.5π + 5πx = 6.25πx

− 5πx − 5πx

12.5π = 1.25πx

12.5π — 1.25π

= 1.25πx

— 1.25π

10 = x

SA = 2πr2 + 2πrh V = πr2h

= 2π(2.5)2 + 2π(2.5)(10) = π(2.5)2(10)

= 12.5π + 50π = π(6.25)(10)

= 62.5π = 62.5π

≈ 62.5(3.1416) ≈ 62.5(3.1416)

= 196.35 = 196.35

So, x = 10, and the surface area is 62.5π, or about 196.35

square centimeters and the volume is 62.5π, or about 196.35 cubic centimeters.

32. 2πr2 + 2πrh = πr2h

2π ( 18 —

5 ) 2 + 2π ( 18

— 5 ) (x) = π ( 18

— 5 ) 2 (x)

2π ( 324 —

25 ) +

36 —

5 πx =

324 —

25 πx

648

— 25

π + 36 —

5 πx =

324 —

25 πx

− 36 —

5 πx − 36

— 5 πx

648

— 25

π = 144

— 25

πx

25

— 144

⋅ 648

— 25

π = 25

— 144

⋅ 144

— 25

πx

9 —

2 π = πx

9 — 2 π — π =

πx —

π

9 —

2 = x

SA = 2 πr2 + 2 πrh V = πr2h

= 2 π ( 18 —

5 ) 2 + 2 π ( 18

— 5 ) ( 9 —

2 ) = π ( 18

— 5 ) 2 ( 9 —

2 )

= 2 π ( 324 —

25 ) + 2 π ( 81

— 5 ) = π ( 324

— 25

) ( 9 — 2 )

= 648 —

25 π +

162 —

5 π = 1458

— 25

π

= 648 —

25 π +

810 —

25 π ≈

1458 —

25 (3.1416)

= 1458 —

25 π ≈ 183.22

≈ 1458

— 25

(3.1416)

≈ 183.22

So, x = 9 —

2 , and the surface area is

1458 —

25 π, or about

183.22 square feet and the volume is 1458

— 25

π, or about

183.22 cubic feet.

33. Words: Cheetah’s

running distance

= 120 feet + Antelope’s

running distance

Variable: Let t be the time (in seconds) the animals are

running.

Equation: 90 ft

— 1 sec

⋅ t sec = 120 ft + 60 ft

— 1 sec

⋅ t sec

90 t = 120 + 60 t

− 60 t − 60 t

30 t = 120

30 t — 30

= 120

— 30

t = 4

The cheetah will catch up to the antelope in 4 seconds.


Chapter 1

34. 90 ft —

1 sec ⋅ t sec = 650 ft +

60 ft —

1 sec ⋅ t sec

90t = 650 + 60t

− 60t − 60t

30t = 650

30t — 30

= 650

— 30

t = 21 2 —

3 sec

no; In order to catch the antelope, the cheetah would have to

be running at top speed for over 20 seconds, so the antelope

is probably safe.

35. a(2x+3) = 9x + 15 + x

a(2x) + a(3) = 10x + 15

2ax + 3a = 10x + 15

Set the coeffi cients of x equal to each other and set the

constant terms equal to each other.

2a = 10 3a = 15

2a — 2 =

10 —

2

3a —

3 =

15 —

3

a = 5 a = 5

If a = 5, then 2ax + 3a = 2(5) x + 3(5) = 10x + 15.

Because the other side of the equal sign is also 10x + 15, the

equation is an identity when a = 5.

36. 8x − 8 + 3ax = 5ax − 2a

8x + 3ax − 8 = 5ax − 2a

(8 + 3a)x − 8 = 5ax − 2a

Set the coeffi cients of x equal to each other and set the

constant terms equal to each other.

8 + 3a = 5a −8 = −2a

− 3a − 3a −8

— −2

= −2a

— −2

8 = 2a 4 = a

8 —

2 =

2a —

2

4 = a If a = 4, then 8x − 8 + 3ax = 8x − 8 + 3(4)x = 8x − 8 +

12x = 20x − 8, and 5ax − 2a = 5(4)x − 2(4) = 20x − 8.

Because the expressions on each side of the equation are the

same, the equation is an identity when a = 4.

37. Words: 2 ⋅ Greater

consecutive

integer

= 3 ⋅ Lesser

consecutive

integer

− 9

Variable: Let n be the lesser consecutive integer. Then n + 1

is the greater consecutive integer.

Equation: 2 ⋅ (n + 1) = 3 ⋅ n − 9

2(n + 1) = 3n − 9

2(n) + 2(1) = 3n − 9

2n + 2 = 3n − 9

− 2n − 2n

2 = n − 9 + 9 + 9 11 = n

n + 1 = 11 + 1 = 12

The integers are 11 and 12.

38. a. After 6 years, there will be equal enrollment in Spanish

and French classes because the graphs meet at this point.

b. The left side, 355 − 9x, represents the predicted Spanish

class enrollment, and the right side, 229 + 12x, represents

the predicted French class enrollment. So, the equation

represents when there will be equal enrollment in Spanish

and French classes. The solution should give the same

result as the graph.

355 − 9x = 229 + 12x

+ 9x + 9x

355 = 229 + 21x

− 229 − 229

126 = 21x

126

— 21

= 21x

— 21

6 = x

The equation confi rms that there will be equal enrollment

in Spanish and French classes after 6 years if the trend

continues as predicted.


Chapter 1

39. a. Sample answer: 2(x + 5) = 2x − 7

2(x) + 2(5) = 2x − 7

2x + 10 = 2x − 7 − 2x − 2x

10 = −7

The statement 10 = −7 is never true. So, the equation

2(x + 5) = 2x − 7 has no solution.

b. Sample answer: 2(3x + 6) = 3(2x + 4)

2(3x)+2(6) = 3(2x) + 3(4)

6x + 12 = 6x + 12

− 6x − 6x

12 = 12

The statement 12=12 is always true. So, the equation

2(3x + 6) = 3(2x + 4) is an identity and has infi nitely

many solutions.

40. Sample answer: The perimeter of the given triangle is

P = (x + 3) + (2x + 1) + 3x = 6x + 4.

Another fi gure with the same perimeter is shown.

22

3x

3x

P = 3x + 2 + 3x + 2 = 6x + 4


41. −4, ∣ 2 ∣ = 2, ∣ −4 ∣ = 4, 5, 9

42. − ∣ 21 ∣ = −21, −16, ∣ −10 ∣ = 10, 22, ∣ −32 ∣ = 32

43. −19, −18, ∣ −18 ∣ = 18, ∣ 22 ∣ = 22, ∣ −24 ∣ = 24

44. − ∣ −3 ∣ = −3, −2, −1, ∣ 0 ∣ = 0, ∣ 2 ∣ = 2

1.1–1.3 What Did You Learn? (p. 25)

1. Sample answer: Let A represent the area of a single

rectangle. Because each rectangle has the same area, and the

area of the square is half the area of a rectangle, use

the expression 4A + 1 — 2 A to represent the total area.

2. Sample answer: A protractor can only measure angles to the

nearest whole degree. It is not a very precise instrument.

3. Sample answer: the defi nition of an identity, which has

infi nitely many solutions

1.1–1.3 Quiz (p. 26)

1. x + 9 = 7 Write the equation.


x = −2 Simplify.

Check: x + 9 = 7

−2 + 9 =?

7

7 = 7 ✓


2. 8.6 = z − 3.8 Write the equation.

+ 3.8 + 3.8 Add 3.8 to each side.

12.4 = z Simplify.

Check: 8.6 = z − 3.8

8.6 =?

12.4 − 3.8

8.6 =?

8.6 ✓

The solution is z = 12.4.

3. 60 = −12r Write the equation.

60 —

−12 = −12r

— −12 Divide each side by −12.

−5 = r Simplify.

Check: 60 = −12r

60 =?

−12(−5)

60 = 60 ✓


4. 3 —

4 p = 18 Write the equation.

4 —

3 ⋅

3 —

4 p =

4 —

3 ⋅ 18 Multiply each side by

4 —

3 .

p = 24 Simplify.

Check: 3 — 4 p = 18

3 —

4 (24) =

? 18

18 = 18 ✓


5. 2m − 3 = 13

+3 +3

2m = 16

2m

— 2 =

16 —

2

m = 8

Check: 2m − 3 = 13

2(8) − 3 =?

13

16 − 3 =?

13

13 = 13 ✓



Chapter 1

6. 5 = 10 − v Check: 5 = 10 − v

5 =?

10 − 5

5 = 5 ✓

− 10 − 10

−5 = −v

−5

— −1

= −v

— −1

5 = v

The solution is v = 5.

7. 5 = 7w + 8w + 2

5 = 15w + 2

− 2 − 2 3 = 15w

3 —

15 =

15w —

15

1 —

5 = w

Check: 5 = 7w + 8w + 2

5 =?

7 ( 1 — 5 ) + 8 ( 1 —

5 ) + 2

5 =?

7 —

5 +

8 —

5 + 2

5 =?

15

— 5 + 2

5 =?

3 + 2

5 = 5 ✓

The solution is w = 1 —

5 .

8. −21a + 28a − 6 = −10.2

7a − 6 = −10.2

+ 6 + 6 7a = −4.2

7a

— 7 =

−4.2 —

7

a = −0.6

Check: −21a + 28a − 6 = −10.2

−21(−0.6) + 28(−0.6) − 6 =?

−10.2

12.6 − 16.8 − 6 =?

−10.2

−4.2 − 6 =?

−10.2

−10.2 = −10.2 ✓

The solution is a = −0.6.

9. 2k − 3(2k − 3) = 45

2k − 3(2k) − 3(−3) = 45

2k − 6k + 9 = 45

−4k + 9 = 45

− 9 − 9 −4k = 36

−4k

— −4

= 36

— −4

k = −9

Check: 2k − 3(2k − 3) = 45

2(−9) − 3[2(−9) − 3] =?

45

−18 − 3(−18 − 3) =?

45

−18 − 3(−21) =?

45

−18 + 63 =?

45

45 = 45 ✓

The solution is k = − 9.

10. 68 = 1 —

5 (20x + 50) + 2

68 = 1 —

5 (20x) +

1 —

5 (50) + 2

68 = 4x + 10 +2

68 = 4x + 12

− 12 − 12

56 = 4x

56

— 4 =

4x —

4

14 = x

Check: 68 = 1 —

5 (20x + 50) + 2

68 =?

1 —

5 [20(14) + 50] + 2

68 =?

1 —

5 (280 + 50) + 2

68 =?

1 —

5 (330) + 2

68 =?

66 + 2

68 = 68 ✓


11. 3c + 1 = c + 1

− c − c 2c + 1 = 1

− 1 − 1 2c = 0

2c — 2 =

0 —

2

c = 0 The solution is c = 0.


Chapter 1

12. −8 − 5n = 64 + 3n

+ 5n + 5n

−8 = 64 + 8n

− 64 − 64

−72 = 8n

−72

— 8 =

8n —

8

−9 = n


13. 2(8q − 5) = 4q

2(8q) − 2(5) = 4q

16q − 10 = 4q

− 16q − 16q

−10 = −12q

−10

— −12

= −12q

— −12

5 —

6 = q

The solution is q = 5 —

6 .

14. 9(y − 4) − 7y = 5(3y − 2)

9(y) − 9(4) − 7y = 5(3y) − 5(2)

9y − 36 − 7y = 15y − 10

2y − 36 = 15y − 10

− 2y − 2y

− 36 = 13y − 10

+ 10 + 10

−26 = 13y

−26

— 13

= 13y

— 13

−2 = y


15. 4(g + 8) = 7 + 4g

4(g) + 4(8) = 7 + 4g

4g + 32 = 7 + 4g

− 4g − 4g

32 = 7


solution.

16. −4(−5h − 4) = 2(10h + 8)

−4(−5h) − 4(−4) = 2(10h) + 2(8)

20h + 16 = 20h + 16

− 20h − 20h

16 = 16



17. Words: Distance

from a

thunderstorm=

Time between

lightning and

thunder÷ 5

Variable: Let s be the time (in seconds) between when

you see lightning and when you hear thunder.

Equation: 2 = s ÷ 5

2 = s —

5

5 ⋅ 2 = 5 ⋅ s —

5

10 = s You would count 10 seconds between when you see

lightning and when you hear thunder for a thunderstorm that

is 2 miles away.

18. Let x be the spacing (in feet) between the posters.

15 = 3 + 2 + x + 2 + x + 2 + 3

15 = 2x + 12

− 12 − 12

3 = 2x

3 —

2 =

2x —

2

3 —

2 = x

There should be 3 —

2 or 1

1 —

2 feet between the posters.


Chapter 1

19. a. Words: Total cost

for Studio A= Total cost for

Studio B

Variable: Let h be the time (in hours) spent painting

at the studio.

Equation: 10 + 8h = 16 + 6h

10 + 8h = 16 + 6h

−6h −6h

10 + 2h = 16

− 10 − 10

2h = 6

2h — 2 =

6 —

2

h = 3

After 3 hours of painting, the total costs will be the same

at both studios.

b. If Studio B increases their studio fee by $2, then both

studios have the same studio fee of $8.

10 + 8h = 16 + 8h

− 8h − 8h

10 = 16

The statement 10 = 16 is never true. So, the equation has

no solution. In other words, because Studio B charges

more for the vase, if their studio fee is the same, their total

costs will never be the same. More specifi cally, Studio B

will always charge more.


1. a. Because ∣ 3 ∣ = 3 and ∣ −3 ∣ = 3, it must be that x + 2

equals either 3 or −3. So,

x + 2 = 3 or x + 2 = −3.

b. x + 2 = 3 or x + 2 = −3

− 2 − 2 − 2 − 2 x = 1 or x = −5

The solutions are x = −5 and x = 1.

c. Sample answer: If an absolute value expression is equal

to a constant, then the expression is equal to the constant

or its opposite. You can write two linear equations, one for

each of these possibilities, and then solve both equations.

2. a. 2 40−2−4−6−8

x + 2 = 0

−2 + 2 =?

0

0 = 0 ✓

b.

2 40−2−4−6−8

−5 1

Both of these values, −5 and 1, are solutions of the

absolute value equation ∣ x + 2 ∣ = 3.

∣ x + 2 ∣ = 3 ∣ x + 2 ∣ = 3

∣ −5 + 2 ∣ =? 3 ∣ 1 + 2 ∣ =? 3

∣ −3 ∣ =? 3 ∣ 3 ∣ =? 3

3 = 3 ✓ 3 = 3 ✓

c. Set the expression inside the absolute value symbol equal

to 0 and solve. Plot the solution on a number line. Then

plot the points that are the constant amount of units from

that point. These last 2 points are the solutions to the

original equation.

3. a. Ax-6-5-4-3-2-1012

B|x + 2|432101234

21

345678910

The solutions given by the spreadsheet are −5 and 1,

because they are the values of x that make ∣ x + 2 ∣ equal

to 3.

b. The spreadsheet method yielded the same solutions, −5

and 1, as the other two methods.

c. Sample answer: Have the spreadsheet calculate the value

of the absolute value expression for many values of x, and

fi nd the ones that give the expected solution.

4. Sample answer: You can solve an absolute value equation

algebraically, graphically, or numerically. For the algebraic

method, write and solve two linear equations, one that has

the expression equal to the constant and one that has the

expression equal to the opposite of the constant. For the

graphical method, fi rst identify the point on the number

line that makes the absolute value expression equal to 0.

Using the constant that the absolute value expression is

equal to, fi nd both of the points that are this many units

from the original point in either direction. These two values

are the solutions. For the numerical method, you can use

a spreadsheet to calculate the values of the absolute value

expression for many values of the variable until you identify

the value or values that make the absolute value expression

equal to the given constant.

Chapter 1


Chapter 1

5. Sample answer: The algebraic method is favorable because

it is the quickest method. The graphical method is also

favorable because it helps to visualize absolute value. The

numerical method is not favorable because setting up the

spreadsheet is time consuming.


1. ∣ x ∣ = 10

x = 10 or x = −10

−10

4 8 120−4−8−12

10

The solutions are x = −10 and x =10.

2. ∣ x − 1 ∣ = 4

x − 1 = 4 or x − 1 = −4

+ 1 + 1 + 1 + 1

x = 5 x = −3

−3

2 4 60−2−4

5


3. ∣ 3 + x ∣ = −3

The absolute value of an expression must be greater than or

equal to 0. The expression ∣ 3 + x ∣ cannot equal −3. So, the

equation has no solution.

4. ∣ x − 2 ∣ + 5 = 9 −5 −5

∣ x − 2 ∣ = 4

x − 2 = 4 or x − 2 = −4

+ 2 + 2 + 2 + 2 x = 6 x = −2

Check:

∣ x − 2 ∣ + 5 = 9 ∣ x − 2 ∣ + 5 = 9

∣ 6 − 2 ∣ + 5 =?

9 ∣ −2 − 2 ∣ + 5 =?

9

∣ 4 ∣ + 5 =?

9 ∣ −4 ∣ + 5 =?

9

4 + 5 =?

9 4 + 5 =?

9

9 = 9 ✓ 9 = 9 ✓


5. 4 ∣ 2x + 7 ∣ = 16

4 ∣ 2x + 7 ∣

— 4 =

16 —

4

∣ 2x + 7 ∣ = 4

2x + 7 = 4 or 2x + 7 = −4

− 7 − 7 − 7 − 7 2x = −3 2x = −11

2x — 2 = − 3 —

2

2x —

2 =

−11 —

2

x = − 3 — 2 x = − 11

— 2

Check:

4 ∣ 2x + 7 ∣ = 16 4 ∣ 2x + 7 ∣ = 16

4 ∣ 2 ( − 3 — 2 ) + 7 ∣ =? 16 4 ∣ 2 ( − 11

— 2 ) + 7 ∣ =? 16

4 ∣ −3 + 7 ∣ =? 16 4 ∣ −11 + 7 ∣ =? 16

4 ∣ 4 ∣ =? 16 4 ∣ −4 ∣ =? 16

4 ⋅ 4 =?

16 4 ⋅ 4 =?

16

16 = 16 ✓ 16 = 16 ✓

The solutions are x = − 11 —

2 and x = − 3 —

2 .

6. −2 ∣ 5x − 1 ∣ − 3 = −11

+ 3 + 3

−2 ∣ 5x − 1 ∣ = −8

−2 ∣ 5x − 1 ∣

— −2

= −8

— −2

∣ 5x − 1 ∣ = 4

5x − 1 = 4 or 5x − 1 = −4

+ 1 + 1 + 1 + 1

5x = 5 5x = −3

5x — 5 =

5 —

5

5x —

5 =

−3 —

5

x = 1 x = − 3 — 5

Check:

−2 ∣ 5x − 1 ∣ − 3 = −11 −2 ∣ 5x − 1 ∣ − 3 = −11

−2 ∣ 5(1) − 1 ∣ − 3 =?

−11 −2 ∣ 5 ( − 3 — 5 ) − 1 ∣ − 3 =

? −11

−2 ∣ 5 − 1 ∣ − 3 =?

−11 −2 ∣ −3 − 1 ∣ − 3 =?

−11

−2 ∣ 4 ∣ − 3 =?

−11 −2 ∣ − 4 ∣ − 3 =?

−11

−2(4) − 3 =?

−11 −2(4) − 3 =?

−11

−8 − 3 =?

−11 −8 − 3 =?

−11

−11 = −11 ✓ −11 = −11 ✓

The solutions are x = − 3 — 5 and x = 1.


Chapter 1

7. 8 8

16 20 24 28 32 3612

The equation is ∣ x − 24 ∣ = 8.

Check: ∣ x − 24 ∣ = 8 ∣ x − 24 ∣ = 8

∣ 16 − 24 ∣ =? 8 ∣ 32 − 24 ∣ =? 8

∣ −8 ∣ =? 8 ∣ 8 ∣ =? 8

8 = 8 ✓ 8 = 8 ✓

8. ∣ x + 8 ∣ = ∣ 2x + 1 ∣ x + 8 = 2x + 1 or x + 8 = −(2x + 1)

− x − x x + 8 = −2x − 1

8 = x + 1 + 2x + 2x

− 1 − 1 3x + 8 = −1

7 = x − 8 − 8 3x = −9

3x

— 3 =

−9 —

3

x = −3

Check: ∣ x + 8 ∣ = ∣ 2x + 1 ∣ ∣ x + 8 ∣ = ∣ 2x + 1 ∣ ∣ 7 + 8 ∣ =? ∣ 2(7) + 1 ∣ ∣ −3 + 8 ∣ =? ∣ 2(− 3) + 1 ∣ ∣ 15 ∣ =? ∣ 14 + 1 ∣ ∣ 5 ∣ =? ∣ −6 + 1 ∣ 15 =

? ∣ 15 ∣ 5 =

? ∣ −5 ∣

15 = 15 ✓ 5 = 5 ✓


9. 3 ∣ x − 4 ∣ = ∣ 2x + 5 ∣ 3(x − 4) = 2x + 5 or 3(x − 4) = −(2x + 5)

3(x) − 3(4) = 2x + 5 3(x) − 3(4) = −2x − 5

3x − 12 = 2x + 5 3x − 12 = −2x − 5

− 2x − 2x + 2x + 2x

x − 12 = 5 5x − 12 = −5

+ 12 + 12 + 12 + 12

x = 17 5x = 7

5x

— 5 =

7 —

5

x = 7 —

5

Check: 3 ∣ x − 4 ∣ = ∣ 2x + 5 ∣ 3 ∣ x − 4 ∣ = ∣ 2x + 5 ∣

3 ∣ 17 − 4 ∣ =? ∣ 2(17) + 5 ∣ 3 ∣ 7 — 5 − 4 ∣ =? ∣ 2 ( 7 —

5 ) + 5 ∣

3 ∣ 13 ∣ =? ∣ 34 + 5 ∣ 3 ∣ 7 — 5 −

20 —

5 ∣ =? ∣ 14

— 5 + 5 ∣

3(13) =?

∣ 39 ∣ 3 ∣ − 13 —

5 ∣ =? ∣ 14

— 5 +

25 —

5 ∣

39 = 39 ✓ 3 ( 13 —

5 ) =? ∣ 39

— 5 ∣

39

— 5 = 39

— 5 ✓

The solutions are x = 7 —

5 and x = 17.

10. ∣ x + 6 ∣ = 2x

x + 6 = 2x or x + 6 = −2x

− x − x − x − x 6 = x 6 = −3x

6 —

−3 =

−3x —

−3

−2 = x

Check: ∣ x + 6 ∣ = 2x ∣ x + 6 ∣ = 2x

∣ 6 + 6 ∣ =? 2(6) ∣ −2 + 6 ∣ =? 2(−2)

∣ 12 ∣ =? 12 ∣ 4 ∣ =? −4

12 = 12 ✓ 4 ≠ −4 ✗

The solution is x = 6. Reject x = −2 because it is extraneous.

11. ∣ 3x − 2 ∣ = x

3x − 2 = x or 3x − 2 = −x

− 3x − 3x − 3x − 3x

−2 = −2x −2 = −4x

−2

— −2

= −2x

— −2

−2

— −4

= −4x

— −4

1 = x 1 —

2 = x

Check: ∣ 3x − 2 ∣ = x ∣ 3x − 2 ∣ = x

∣ 3(1) − 2 ∣ =? 1 ∣ 3 ( 1 — 2 ) − 2 ∣ =?

1 —

2

∣ 3 − 2 ∣ =? 1 ∣ 3 — 2 − 2 ∣ =?

1 —

2

∣ 1 ∣ =? 1 ∣ 3 — 2 −

4 —

2 ∣ =?

1 —

2

1 = 1 ✓ ∣ − 1 —

2 ∣ =?

1 —

2

1 —

2 =

1 —

2 ✓

The solutions are x = 1 —

2 and x = 1.


Chapter 1

12. ∣ 2 + x ∣ = ∣ x − 8 ∣ 2 + x = x − 8 or 2 + x = −(x − 8)

− x − x 2 + x = −x + 8

2 = −8 + x + x 2 + 2x = 8

− 2 − 2 2x = 6

2x

— 2 = 6 —

2

x = 3

Check: ∣ 2 + x ∣ = ∣ x − 8 ∣ ∣ 2 + 3 ∣ =? ∣ 3 − 8 ∣ ∣ 5 ∣ =? ∣ −5 ∣ 5 = 5 ✓

The solution is x =3.

13. ∣ 5x − 2 ∣ = ∣ 5x + 4 ∣ 5x − 2 = 5x + 4 or 5x − 2 = −(5x + 4)

− 5x − 5x 5x − 2 = −5x − 4

−2 = 4 + 5x + 5x

10x − 2 = −4

+2 +2

10x = −2

10x

— 10

= −2

— 10

x = − 1 — 5

Check: ∣ 5x − 2 ∣ = ∣ 5x + 4 ∣

∣ 5 ( − 1 — 5 ) − 2 ∣ =? ∣ 5 ( − 1 —

5 ) + 4 ∣

∣ −1 − 2 ∣ =? ∣ −1 + 4 ∣ ∣ −3 ∣ =? ∣ 3 ∣ 3 = 3 ✓

The solution is x = − 1 — 5 .



1. An extraneous solution is an apparent solution that must be

rejected because it does not satisfy the original equation.

2. The absolute value of an expression must be greater than or

equal to 0. So, the expression ∣ 4x − 7 ∣ cannot equal −1, and

the equation has no solution.


3. ∣ −9 ∣ = 9 4. − ∣ 15 ∣ = −15

5. ∣ 14 ∣ − ∣ −14 ∣ = 14 − 14 = 0

6. ∣ −3 ∣ + ∣ 3 ∣ = 3 + 3 = 6

7. − ∣ −5 ∙ (−7) ∣ = − ∣ 35 ∣ = −35

8. ∣ −0.8 ∙ 10 ∣ = ∣ −8 ∣ = 8

9. ∣ 27 —

−3 ∣ = ∣ −9 ∣ = 9

10. ∣ − −12 —

4 ∣ = ∣ −(−3) ∣ = ∣ 3 ∣ =3

11. ∣ w ∣ = 6

w = 6 or w = −6

2 4 60−2−4−6

The solutions are w = −6 and w = 6.

12. ∣ r ∣ = −2

The absolute value of a number must be greater than or equal

to 0 and cannot be equal to −2. So, the equation has no

solution.

13. ∣ y ∣ = −18

The absolute value of a number must be greater than or equal

to 0 and cannot be equal to −18. So, the equation has no

solution.

14. ∣ x ∣ = 13

x = 13 or x = –13

−13

4 8 120−4−8−12

13

The solutions are x = –13 and x = 13.

15. ∣ m + 3 ∣ = 7m + 3 = 7 or m + 3 = −7

− 3 − 3 − 3 − 3

m = 4 m = −10

2 40−2−4−6−8−10

The solutions are m = −10 and m = 4.

16. ∣ q − 8 ∣ = 14

q − 8 = 14 or q − 8 = −14

+ 8 + 8 + 8 + 8

q = 22 q = −6

−6

8 16 240−8

22

The solutions are q = −6 and q = 22.

The statement

2 = –8 is false.

So, the original

equation has only

one solution.

The statement –2 = 4

is false. So, the original

equation has only one

solution.


Chapter 1

17. ∣ −3d ∣ = 15

−3d = 15 or −3d = −15

−3d —

−3 =

15 —

−3 −3d

— −3

= −15

— −3

d = −5 d = 5

−5

2 4 60−2−4−6

5

The solutions are d = −5 and d = 5.

18. ∣ t — 2 ∣ = 6

t —

2 = 6 or

t —

2 = −6

2⋅ t —

2 = 2⋅ 6 2⋅

t —

2 = 2(−6)

t = 12 t = −12

4 8 120−4−8−12

The solutions are t = −12 and t = 12.

19. ∣ 4b − 5 ∣ = 19

4b − 5 = 19 or 4b − 5 = −19

+ 5 + 5 + 5 + 5

4b = 24 4b = −14

4b

— 4 =

24 —

4

4b —

4 =

−14 —

4

b = 6 b = − 7 — 2

−3.5

2 4 60−2−4

The solutions are b = − 7 — 2 and b = 6.

20. ∣ x − 1 ∣ + 5 = 2

− 5 − 5∣ x − 1 ∣ = −3


equal to 0. The expression ∣ x − 1 ∣ cannot equal −3. So, the


21. −4 ∣ 8 − 5n ∣ = 13

−4 ∣ 8 − 5n ∣ — −4

= 13

— −4

∣ 8 − 5n ∣ = − 13 —

4


equal to 0. The expression ∣ 8 − 5n ∣ cannot equal − 13 — 4 . So,

the equation has no solution.

22. −3 ∣ 1 − 2 — 3 v ∣ = −9

−3 ∣ 1 − 2 — 3 v ∣ —

−3 =

−9 —

−3

∣ 1 − 2 —

3 v ∣ = 3

1 − 2 — 3 v = 3 or 1 − 2 —

3 v = −3

− 1 − 1 − 1 − 1

− 2 —

3 v = 2 − 2 —

3 v = −4

− 3 — 2 ( − 2 —

3 v ) = − 3 —

2 ⋅ 2 − 3 —

2 ( − 2 —

3 v ) = −

3 —

2 (−4)

v = −3 v = 6

−3

2 4 60−2−4

The solutions are v = −3 and v = 6.

23. 3 = −2 ∣ 1 — 4 s − 5 ∣ + 3

− 3 − 3

0 = −2 ∣ 1 — 4 s − 5 ∣

0 —

−2 =

−2 ∣ 1 — 4 s −5 ∣ ——

−2

0 = ∣ 1 — 4 s − 5 ∣

0 = 1 —

4 s − 5

+ 5 + 5

5 = 1 —

4 s

4(5) = 4⋅ 1 — 4 s

20 = s

10 200



Chapter 1

24. 9 ∣ 4p + 2 ∣ + 8 = 35

− 8 − 8 9 ∣ 4p + 2 ∣ = 27

9 ∣ 4p + 2 ∣

— 9

= 27

— 9

∣ 4p + 2 ∣ = 3

4p + 2 = 3 or 4p + 2 = −3

− 2 − 2 − 2 − 2 4p = 1 4p = −5

4p — 4 =

1 —

4

4p —

4 =

−5 —

4

p = 1 —

4 p = − 5 —

4

0−−1 12

12

− 14

54

− 32

The solutions are p = − 5 — 4 and p = 1 —

4 .

25. a. 91.4

91 92 93 94 95

94.5

b. Let d be the distance (in millions of miles) from Earth to

the Sun.

94.5 − 91.4

— 2 =

3.1 —

2 = 1.55

91.4 + 1.55 = 92.95

The equation is ∣ d − 92.95 ∣ = 1.55.

Check: ∣ d − 92.95 ∣ = 1.55 ∣ d − 92.95 ∣ = 1.55

∣ 91.4 − 92.95 ∣ =? 1.55 ∣ 94.5 − 92.95 ∣ =? 1.55

∣ −1.55 ∣ =? 1.55 ∣ 1.55 ∣ =? 1.55

1.55 = 1.55 ✓ 1.55 = 1.55 ✓

26. a.

10 12 14 16

15

b. Let h be the shoulder height (in inches).

15 − 10

— 2 =

5 —

2 = 2.5 10 + 2.5 = 12.5

The equation is ∣ h − 12.5 ∣ = 2.5.

Check: ∣ h − 12.5 ∣ = 2.5 ∣ h − 12.5 ∣ = 2.5

∣ 10 − 12.5 ∣ =? 2.5 ∣ 15 − 12.5 ∣ =? 2.5

∣ −2.5 ∣ =? 2.5 ∣ 2.5 ∣ =? 2.5

2.5 = 2.5 ✓ 2.5 = 2.5 ✓

27. B; The halfway point is −2, and the distance from the halfway

point to the minimum and maximum respectively is 4.

28. D; The halfway point is 4, and the distance from the halfway


29. C; The halfway point is 2, and the distance from the halfway


30. A; The halfway point is −4, and the distance from the halfway


31. 18 − 8 —

2 =

10 —

2 = 5 8 + 5 = 13


Check: ∣ x − 13 ∣ = 5 ∣ x − 13 ∣ = 5

∣ 8 − 13 ∣ =? 5 ∣ 18 − 13 ∣ =? 5

∣ − 5 ∣ =? 5 ∣ 5 ∣ =? 5

5 = 5 ✓ 5 = 5 ✓

32. 10 − (−6) —

2 =

16 —

2 = 8 −6 + 8 = 2


Check: ∣ x − 2 ∣ = 8 ∣ x − 2 ∣ = 8

∣ −6 − 2 ∣ =? 8 ∣ 10 − 2 ∣ =? 8

∣ −8 ∣ =? 8 ∣ 8 ∣ =? 8

8 = 8 ✓ 8 = 8 ✓

33. 9 − 2 —

2 =

7 —

2 = 3.5 2 + 3.5 = 5.5

The equation is ∣ x − 5.5 ∣ = 3.5.

Check: ∣ x − 5.5 ∣ = 3.5 ∣ x − 5.5 ∣ = 3.5

∣ 2 − 5.5 ∣ =? 3.5 ∣ 9 − 5.5 ∣ =? 3.5

∣ −3.5 ∣ =? 3.5 ∣ 3.5 ∣ =? 3.5

3.5 = 3.5 ✓ 3.5 = 3.5 ✓

34. −5 − (−10) ——

2 =

−5 + 10 —

2 =

5 —

2 = 2.5 −10 + 2.5 = −7.5

The equation is ∣ x + 7.5 ∣ = 2.5.

Check: ∣ x + 7.5 ∣ = 2.5 ∣ x + 7.5 ∣ = 2.5

∣ −10 + 7.5 ∣ =? 2.5 ∣ −5 + 7.5 ∣ =? 2.5

∣ −2.5 ∣ =? 2.5 ∣ 2.5 ∣ =? 2.5

2.5 = 2.5 ✓ 2.5 = 2.5 ✓


Chapter 1

35. ∣ 4n − 15 ∣ = ∣ n ∣ 4n − 15 = n or 4n − 15 = −n

− 4n − 4n − 4n − 4n

−15 = −3n −15 = −5n

−15 — −3 =

−3n —

−3

−15 —

−5 =

−5n —

−5

5 = n 3 = n

Check: ∣ 4n − 15 ∣ = ∣ n ∣ ∣ 4n − 15 ∣ = ∣ n ∣ ∣ 4(5) − 15 ∣ =? ∣ 5 ∣ ∣ 4(3) − 15 ∣ =? ∣ 3 ∣ ∣ 20 − 15 ∣ =? 5 ∣ 12 − 15 ∣ =? 3

∣ 5 ∣ =? 5 ∣ −3 ∣ =? 3

5 = 5 ✓ 3 = 3 ✓

The solutions are n = 3 and n = 5.

36. ∣ 2c + 8 ∣ = ∣ 10c ∣ 2c + 8 = 10c or 2c + 8 = −10c

− 2c − 2c − 2c − 2c

8 = 8c 8 = −12c

8 —

8 = 8c

— 8

8 —

−12 =

−12c —

−12

1 = c − 2 — 3 = c

Check: ∣ 2c + 8 ∣ = ∣ 10c ∣ ∣ 2c + 8 ∣ = ∣ 10c ∣

∣ 2(1) + 8 ∣ =? ∣ 10(1) ∣ ∣ 2 ( − 2 — 3 ) + 8 ∣ =? ∣ 10 ( − 2 —

3 ) ∣

∣ 2 + 8 ∣ =? ∣ 10 ∣ ∣ − 4 —

3 + 8 ∣ =? ∣ −

20 —

3 ∣

∣ 10 ∣ =? 10 ∣ − 4 —

3 +

24 —

3 ∣ =?

20 —

3

10 = 10 ✓ ∣ 20 —

3 ∣ =?

20 —

3

20

— 3 =

20 —

3 ✓

The solutions are c = − 2 — 3 and c = 1.

37. ∣ 2b − 9 ∣ = ∣ b − 6 ∣ 2 b − 9 = b − 6 or 2 b − 9 = −(b − 6)

− b − b 2b − 9 = −b + 6

b − 9 = −6 + b + b + 9 + 9 3b − 9 = 6

b = 3 + 9 + 9

3b = 15

3b

— 3 =

15 —

3

b = 5

Check: ∣ 2b − 9 ∣ = ∣ b − 6 ∣ ∣ 2b − 9 ∣ = ∣ b − 6 ∣

∣ 2(3) − 9 ∣ =? ∣ 3 − 6 ∣ ∣ 2(5) − 9 ∣ =? ∣ 5 − 6 ∣

∣ 6 − 9 ∣ =? ∣ −3 ∣ ∣ 10 − 9 ∣ =? ∣ −1 ∣ ∣ −3 ∣ =? 3 ∣ 1 ∣ =? 1

3 = 3 ✓ 1 = 1 ✓

The solutions are b = 3 and b = 5.

38. ∣ 3k − 2 ∣ = 2 ∣ k + 2 ∣ 3k − 2 = 2(k + 2) or 3k − 2 = 2[−(k + 2)]

3k − 2 = 2(k) + 2(2) 3k − 2 = 2(−k − 2)

3k − 2 = 2k + 4 3k − 2 = 2(−k) − 2(2)

− 2k − 2k 3k − 2 = −2k − 4

k − 2 = 4 + 2k + 2k

+ 2 + 2 5k − 2 = −4

k = 6 + 2 + 2 5k = −2

5k

— 5 =

−2 —

5

k = − 2 — 5

Check: ∣ 3k − 2 ∣ = 2 ∣ k + 2 ∣ ∣ 3k − 2 ∣ = 2 ∣ k + 2 ∣

∣ 3(6) − 2 ∣ =? 2 ∣ 6 + 2 ∣ ∣ 3 ( − 2 —

5 ) − 2 ∣ =? 2 ∣ −

2 —

5 + 2 ∣

∣ 18 − 2 ∣ =? 2 ∣ 8 ∣ ∣ − 6 —

5 − 2 ∣ =? 2 ∣ −

2 —

5 +

10 —

5 ∣

∣ 16 ∣ =? 2(8) ∣ − 6 —

5 −

10 —

5 ∣ =? 2 ∣ 8 —

5 ∣

16 = 16 ✓ ∣ − 16

— 5 ∣ =? 2 ( 8 —

5 )

16

— 5 =

16 —

5 ✓

The solutions are k = − 2 — 5 and k = 6.


Chapter 1

39. 4 ∣ p − 3 ∣ = ∣ 2p + 8 ∣ 4(p − 3) = 2p + 8 or 4(p − 3) = −(2p + 8)

4(p) − 4(3) = 2p + 8 4(p) − 4(3) = −2p − 8

4p − 12 = 2p + 8 4p − 12 = −2p − 8

− 2p − 2p + 2p + 2p

2p − 12 = 8 6p − 12 = −8

+ 12 + 12 + 12 + 12

2p = 20 6p = 4

2p — 2 = 20

— 2

6p —

6 = 4 —

6

p = 10 p = 2 — 3

Check: 4 ∣ p − 3 ∣ = ∣ 2p + 8 ∣ 4 ∣ p − 3 ∣ = ∣ 2p + 8 ∣

4 ∣ 10 − 3 ∣ =? ∣ 2(10) + 8 ∣ 4 ∣ 2 — 3 − 3 ∣ =? ∣ 2 ( 2 —

3 ) + 8 ∣

4 ∣ 7 ∣ =? ∣ 20 + 8 ∣ 4 ∣ 2 — 3 −

9 —

3 ∣ =? ∣ 4 —

3 + 8 ∣

4(7) =?

∣ 28 ∣ 4 ∣ − 7 —

3 ∣ =? ∣ 4 —

3 +

24 —

3 ∣

28 = 28 ✓ 4 ( 7 — 3 ) =? ∣ 28

— 3 ∣

28

— 3 =

28 —

3 ✓

The solutions are p = 2 —

3 and p = 10.

40. 2 ∣ 4w − 1 ∣ = 3 ∣ 4w + 2 ∣ 2(4w − 1) = 3(4w + 2)

2(4w) − 2(1) = 3(4w) + 3(2)

8w − 2 = 12w + 6

− 8w − 8w

−2 = 4w + 6

− 6 − 6 −8 = 4w

−8

— 4 =

4w —

4

−2 = w

or

2(4w − 1) = 3 [ −(4w + 2) ]

2(4w) − 2(1) = 3(−4w − 2)

8w − 2 = 3(−4w) − 3(2)

8w − 2 = −12w − 6

+ 12w + 12w

20w − 2 = −6

+ 2 + 2 20w = −4

20w — 20

= −4

— 20

w = − 1 —

5

Check: 2 ∣ 4w − 1 ∣ = 3 ∣ 4w + 2 ∣

2 ∣ 4(−2) − 1 ∣ =? 3 ∣ 4(−2) + 2 ∣

2 ∣ −8 − 1 ∣ =? 3 ∣ −8 + 2 ∣

2 ∣ −9 ∣ =? 3 ∣ −6 ∣

2(9) =?

3(6)

18 = 18 ✓

and

2 ∣ 4w − 1 ∣ = 3 ∣ 4w + 2 ∣

2 ∣ 4 ( − 1 — 5 ) − 1 ∣ =? 3 ∣ 4 ( −

1 — 5 ) + 2 ∣

2 ∣ − 4 — 5 − 1 ∣ =? 3 ∣ −

4 — 5 + 2 ∣

2 ∣ − 4 — 5 −

5 —

5 ∣ =? 3 ∣ −

4 — 5 +

10 —

5 ∣

2 ∣ − 9 — 5 ∣ =? 3 ∣ 6 —

5 ∣

2 ( 9 — 5 ) =? 3 ( 6 —

5 )

18

— 5 =

18 —

5 ✓

The solutions are w = −2 and w = − 1 — 5 .

41. ∣ 3h + 1 ∣ = 7h

3h + 1 = 7h or 3h + 1 = −7h

− 3h − 3h − 3h − 3h

1 = 4h 1 = −10h

1 — 4 =

4h —

4

1 —

−10 =

−10h —

−10

1 —

4 = h − 1 —

10 = h

Check: ∣ 3h + 1 ∣ = 7h ∣ 3h + 1 ∣ = 7h

∣ 3 ( 1 — 4 ) + 1 ∣ =? 7 ( 1 —

4 ) ∣ 3 ( −

1 —

10 ) + 1 ∣ =? 7 ( −

1 —

10 )

∣ 3 — 4 + 1 ∣ =?

7 —

4 ∣ −

3 —

10 + 1 ∣ =? − 7 —

10

∣ 3 — 4 +

4 —

4 ∣ =?

7 —

4 ∣ −

3 —

10 +

10 —

10 ∣ =? − 7 —

10

∣ 7 — 4 ∣ =?

7 —

4 ∣ 7 —

10 ∣ =? − 7 —

10

7 —

4 =

7 —

4

7 —

10 = − 7 —

10 ✘

The solution is h = 1 — 4 . Reject h = −

1 — 10 because it is

extraneous.


Chapter 1

42. ∣ 6a − 5 ∣ = 4a

6a − 5 = 4a

− 6a − 6a

−5 = −2a

−5

— −2

= −2a

— −2

5 —

2 = a

or 6a − 5 = −4a

− 6a − 6a

−5 = −10a

−5

— −10

= −10a —

−10

1 —

2 = a

Check: ∣ 6a − 5 ∣ = 4a ∣ 6a − 5 ∣ = 4a

∣ 6 ( 5 — 2 ) − 5 ∣ =? 4 ( 5 —

2 ) ∣ 6 ( 1 —

2 ) − 5 ∣ =? 4 ( 1 —

2 )

∣ 15 − 5 ∣ =? 10 ∣ 3 − 5 ∣ =? 2

∣ 10 ∣ =? 10 ∣ −2 ∣ =? 2

10 = 10 ✓ 2 = 2 ✓

The solutions are a = 1 —

2 and a =

5 —

2 .

43. ∣ f − 6 ∣ = ∣ f + 8 ∣ f − 6 = f + 8 or f − 6 = −(f + 8)

− f − f f − 6 = −f − 8

−6 = 8 + f + f The statement

−6 = 8 is false.

So, the original

equation has only

one solution.

2f − 6 = −8

+ 6 + 6

2f = −2

2f

— 2 =

−2 —

2

f = −1

Check: ∣ f − 6 ∣ = ∣ f + 8 ∣ ∣ −1 − 6 ∣ =? ∣ −1 + 8 ∣ ∣ − 7 ∣ =? ∣ 7 ∣ 7 = 7 ✓

The solution is f = −1.

44. ∣ 3x − 4 ∣ = ∣ 3x − 5 ∣ 3x − 4 = 3x − 5 or 3x − 4 = −(3x − 5)

− 3x − 3x 3x − 4 = −3x + 5

−4 = −5 + 3x + 3x

The statement

−4 = −5 is

false. So, the

original equation

has only one

solution.

6x − 4 = 5

+ 4 + 4 6x = 9

6x

— 6 =

9 —

6

x = 3 —

2

Check: ∣ 3x − 4 ∣ = ∣ 3x − 5 ∣

∣ 3 ( 3 — 2 ) − 4 ∣ =? ∣ 3 ( 3 —

2 ) − 5 ∣

∣ 9 — 2 − 4 ∣ =? ∣ 9 —

2 − 5 ∣

∣ 9 — 2 −

8 —

2 ∣ =? ∣ 9 —

2 −

10 —

2 ∣

∣ 1 — 2 ∣ =? ∣ − 1 —

2 ∣

1 —

2 =

1 —

2 ✓


2 .

45. d = ∣ 300 − 48t ∣ 60 = ∣ 300 − 48t ∣ 60 = 300 − 48t or −60 = 300 − 48t

−300 − 300 − 300 − 300

−240 = −48t −360 = −48t

−240

— −48

= −48t

— −48

−360

— −48

= −48t

— −48

5 = t 7.5 = t

Check: 60 = ∣ 300 − 48t ∣ 60 = ∣ 300 − 48t ∣ 60 =

? ∣ 300 − 48(5) ∣ 60 =

? ∣ 300 − 48(7.5) ∣

60 =?

∣ 300 − 240 ∣ 60 =?

∣ 300 − 360 ∣ 60 =

? ∣ 60 ∣ 60 =

? ∣ −60 ∣

60 = 60 ✓ 60 = 60 ✓

The car is 60 feet from you after 5 seconds and again after

7.5 seconds.

46. ∣ 3x + 8 ∣ − 9 = −5

+ 9 + 9 ∣ 3x + 8 ∣ = 4

no; When you isolate the absolute value expression on

one side of the equation, the constant on the other side is

positive. So, the equation has two solutions.


Chapter 1

47. a. ∣ x − 32 ∣ = 5

x − 32 = 5 or x − 32 = −5

+ 32 + 32 + 32 + 32

x = 37 x = 27

Check: ∣ x − 32 ∣ = 5 ∣ x − 32 ∣ = 5

∣ 37 − 32 ∣ =? 5 ∣ 27 − 32 ∣ =? 5

∣ 5 ∣ =? 5 ∣ −5 ∣ =? 5

5 = 5 ✓ 5 = 5 ✓

The solutions are x = 27 and x = 37.

b. no; Because 1 —

3 ≈ 33% and 33% is in the range of 27%

to 37%, it would be accurate to say that about 1 —

3 of the

student body is in favor of year-round school.

48. a. Words: ∣ Weight of a

soccer ball

− Average

weight ∣ = Weight variation

allowed

Variable: Let x be the weight of a soccer ball.

Equation: ∣ x − 430 ∣ = 20

∣ x − 430 ∣ = 20

x − 430 = 20 or x − 430 = −20

+ 430 + 430 + 430 + 430

x = 450 x = 410

The minimum and maximum acceptable soccer ball

weights are 410 grams and 450 grams, respectively.

b. 423 − 16 = 407

Because the soccer ball’s weight of 407 grams is less than

the minimum acceptable weight of 410 grams, the soccer

ball’s weight is not acceptable.

49. Because the value of an expression must be greater than or

equal to 0, the expression ∣ 2x − 1 ∣ cannot equal −9. So, the


50. You have to check for extraneous solutions when solving

absolute value equations with variables on both sides.

5x + 8 = x or 5x + 8 = −x

− 5x − 5x − 5x − 5x

8 = −4x 8 = −6x

8 —

−4 =

−4x —

−4

8 —

−6 =

−6x —

−6

−2 = x − 4 — 3 = x

Check: ∣ 5x + 8 ∣ = x ∣ 5x + 8 ∣ = x

∣ 5(−2) + 8 ∣ =? −2 ∣ 5 ( − 4 — 3 ) + 8 ∣ =? − 4 —

3

∣ −10 + 8 ∣ =? −2 ∣ −20 —

3 +

24 —

3 ∣ =? − 4 —

3

∣ −2 ∣ =? −2 ∣ 4 — 3 ∣ =? − 4 —

3

2 = −2 ✗ 4 —

3 = − 4 —

3 ✗

Because neither 2 = −2 nor 4 —

3 = − 4 — 3 is true, reject both

solutions. So, this equation has no solution.

51. First, isolate the absolute value expression on one side.

∣ x − 2 ∣ + 6 = 0 ∣ x + 3 ∣ − 1 = 0

− 6 − 6 + 1 + 1 ∣ x + 2 ∣ = −6 ∣ x + 3 ∣ = 1

∣ x + 8 ∣ + 2 = 7 ∣ x − 1 ∣ + 4 = 4

− 2 − 2 − 4 − 4

∣ x + 8 ∣ = 5 ∣ x − 1 ∣ = 0

∣ x − 6 ∣ − 5 = −9 ∣ x + 5 ∣ − 8 = −8

+ 5 + 5 + 8 + 8 ∣ x − 6 ∣ = −4 ∣ x + 5 ∣ = 0

If an absolute value expression is equal to a number greater

than 0, then the equation has 2 solutions. If an absolute value

expression is equal to 0, then the equation has 1 solution. If

an absolute value expression is equal to a number less than 0,

then the equation has no solutions.

No solution One solution Two solutions

∣ x − 2 ∣ + 6 = 0 ∣ x − 1 ∣ + 4 = 4 ∣ x + 8 ∣ + 2 = 7

∣ x − 6 ∣ − 5 = −9 ∣ x + 5 ∣ − 8 = −8 ∣ x + 3 ∣ − 1 = 0

52. b is the halfway point and d is the distance from the halfway

point. So, the equation is ∣ x − b ∣ = d.

53. If x2 = a2, then ∣ x ∣ is always equal to ∣ a ∣ . Sample answer: Square roots of the same number have the

same absolute value.

54. If a and b are real numbers, then ∣ a − b ∣ is always equal to ∣ b − a ∣ .

Sample answer: a − b and b − a are opposites, so they have

the same absolute value.

55. For any real number p, the equation ∣ x − 4 ∣ = p will

sometimes have two solutions.

Sample answer: The equation ∣ x − 4 ∣ = p will have two

solutions for all values of p that are greater than 0.

56. For any real number p, the equation ∣ x − p ∣ = 4 will always have two solutions.

Sample answer: Because the absolute value expression is

equal to 4, which is greater than 0, the equation has two

solutions for any real number p.

57. Sample answer: If an absolute value expression is equal to

a value greater than 0, then the expression can equal that

value or its opposite. So, the equation has two solutions.

For example, ∣ x − 7 ∣ = 5 has two solutions, x = 2 and

x = 12. If an absolute value expression is equal to 0, then

the equation has one solution, because 0 does not have an

opposite, and the only number that has an absolute value of

0 is 0. For example, ∣ 2x + 6 ∣ = 0 has one solution, x = −3.

If an absolute value expression is equal to a value less than 0,

the equation has no solution because absolute value always

indicates a number that is not negative. For example,

∣ x + 5 ∣ = −2 has no solution.


Chapter 1

58. Sample answer: Let x be the high and low temperatures (in

degrees Fahrenheit) of a California town today.

72 − 62

— 2 =

10 —

2 = 5 62 + 5 = 67

If the equation that represents the day’s temperature

extremes is ∣ x − 67 ∣ = 5, then the solutions are x = 62 and

x = 72. So, this equation could be used on a day when the

low temperature is 62°F and the high temperature is 72°F.

59. 8 ∣ x + 2 ∣ − 6 = 5 ∣ x + 2 ∣ + 3

Let y = ∣ x + 2 ∣ . Solve 8y − 6 = 5y + 3.

8y − 6 = 5y + 3

− 5y − 5y

3y − 6 = 3

+ 6 + 6

3y = 9

3y — 3 =

9 —

3

y = 3

So, y = 3 = ∣ x + 2 ∣ . 3 = x + 2 or −3 = x + 2

− 2 − 2 − 2 − 2 1 = x −5 = x

Check: 8 ∣ x + 2 ∣ − 6 = 5 ∣ x + 2 ∣ + 3

8 ∣ 1 + 2 ∣ − 6 =?

5 ∣ 1 + 2 ∣ + 3

8 ∣ 3 ∣ − 6 =?

5 ∣ 3 ∣ + 3

8(3) − 6 =?

5(3) + 3

24 − 6 =?

15 + 3

18 = 18 ✓

8 ∣ x + 2 ∣ − 6 = 5 ∣ x + 2 ∣ + 3

8 ∣ −5 + 2 ∣ − 6 =?

5 ∣ −5 + 2 ∣ + 3

8 ∣ −3 ∣ − 6 =?

5 ∣ −3 ∣ + 3

8(3) − 6 =?

5(3) + 3

24 − 6 =?

15 + 3

18 = 18 ✓


60. a. Republican: 42% − 2% = 40% and 42% + 2% = 44%

Green: 2% − 2% = 0% and 2% + 2% = 4%

b. Republican: If ∣ x − 42 ∣ = 2, the solutions are x = 40%

and x = 44%.

Green: If ∣ y − 2 ∣ = 2, the solutions are y = 0% and y = 4%.

c. the Republican party; The maximum percent of the

vote that a candidate from the Republican party can get,

according to the survey, is 44%, and 44% does not fall

in the range of possible percentages for any of the other

parties. So, the candidate must be Republican.

61. When a > 0 and c = d, d − c = 0.

a ∣ x + b ∣ + c = d

− c − c a ∣ x + b ∣ = 0

a ∣ x + b ∣

— a =

0 —

a

∣ x + b ∣ = 0

So, the equation has one solution.

When a < 0 and c > d, 0 > d − c.

a ∣ x + b ∣ + c = d

− c − c a ∣ x + b ∣ = d − c

a ∣ x + b ∣ < 0

Because a < 0, when you divide each side of the

inequality by a, the inequality symbol changes direction.

So, ∣ x + b ∣ > 0, which means that the equation has two

solutions.


62. Addition Property of Equality; If you add 1 to each side of

Equation 1, you get Equation 2.

63. Division Property of Equality; If you divide each side of

Equation 1 by 4, you get Equation 2.

64. A = s2

81 = s2

√—

81 = √—

s2

9 = s

Each side of the square is 9 meters.

65. A = πr2

36π = πr2

36π — π

= πr2

— π

36 = r2

√—

36 = √—

r2 6 = r The radius of the circle is 6 inches.

66. A = 1 —

2 bh

48 = 1 —

2 b(8)

48 = 4b

48

— 4 =

4b —

4

12 = b

The base of the triangle is 12 feet.


Chapter 1

67. P = 2ℓ+ 2w

26 = 2ℓ+ 2(4)

26 = 2ℓ+ 8

− 8 − 8 18 = 2ℓ

18

— 2 =

2ℓ — 2

9 =ℓ The length of the rectangle is 9 centimeters.


1. a. A = bh, where A is the area, b is the length of the base,

and h is the height.

b. A = bh Write the formula.

30 = b(5) Substitute 30 for A and 5 for h.

30

— 5 =

b(5) —


6 = b Simplify.

So, b = 6, and the base of the parallelogram is 6 inches.

c. A = bh Write the formula.

A — h =

bh —

h Divide each side by h.

A — h = b Simplify.

d. Sample answer: In both cases, you use the Division

Property of Equality to undo multiplication. The second

process involved all variables.

2. a. A = 1 —

2 h(b1 + b2)

2 ⋅ A = 2 ⋅ 1 —

2 h(b1 + b2)

2A = h(b1 + b2)

2A —

(b1 + b2) =

h(b1 + b2) — (b1 + b2)

2A —

b1 + b2

= h

h = 2A —

b1 + b2

= 2(63)

— 8 + 10

= 126

— 18

= 7

The height of the trapezoid is 7 centimeters.

b. C = 2πr

C — 2π

= 2πr

— 2π

C — 2π

= r

r = C

— 2π

= 24π — 2π

= 12

The radius of the circle is 12 feet.

c. V = Bh

V — B

= Bh

— B

V — B

= h

h = V

— B

= 75

— 15

= 5

The height of the rectangular prism is 5 yards.

d. V = 1 —

3 Bh

3 ⋅ V = 3 ⋅ 1 —

3 Bh

3V = Bh

3V — B

= Bh

— B

3V

— B

= h

h = 3V

— B

= 3(24π)

— 12π

= 72π — 12π

= 6

The height of the cone is 6 meters.

3. Sample answer: You can solve a given formula for a different

variable to form a new formula that can be used to solve for

the variable. For instance, suppose a rectangle has a length

of 15 inches and an area of 60 square inches. You can solve

the formula for the area of a rectangle for w, and use the new

formula to fi nd the width.

A = ℓw

A

— ℓ = ℓw

— ℓ

A

— ℓ = w

w = A

— ℓ = 60

— 15

= 4

The width of the rectangle is 4 inches.


1. 3y − x = 9

3y − x + x = 9 + x

3y = 9 + x

3y — 3 =

9 + x —

3

y = 3 + 1 —

3 x

The rewritten literal equation is y = 3 + 1 —

3 x.


Chapter 1

2. 2x − 2y = 5

2x − 2x − 2y = 5 − 2x

−2y = 5 − 2x

−2y — −2

= 5 − 2x

— −2

y = − 5 — 2 + x

The rewritten literal equation is y = − 5 —

2 + x.

3. 20 = 8x + 4y

20 − 8x = 8x − 8x + 4y

20 − 8x = 4y

20 − 8x —

4 =

4y —

4

5 − 2x = y

The rewritten literal equation is y = 5 − 2x.

4. y = 5x − 4x

y = x(5 − 4)

y = x(1)

y = x

The rewritten literal equation is x = y.

5. 2x + kx = m

x(2 + k) = m

x(2 + k)

— (2 + k)

= m —

(2 + k)

x = m —

2 + k

The rewritten literal equation is x = m —

2 + k .

6. 3 + 5x − kx = y

3 − 3 + 5x − kx = y − 3

5x − kx = y − 3

x(5 − k) = y − 3

x(5 − k)

— (5 − k)

= y − 3

— (5 − k)

x = y − 3

— 5 − k

The rewritten literal equation is x = y − 3

— 5 − k

.

7. A = 1 —

2 bh

2 ⋅ A = 2 ⋅ 1 —

2 bh

2A = bh

2A — b =

bh —

b

2A — b = h

When you solve for h, you obtain h = 2A

— b .

8. S = πr2 + πrℓ S − πr2 = πr2 − πr2 + πrℓ S − πr2 = πrℓ

S − πr2

— πr

= πrℓ — πr

S − πr2

— πr

= ℓ

When you solve for ℓ, you obtain ℓ = S − πr2

— πr

.

9. F = 9 —

5 C + 32 =

9 —

5 (37) + 32 = 66.6 + 32 = 98.6

Because 98.6°F is less than 100°F, your friend does not have

a fever.

10. I = Prt

I — rt

= Prt

— rt

I — rt

= P

P = I —

rt =

500 —

(0.04)(5) =

500 —

0.2 = 2500

You must deposit $2500.

11. d = rt

d —

r =

rt —

r

d —

r = t

So, d —

60 is how long (in hours) it takes for the truck driver to

deliver the freight, and d —

45 is the driving time (in hours) of

the return trip.

d —

60 +

d —

45 = 7

3 ⋅ d

— 3 ⋅ 60

+ 4 ⋅ d

— 4 ⋅ 45

= 7

3d + 4d —

180 = 7

7d — 180

= 7

180

— 7 ⋅

7d —

180 =

180 —

7 ⋅ 7

d = 180

So, it takes d — 60

= 180

— 60

= 3 hours to deliver the freight, and

the return trip takes d — 45

= 180

— 45

= 4 hours.



1. no; The equation 9r + 16 = π

— 5 is not a literal equation

because it has only one variable.


Chapter 1

2. “Solve 6y = 24 − 3x for y in terms of x” is different because

it asks to solve for y, whereas the other three questions ask to

solve for x.

6y = 24 − 3x 3x + 6y = 24

6y — 6 =

24 − 3x —

6

3x + 6y − 6y = 24 − 6y

y = 4 − 1 —

2 x

3x = 24 − 6y

3x

— 3 =

24 − 6y —

3

x = 8 − 2y


3. y − 3x = 13

y − 3x + 3x = 13 + 3x

y = 13 + 3x

The rewritten literal equation is y = 13 + 3x.

4. 2x + y = 7

2x − 2x + y = 7 − 2x

y = 7 − 2x


5. 2y − 18x = −26

2y − 18x + 18x = −26 + 18x

2y = −26 + 18x

2y — 2 =

−26 + 18x —

2

y = −13 + 9x

The rewritten literal equation is y = −13 + 9x.

6. 20x + 5y = 15

20x − 20x + 5y = 15 − 20x

5y = 15 − 20x

5y — 5 =

15 − 20x —

5

y = 3 − 4x


7. 9x − y = 45

9x − 9x − y = 45 − 9x

−y = 45 − 9x

−y — −1 =

45 − 9x —

−1

y = − 45 + 9x

The rewritten literal equation is y = −45 + 9x.

8. 6x − 3y = −6

6x − 6x − 3y = −6 − 6x

− 3y = −6 − 6x

−3y — −3 =

−6 − 6x —

−3

y = 2 + 2x

The rewritten literal equation is y = 2 + 2x.

9. 4x − 5 = 7 + 4y

4x − 5 − 7 = 7 − 7 + 4y

4x − 12 = 4y

4x − 12

— 4 =

4y —

4

x − 3 = y

The rewritten literal equation is y = x − 3.

10. 16x + 9 = 9y − 2x

16x + 2x + 9 = 9y − 2x + 2x

18x + 9 = 9y

18x + 9

— 9 =

9y —

9

2x + 1 = y

The rewritten literal equation is y = 2x + 1.

11. 2 + 1 —

6 y = 3x + 4

2 − 2 + 1 —

6 y = 3x + 4 − 2

1 —

6 y = 3x + 2

6⋅ 1

— 6 y = 6 ⋅ (3x + 2)

y = 6(3x) + 6(2)

y = 18x + 12

The rewritten literal equation is y = 18x + 12.

12. 11 − 1 —

2 y = 3 + 6x

11 − 11 − 1 —

2 y = 3 − 11 + 6x

− 1 —

2 y = −8 + 6x

−2 ⋅ ( − 1 —

2 y ) = −2 ⋅ (−8 + 6x)

y = −2(−8) − 2(6x)

y = 16 − 12x


13. y = 4x + 8x

y = x (4 + 8)

y = x (12)

y — 12

= x(12)

— 12

y — 12

= x

The rewritten literal equation is x = y —

12 .


Chapter 1

14. m = 10x − x

m = x(10 − 1)

m = x(9)

m — 9 =

9x —

9

m — 9 = x

The rewritten literal equation is x = m

— 9 .

15. a = 2x + 6xz

a = x(2 + 6z)

a —

(2 + 6z) =

x(2 + 6z) —

(2 + 6z)

a —

2 + 6z = x

The rewritten literal equation is x = a —

2 + 6z .

16. y = 3bx − 7x

y = x (3b − 7)

y —

(3b − 7) =

x(3b − 7) —

(3b − 7)

y —

3b − 7 = x

The rewritten literal equation is x = y —

3b − 7 .

17. y = 4x + rx + 6

y − 6 = 4x + rx + 6 − 6

y − 6 = 4x + rx

y − 6 = x(4 + r)

y − 6

— (4 + r)

= x(4 + r)

— (4 + r)

y − 6

— 4 + r

= x

The rewritten literal equation is x = y − 6

— 4 + r

.

18. z = 8 + 6x − px

z − 8 = 8 − 8 + 6x − px

z − 8 = 6x − px

z − 8 = x(6 − p)

z − 8

— (6 − p)

= x(6 − p)

— (6 − p)

z − 8

— 6 −p

= x

The rewritten literal equation is x = z − 8

— 6 − p

.

19. sx + tx = r

x(s + t) = r

x(s + t)

— (s + t)

= r —

(s + t)

x = r —

s + t

The rewritten literal equation is x = r —

s + t.

20. a = bx + cx + d

a − d = bx + cx + d − d

a − d = bx + cx

a − d = x(b + c)

a − d —

(b + c) =

x(b + c) —

(b + c)

a − d — b + c

= x

The rewritten literal equation is x = a − d

— b + c

.

21. 12 − 5x − 4kx = y

12 − 12 − 5x − 4kx = y − 12

−5x − 4kx = y − 12

x(−5 − 4k) = y − 12

x(−5 − 4k) —

(−5 − 4k) =

y − 12 —

(−5 − 4k)

x = y − 12

— −5 − 4k

x = −1(y − 12)

—— −1(−5 − 4k)

x = −y + 12

— 5 + 4k

x = 12 − y

— 5 + 4k

The rewritten literal equation is x = 12 − y

— 5 + 4k

.

22. x − 9 + 2wx = y

x − 9 + 9 + 2wx = y + 9

x + 2wx = y + 9

x(1 + 2w) = y + 9

x(1 + 2w)

— (1 + 2w)

= y + 9

— (1 + 2w)

x = y + 9

— 1 + 2w

The rewritten literal equation is x = y + 9

— 1 + 2w

.

23. a. C = 85x + 60

C − 60 = 85x + 60 − 60

C − 60 = 85x

C − 60

— 85

= 85x

— 85

C − 60

— 85

= x

The rewritten literal equation is x = C − 60

— 85

.

b. x = C − 60

— 85

= 315 − 60

— 85

= 255

— 85

= 3

x = C − 60

— 85

= 485 − 60

— 85

= 425

— 85

= 5

It will cost $315 to take 3 ski trips and $485 to take

5 trips.


Chapter 1

24. a. d = 4n − 2

d + 2 = 4n − 2 + 2

d + 2 = 4n

d + 2

— 4 =

4n —

4

d + 2

— 4 = n

b. n = d + 2

— 4 =

3 + 2 —

4 =

5 —

4 or 1

1 —

4

n = d + 2

— 4 =

6 + 2 —

4 =

8 —

4 = 2

n = d + 2

— 4 =

10 + 2 —

4 =

12 —

4 = 3

Nails with penny sizes 3, 6, and 10 are 1 1 —

4 inches,

2 inches, and 3 inches, respectively.

25. The equation still has an x-term on each side.

12 − 2x = −2(y −x)

12 − 2x = −2(y) − 2(−x)

12 − 2x = −2y + 2x

12 − 2x + 2x = −2y + 2x + 2x

12 = −2y + 4x

12 + 2y = −2y + 2y + 4x

12 + 2y = 4x

12 + 2y —

4 =

4x —

4

3 + 1 —

2 y = x

When you solve the equation for x, you obtain x = 3 + 1 —

2 y.

26. The Distributive Property should not have been used because

only one term has an x.

10 = ax − 3b

10 + 3b = ax − 3b + 3b

10 + 3b = ax

10 + 3b —

a =

ax —

a

10 + 3b —

a = x

When you solve the equation for x, you obtain x = 10 + 3b

— a .

27. P = R − C P−R = R − R − C P−R = −C

P−R — −1

= −C

— −1

−P + R = C

R − P = C

When you solve the formula for C, you obtain C = R − P.

28. S = 2πr2 + 2πrh

S − 2πr2 = 2πr2 − 2πr2 + 2πrh

S − 2πr2 = 2πrh

S − 2πr2

— 2πr

= 2πrh

— 2πr

S − 2πr2

— 2πr

= h

When you solve the formula for h, you obtain h = S − 2πr2

— 2πr

.

29. A = 1 —

2 h (b1 + b2)

2 ⋅ A = 2 ⋅ 1 —

2 h(b1 + b2)

2A = h(b1 + b2)

2A = hb1 + hb2

2A − hb1 = hb1 −hb1 + hb2

2A − hb1 = hb2

2A − hb1 — h =

hb2 — h

2A

— h − b1 = b2

When you solve the formula for b2, you obtain

b2 = 2A

— h − b1.

30. a = v1 − v0 —

t

t ⋅ a = t ⋅ v1 − v0 —

t

at = v1 − v0

at + v0 = v1 − v0 + v0

at + v0 = v1

When you solve the formula for v1, you obtain v1 = at + v0.

31. R = 5 ( C — A

− 0.3 )

R

— 5 = 5 ( C —

A − 0.3 )

5

R

— 5 =

C —

A − 0.3

R

— 5 + 0.3 =

C —

A − 0.3 + 0.3

R

— 5 + 0.3 =

C —

A

A ( R — 5 + 0.3 ) = A ⋅

C —

A

A ( R — 5 + 0.3 ) = C

When you solve the formula for C, you obtain

C = A ( R — 5 + 0.3 ) .


Chapter 1

32. F = G ( m1m2 — d2

)

F — G

=

G ( m1m2 — d2

) —

G

F — G

= m1m2 —

d2

d2 ⋅ F

— G

= d2 ⋅ m1m2 —

d2

d2F — G

= m1m2

d2F — G

⋅ 1 —

m2

= m1m2 ⋅ 1 —

m2

d2F — Gm2

= m1

When you solve the formula for m1, you obtain m1 = d2F

— Gm2

.

33. a. S = L − rL

S − L = L − L − rL

S − L = −rL

S − L — −L

= −rL

— −L

− S — L

+ 1 = r

1 − S — L

= r

When you solve the formula for r, you obtain r = 1 − S — L

.

b. r = 1 − S — L

= 1 − 18 —

30 = 1 − 0.6 = 0.4

The discount rate is 0.4, or 40%.

34. a. d = m

— V

V ⋅ d = V ⋅ m

— V

dV = m

When you solve the formula for m, you obtain m = dV.

b. m = dV = (5.01)(1.2) = 6.012.

The mass of the pyrite sample is 6.012 grams.

35. I = Prt

I —

Pr =

Prt —

Pr

I —

Pr = t

t = I —

Pr =

500 ——

(2000)(0.04) = 6.25 = 6

1 —

4

You must leave the money in the account for 6.25 years, or

6 years and 3 months.

36. d = rt

d —

r =

rt —

r

d —

r = t

So, d —

460 is how long the fi rst fl ight takes, and

d —

500 is how long

the return fl ight takes. Add these expressions and solve for

the one-way distance:

d —

460 +

d —

500 = 4.8

25 ⋅ d —

25 ⋅ 460 +

23 ⋅ d —

23 ⋅ 500 = 4.8

25d + 23d

— 11,500

= 4.8

48d

— 11,500

= 4.8

11,500

— 48

⋅ 48d —

11,500 =

11,500 —

48 ⋅ 4.8

d = 1150

Substitute this distance into each of the original expressions.

So, the fi rst fl ight takes d —

460 =

1150 —

460 = 2.5 hours, and the

return fl ight takes d —

500 =

1150 —

500 = 2.3 hours.

37. a. P = 2x + 2 ( 1 — 2 ⋅ C )

P = 2x + C

P = 2x + 2πr

b. P = 2x + 2πr

P − 2πr = 2x

P − 2πr —

2 = x

c. x = P − 2πr

— 2 =

660 − 2π(50) ——

2 =

660 − 100π — 2 ≈ 173

So, the length of the rectangular portion of the track is

about 173 feet.

38. a. Because d = 55t and d = 20g,

55t = 20g.

b. 55t = 20g

55t — 20

= 20g

— 20

2.75t = g

c. The amount of gasoline used can be found using the

formula from part (b). Either of the original formulas can

be used to fi nd the distance. If you travel for 6 hours, you

will use g = 2.75t = 2.75(6) = 16.5 gallons of gas, and

you will go d = 55t = 55(6) = 330 miles.


Chapter 1

39. a. C = 2πr

C

— 2π

= 2πr

— 2π

C

— 2π

= r

So, the radius of a column is given by the rewritten

formula r = C

— 2π

.

b. r = C

— 2π

= 7 —

2π ≈ 1.1 ft

r = C

— 2π

= 8 —

2π ≈ 1.3 ft

r = C

— 2π

= 9 —

2π ≈ 1.4 ft

So, the radius of a column that has a circumference of

7 feet, 8 feet, or 9 feet is 1.1 feet, 1.3 feet, or 1.4 feet,

respectively.

c. Sample answer: First you can use the formula r = C

— 2π

to

fi nd the radius of the cross section. Then you can use the

value of the radius and the formula A = πr2 to calculate

the area of the cross section.

40. a. The rectangular prism has two square bases with side

length b and four rectangular sides that are b units wide

andℓ units long.

S = 2b2 + 4bℓ b. Sample answer: Choose sideℓbecause this variable is only

in the formula one time. So, it will be easier to isolate.

41. F = 9 —

5 C + 32 =

9 —

5 (20) + 32 = 36 + 32 = 68

no; Because 68°F is less than 70°F, Thermometer A displays

a lesser temperature than Thermometer B.

42. Sample answer: One possible value for h is 8. Then, using

the formula from Exercise 29, the missing base is

b2 = 2A

— b − b1

= 2(40)

— 8 − 8

= 80

— 8 − 8

= 10 − 8

= 2 centimeters.

h = 8 cm

8 cm

2 cm

A = 40 cm2

43.

h b

A = 5 ( 1 — 2 bh )

A = 5bh

— 2

2A = 2 ⋅ 5bh —

2

2A = 5bh

2A

— 5b

= 5bh

— 5b

2A

— 5b

= h

So, the height, h, is given by h = 2A

— 5b

.

44.

h

b

A = 8 ( 1 — 2 bh )

A = 4bh

A

— 4b

= 4bh

— 4b

A

— 4b

= h

So the height, h, is given by h = A

— 4b

.

45. x = a + b + c

— ab

ab ⋅ x = ab ⋅ a + b + c —

ab

abx = a + b + c

abx − a = a − a + b + c

abx − a = b + c

a(bx − 1) = b + c

a(bx − 1)

— (bx − 1)

= b + c

— (bx − 1)

a = b + c

— bx − 1

So, the rewritten literal equation is a = b + c

— bx − 1

.


Chapter 1

46. y = x ( ab —

a − b )

y = abx

— a − b

y ⋅ (a − b) = abx

— a − b

⋅ (a − b)

y(a) − y(b) = abx

ay − by = abx

ay − by + by = abx + by

ay = abx + by

ay − abx = abx − abx + by

ay − abx = by

a(y − bx) = by

a(y − bx)

— (y − bx)

= by —

(y − bx)

a = by —

y − bx

So, the rewritten literal equation is a = by —

y − bx .


47. 15 − 5 + 52 = 15 − 5 + 25

= 10 + 25

= 35

48. 18 ⋅ 2 − 42 ÷ 8 = 18 ⋅ 2 − 16 ÷ 8

= 36 − 16 ÷ 8

= 36 − 2

= 34

49. 33 + 12 ÷ 3 ⋅ 5 = 27 + 12 ÷ 3 ⋅ 5 = 27 + 4 ⋅ 5 = 27 + 20

= 47

50. 25 (5 − 6) + 9 ÷ 3 = 25(−1) + 9 ÷ 3

= 32(−1) + 9 ÷ 3

= −32 + 9 ÷ 3

= −32 + 3

= −29

51. ∣ x − 3 ∣ + 4 = 9

− 4 − 4 ∣ x − 3 ∣ = 5

x − 3 = 5 or x − 3 = −5

+ 3 + 3 + 3 + 3

x = 8 x = −2

−2

4 80−4


52. ∣ 3y − 12 ∣ − 7 = 2

+ 7 +7

∣ 3y − 12 ∣ = 9

3y − 12 = 9 or 3y − 12 = −9

+ 12 + 12 + 12 + 12

3y = 21 3y = 3

3y — 3 = 21

— 3

3y —

3 =

3 —

3

y = 7 y = 1

1 7

0 2 4 6 8

The solutions are y = 1 and y = 7.

53. 2 ∣ 2r + 4 ∣ = −16

2 ∣ 2r + 4 ∣

— 2 =

−16 —

2

∣ 2r + 4 ∣ = −8

Because an absolute value expression must be greater than or

equal to 0, the expression ∣ 2r + 4 ∣ cannot equal −8. So, the


54. −4 ∣ s + 9 ∣ = −24

−4 ∣ s + 9 ∣

— −4

= −24

— −4

∣ s + 9 ∣ = 6

s + 9 = 6 or s + 9 = −6

− 9 − 9 − 9 − 9 s = −3 s = −15

−15

−16 −14 −12 −10 −8 −6 −4 −2 0

−3

The solutions are s = −15 and s = −3.

1.4–1.5 What Did You Learn? (p. 43)

1. Sample answer: The fi rst step is to add 9 to each side in order

to isolate the absolute value expression. After this step, the

absolute value expression is equal to a positive value. So,

you know that the equation has two solutions.

2. Sample answer: The absolute value expressions were the

same on each side. So, use another variable to represent the

absolute value expression and solve the new equation for the

value of this variable. Because the absolute value expression

is equal to this value, write two linear equations, one of

which has the expression equal to the value and one of which

has the expression equal to its opposite. These equations give

two solutions for the original equation.

3. Sample answer: First, partition each shape into congruent

triangles by connecting the center to each of the vertices.

Then multiply the number of triangles by the formula for the

area of a triangle. This gives the area of the original shape.


Chapter 1

Chapter 1 Review (pp. 44–46)

1. z + 3 = −6 Write the equation.


z = −9 Simplify.

Check: z + 3 = −6

−9 + 3 =?

−6

−6 = −6 ✓


2. 2.6 = −0.2t Write the equation.

2.6

— −0.2

= −0.2t

— −0.2

Divide each side by −0.2.

−13 = t Simplify.

Check: 2.6 = −0.2t

2.6 =?

−0.2(−13)

2.6 = 2.6 ✓

3. − n — 5

= −2 Write the equation.

−5 ⋅ ( − n — 5 ) = −5 ⋅ (−2) Multiply each side by −5.

n = 10 Simplify.

Check: − n — 5 = −2

− 10 —

5 =

? −2

−2 = −2 ✓


4. 3y + 11 = −16

− 11 − 11

3y = −27

3y — 3 =

−27 —

3

y = −9

Check: 3y + 11 = −16

3(−9) + 11 =?

−16

−27 + 11 =?

−16

−16 = −16 ✓


5. 6 = 1 − b − 1 − 1 5 = −b

5 —

−1 =

−b —

−1

−5 = b

Check: 6 = 1 −b

6 =?

1 −(−5)

6 =?

1 + 5

6 = 6 ✓

The solution is b = −5.

6. n + 5n + 7 = 43

6n + 7 = 43

− 7 − 7 6n = 36

6n — 6 =

36 —

6

n = 6

Check: n + 5n + 7 = 43

6 + 5(6) + 7 =?

43

6 + 30 + 7 =?

43

36 + 7 =?

43

43 = 43 ✓


7. −4(2z + 6) − 12 = 4

−4(2z) − 4(6) − 12 = 4

−8z − 24 − 12 = 4

−8z − 36 = 4

+ 36 + 36

−8z = 40

−8z — −8 =

40 —

−8

z = −5

Check: −4(2z + 6) − 12 = 4

−4[2(−5) + 6] − 12 =?

4

−4(−10 + 6) − 12 =?

4

−4(−4) − 12 =?

4

16 − 12 =?

4

4 = 4 ✓



Chapter 1

8. 3 —

2 (x − 2) − 5 = 19

3 —

2 (x) −

3 —

2 (2) − 5 = 19

3 —

2 x − 3 − 5 = 19

3 —

2 x − 8 = 19

+ 8 + 8

3 —

2 x = 27

2 —

3 ⋅

3 —

2 x =

2 —

3 ⋅ 27

x = 18

Check: 3 — 2 (x − 2) − 5 = 19

3 —

2 (18 − 2) − 5 =

? 19

3 —

2 (16) − 5 =

? 19

24 − 5 =?

19

19 = 19 ✓


9. 6 = 1 —

5 w +

7 —

5 w − 4

6 = 8 —

5 w − 4

+ 4 + 4

10 = 8 —

5 w

5 —

8 ⋅ 10 =

5 —

8 ⋅

8 —

5 w

25

— 4 = w

Check: 6 = 1 —

5 w +

7 —

5 w − 4

6 =?

1 —

5 ( 25

— 4 ) +

7 —

5 ( 25

— 4 ) − 4

6 =?

5 —

4 +

35 —

4 − 4

6 =?

40

— 4 − 4

6 =?

10 − 4

6 = 6 ✓

The solution is w = 25

— 4 .

10. 5x + 2x + 110 = 180

7x + 110 = 180

− 110 − 110

7x = 70

7x — 7 =

70 —

7

x = 10

So, x = 10, and the measures of the angles of the triangle are

5x = 5(10) = 50°, 2x = 2(10) = 20°, and 110°.

11. 2(x) + 3(x − 30) = 540

2x + 3(x) − 3(30) = 540

2x + 3x − 90 = 540

5x − 90 = 540

+ 90 + 90

5x = 630

5x — 5 =

630 —

5

x = 126

So, x = 126, and the pentagon has two angles whose

measures are each x = 126° and three angles whose

measures are each x − 30 = 126 − 30 = 96°.

12. 3n − 3 = 4n + 1

− 3n − 3n

−3 = n + 1

− 1 − 1 −4 = n


13. 5(1 + x) = 5x + 5

5(1) + 5(x) = 5x + 5

5 + 5x = 5x + 5

− 5x − 5x

5 = 5

Because the statement 5 = 5 is always true, the equation is


14. 3(n + 4) = 1 —

2 (6n + 4)

3(n) + 3(4) = 1 —

2 (6n) +

1 —

2 (4)

3n + 12 = 3n + 2

− 3n − 3n

12 = 2

Because the statement 12 = 2 is never true, the equation has

no solution.


Chapter 1

15. ∣ y + 3 ∣ = 17

y + 3 = 17 or y + 3 = −17

− 3 − 3 − 3 − 3 y = 14 y = −20

Check: ∣ y + 3 ∣ = 17 ∣ y + 3 ∣ = 17

∣ 14 + 3 ∣ =? 17 ∣ −20 + 3 ∣ =? 17

∣ 17 ∣ =? 17 ∣ −17 ∣ =? 17

17 = 17 ✓ 17 = 17 ✓

The solutions are y = −20 and y = 14.

16. −2 ∣ 5w − 7 ∣ + 9 = −7

− 9 − 9 −2 ∣ 5w − 7 ∣ = −16

−2 ∣ 5w − 7 ∣

—— −2

= −16

— −2

∣ 5w − 7 ∣ = 8

5w − 7 = 8 or 5w − 7 = −8

+ 7 + 7 + 7 + 7 5w = 15 5w = −1

5w

— 5 =

15 —

5

5w —

5 =

−1 —

5

w = 3 w = − 1 —

5

Check: −2 ∣ 5w − 7 ∣ + 9 = −7 −2 ∣ 5w − 7 ∣ + 9 = −7

−2 ∣ 5(3) − 7 ∣ + 9 =?

−7 −2 ∣ 5 ( − 1 —

5 ) − 7 ∣ + 9 =

? −7

−2 ∣ 15 − 7 ∣ + 9 =?

−7 −2 ∣ −1 − 7 ∣ + 9 =?

−7

−2 ∣ 8 ∣ + 9 =?

−7 −2 ∣ −8 ∣ + 9 =?

−7

−2(8) + 9 =?

−7 −2(8) + 9 =?

−7

−16 + 9 =?

−7 −16 + 9 =?

−7

−7 = −7 ✓ −7 = −7 ✓

The solutions are w = − 1 —

5 and w = 3.

17. ∣ x − 2 ∣ = ∣ 4 + x ∣ x − 2 = 4 + x or x − 2 = −(4 + x)

− x − x x − 2 = −4 − x −2 = 4 + x + x 2x − 2 = −4

+ 2 + 2 2x = −2

2x

— 2 =

−2 —

2

x = −1

Check: ∣ x − 2 ∣ = ∣ 4 + x ∣ ∣ −1 − 2 ∣ =? ∣ 4 + (−1) ∣ ∣ −3 ∣ =? ∣ 3 ∣ 3 = 3 ✓


18. 95 − 74 —

2 = 10.5 74 + 10.5 = 84.5

The equation that represents the minimum and maximum

wind speeds is ∣ v − 84.5 ∣ = 10.5.

Check: ∣ v − 84.5 ∣ = 10.5 ∣ v − 84.5 ∣ = 10.5

∣ 74 − 84.5 ∣ =? 10.5 ∣ 95 − 84.5 ∣ = 10.5

∣ −10.5 ∣ =? 10.5 ∣ 10.5 ∣ = 10.5

10.5 = 10.5 ✓ 10.5 = 10.5 ✓

19. 2x − 4y = 20

2x − 2x − 4y = 20 − 2x

−4y = 20 − 2x

−4y — −4 =

20 − 2x —

−4

y = −5 + 1 —

2 x

The rewritten literal equation is y = −5 + 1

— 2 x.

20. 8x − 3 = 5 + 4y

8x − 3 − 5 = 5 − 5 + 4y

8x − 8 = 4y

8x − 8

— 4 =

4y —

4

2x − 2 = y

The rewritten literal equation is y = 2x − 2.

Because the

statement −2 = 4

is never true, the

equation only has

one solution.


Chapter 1

21. a = 9y + 3yx

a = y(9 + 3x)

a —

(9 + 3x) =

y(9 + 3x) —

(9 + 3x)

a —

9 + 3x = y

The rewritten literal equation is y = a —

9 + 3x .

22. a. V = 1 —

3 Bh

3 ⋅ V = 3 ⋅ 1 —

3 Bh

3V = Bh

3V — B

= Bh

— B

3V — B

= h

When you solve the formula for h, you obtain h = 3V

— B

.

b. h = 3V

— B

= 3(216)

— 36

= 648

— 36

= 18

So, the height of the pyramid is 18 centimeters.

23. a. F = 9 —

5 (K − 273.15) + 32

F − 32 = 9 —

5 (K − 273.15)

5 —

9 (F − 32) = K − 273.15

5 — 9 (F − 32) + 273.15 = K

When you solve the formula for K, you obtain

K = 5 —

9 (F − 32) + 273.15.

b. K = 5 —

9 (F − 32) + 273.15 =

5 —

9 (180 − 32) + 273.15

= 5 —

9 (148) + 273.15 ≈ 355.37

So, 180°F is about 355.37 kelvin.

Chapter 1 Test (p. 47)

1. x − 7 = 15 Write the equation.


x = 22 Simplify.

Check: x − 7 = 15

22 − 7 =?

15

15 = 15 ✓


2. 2 —

3 x + 5 = 3 Write the equation.


2 —

3 x = −2 Simplify.

3 — 2 ∙ 2 —

3 x = 3 —

2 ⋅ (−2) Multiply each side by 3 —

2 .

x = −3 Simplify.

Check: 2 —

3 x + 5 = 3

2 — 3 (−3) + 5 =

? 3

−2 + 5 =?

3

3 = 3 ✓


3. 11x + 1 = −1 + x Write the equation.

− x − x Subtract x from each side.

10x + 1 = −1 Simplify.


10x = −2 Simplify.

10x

— 10

= −2

— 10

Divide each side by 10.

x = − 1 — 5

Check: 11x + 1 = −1 + x

11 ( − 1 — 5 ) + 1 =

? −1 + ( − 1 —

5 )

− 11 —

5 + 1 =

? −

5 —

5 −

1 —

5

− 11 —

5 +

5 —

5 =

? −

6 —

5

− 6 — 5 = − 6 —

5 ✓

The solution is x = − 1 —

5 .

4. 2 ∣ x − 3 ∣ − 5 = 7

+ 5 + 5 2 ∣ x − 3 ∣ = 12

2 ∣ x − 3 ∣

— 2 =

12 —

2

∣ x − 3 ∣ = 6

x − 3 = 6 or x − 3 = −6

+ 3 + 3 + 3 + 3 x = 9 x = −3



Chapter 1

5. ∣ 2x − 19 ∣ = 4x + 1

2x − 19 = 4x + 1 or 2x − 19 = −(4x + 1)

− 2x − 2x 2x − 19 = −4x − 1

−19 = 2x + 1 + 4x + 4x

− 1 − 1 6x − 19 = −1

−20 = 2x + 19 + 19

− 20 —

2 = 2x

— 2 6x = 18

−10 = x 6x

— 6 =

18 —

6

x = 3

Check: ∣ 2x − 19 ∣ = 4x + 1 ∣ 2x − 19 ∣ = 4x + 1

∣ 2(−10) − 19 ∣ =? 4(−10) + 1 ∣ 2(3) − 19 ∣ =? 4(3) + 1

∣ −20 − 19 ∣ =? −40 + 1 ∣ 6 − 19 ∣ =? 12 + 1

∣ −39 ∣ =? −39 ∣ −13 ∣ =? 13

39 = −39 ✗ 13 = 13 ✓

The solution is x = 3. Reject x = −10 because it is extraneous.

6. −2 + 5x − 7 = 3x − 9 + 2x

5x − 9 = 5x − 9

− 5x − 5x

−9 = −9

The statement −9 = −9 is always true. So, the equation is


7. 3(x + 4) − 1 = −7

3(x) + 3(4) − 1 = −7

3x + 12 − 1 = −7

3x + 11 = −7

− 11 − 11

3x = −18

3x

— 3 =

−18 —

3

x = −6


8. ∣ 20 + 2x ∣ = ∣ 4x + 4 ∣ 20 + 2x = 4x + 4 or 20 + 2x = −(4x + 4)

− 2x − 2x 20 + 2x = −4x − 4

20 = 2x + 4 + 4x + 4x

− 4 − 4 20 + 6x = −4

16 = 2x − 20 − 20

16

— 2 =

2x —

2 6x = −24

8 = x 6x

— 6 =

−24 —

6

x = −4

Check: ∣ 20 + 2x ∣ = ∣ 4x + 4 ∣ ∣ 20 + 2x ∣ = 4x + 4

∣ 20 + 2(8) ∣ =? 4(8) + 4 ∣ 20 + 2(−4) ∣ =? ∣ 4(−4) + 4 ∣ ∣ 20 + 16 ∣ =? ∣ 32 + 4 ∣ ∣ 20 − 8 ∣ =? ∣ −16 + 4 ∣ ∣ 36 ∣ =? ∣ 36 ∣ ∣ 12 ∣ =? ∣ −12 ∣ 36 = 36 ✓ 12 = 12 ✓


9. 1 —

3 (6x + 12) − 2(x − 7) = 19

1 —

3 (6x) +

1 —

3 (12) − 2(x) − 2(−7) = 19

2x + 4 − 2x + 14 = 19

18 = 19


solution.

10. 3x − 5 = 3x − c

− 3x − 3x

−5 = −c

− 5 = − c −1 = −1

5 = c

If c = 5, the equation is an identity. So, for c ≠ 5, the


11. ∣ x − 7 ∣ = c

Because absolute value expressions must be equal to a value

greater than or equal to 0, the equation will have no solution

for c < 0.

12. 38 − 30 —

2 =

8 —

2 = 4 30 + 4 = 34

The equation that represents the minimum and maximum

hand rail heights is ∣ x − 34 ∣ = 4.

Check: ∣ x − 34 ∣ = 4 ∣ x − 34 ∣ = 4

∣ 30 − 34 ∣ =? 4 ∣ 38 − 34 ∣ =? 4

∣ −4 ∣ =? 4 ∣ 4 ∣ =? 4

4 = 4 ✓ 4 = 4 ✓


Chapter 1

13. a. P = 2ℓ+ 2w

P − 2ℓ = 2ℓ − 2ℓ + 2w

P − 2ℓ = 2w

P − 2ℓ —

2 =

2w —

2

P − 2ℓ —

2 = w

When you solve the formula for w, you obtain

w = P − 2ℓ —

2 .

b. w = P − 2ℓ —

2

= 330 − 2(100)

—— 2

= 330 − 200

— 2

= 130

— 2

= 65 yd

The fi eld is 65 yards wide.

c. A =ℓw A = πr2 314.16

— 6500

≈ 0.048, or 4.8%

A = (100)(65) A = π(10)2

A = 6500 A ≈ 314.16

About 4.8% of the fi eld is inside the circle.

14. a. Words: Dealership

total cost =Local mechanic

total cost

Variable: Let h be the time (in hours) it takes to do the

repairs.

Equation: 24 + 99 ⋅ h = 45 + 89 ⋅ h

24 + 99h = 45 + 89h

− 89h − 89h

24 + 10h = 45

− 24 − 24

10h = 21

10h

— 10

= 21

— 10

h = 2.1, or 2 hours and 0.1(60) = 6 minutes

After 2.1 hours, or 2 hours and 6 minutes, of work the

total costs are the same at both places.

b. For less than 2.1 hours of work, the repairs cost less at the

dealership because the parts cost less there, and for more

than 2.1 hours of work, the repairs cost less at the local

mechanic.

15. Sample answer: If x = −2, then the right side of the

equation, 6x, is negative, and the absolute value expression

cannot equal a negative value. So, x = −2 cannot make the

equation true and must be an extraneous solution.

16. Sample answer: The variables cancelled out of the equation,

and the resulting statement, −8 = −8, is always true. So, the

equation is an identity and has infi nitely many solutions.

Chapter 1 Standards Assessment (pp. 48–49)

1. B; 37.5% of 48 = 0.375 ⋅ 48 = 18 beginner trails

48 − 18

— 2 =

30 —

2 = 15 each of intermediate and expert trails

2. cx − a + b = 2b, x = a + b

— c , b + a = cx;

cx − a + b = 2b

− b − bcx − a = b ✓

0 = cx − a + b

− b − b−b = cx − a ✗

2cx − 2a = b —

2

2(cx − a) = b —

2

1 —

2 ⋅ 2(cx − a) =

1 —

2 ⋅

b —

2

cx − a = b —

4 ✗

x − a = b —

c x = a + b —

c b + a = cx

c ⋅ (x − a) = c⋅ b — c c⋅ x = c⋅

a + b —

c cx = b + a

c(x) − c(a) = b cx = a + b cx − a = b + a − a

cx − ca = b ✗ cx − a = a − a + b cx − a = b ✓

cx − a = b ✓

3. 3(x − a) = 3x − 6

3(x) − 3(a) = 3x − 6

3x − 3a = 3x − 6

− 3x − 3x

−3a = −6

a. −3a = −6 b. −3a = −6 c. −3a = −6

−3(3) =?

−6 −3(−3) =?

−6 −3(2) =?

−6

−9 = −6 ✗ 9 = −6 ✗ −6 = −6 ✓

When a = 3, N < 1. When a = −3, N < 1. When a = 2, N > 1.

d. −3a = −6 e. −3a = −6 f. −3a = −6

−3(−2) =?

−6 −3(x) = −6 −3(−x) = −6

6 = −6 ✗ −3x = −6 3x = −6

When a = −2, N < 1. −3x —

−3 =

−6 —

−3

3x — 3 =

−6 —

3

x = 2 x = −2

When a = x, N = 1. When a = −x, N = 1.


Chapter 1

4. a.

Words:Total

cost =

Cost per

can of

white

paint

⋅Number

of cans

of white

paint

+

Cost

per can

of blue

paint

⋅Number

of cans

of blue

paint

Variable: Let x be the number of cans of white paint. Then

(5 − x) is the number of cans of blue paint.

Equation: 132 = 24 ⋅ x + 28 ⋅ (5 − x)

132 = 24x + 28(5 − x)

b. 132 = 24x + 28(5 − x)

132 = 24x + 28(5) − 28(x)

132 = 24x + 140 − 28x

132 = −4x + 140

− 140 − 140

− 8 = −4x

−8

— −4

= −4x

— −4

2 = x and 5 − x = 5 − 2 = 3

So, you bought 2 cans of white paint and 3 cans of blue

paint. However, if you switched the colors and bought 3 cans

of white paint and 2 cans of blue paint, you would have spent

24(3) + 28(2) = $128, which is $132 − $128 = $4 less.

5. The equations 8x + 6 = −2x − 14 and 5x + 3 = −7 are

equivalent because they have the same solution, x = −2.

6x + 6 = −14

− 6 − 66x = −20

6x

— 6 =

−20 —

6

x = − 10 —

3

8x + 6 = −2x − 14

+ 2x + 2x

10x + 6 = −14

−6 −6

10x = −20

10x

— 10

= −20

— 10

x = −2

7x + 3 = 2x − 13 5x + 3 = −7

− 2x − 2x − 3 − 3 5x + 3 = 13 5x = −10

− 3 − 3 5x

— 5 =

−10 —

5

5x = 10 x = −2

5x — 5 =

10 —

5

x = 2

6. B; (x − 5) + x —

2 + 6 = 13

x + 1 —

2 x − 5 + 6 = 13

3 —

2 x + 1 = 13

− 1 − 1

3 —

2 x = 12

2 —

3 ⋅

3 —

2 x =

2 —

3 ⋅ 12

x = 8

So, the sides are 6 inches, x − 5 = 8 − 5 = 3 inches,

and x —

2 =

8 —

2 = 4 inches, and the shortest side is 3 inches.

7. a.

Words:

Cable

TV

monthly

cost

⋅Number

of

months=

Cost of

satellite

TV

receiver

+

Satelite

TV

monthly

cost

⋅Number

of

months

Variable: Let m be the number of months.

Equation: 45 ⋅ m = 99 + 36 ⋅ m

45m = 99 + 36m

− 36m − 36m

9m = 99

9m — 9 =

99 —

9

m = 11

After 11 months, you and your friend will have paid the

same amount for TV services.

b. yes; After 11 months, you will have paid the same

amount, and because your friend’s monthly cost is less, he

will pay less in the 12th month and therefore less overall

for the year.


Chapter 1

8. ∣ 8x + 3 ∣ = 0

8x + 3 = 0

− 3 − 38x = −3

8x

— 8 =

−3 —

8

x = − 3 — 8

−6 = 5x − 9

+ 9 + 93 = 5x

3 —

5 =

5x —

5

3 —

5 = x

3x − 12 = 3(x − 4) + 1

3x − 12 = 3(x) − 3(4) + 1

3x − 12 = 3x − 12 + 1

3x − 12 = 3x − 11

− 3x − 3x

−12 = −11 ✗

−2x + 4 = 2x + 4

+ 2x + 2x

4 = 4x + 4

− 4 − 40 = 4x

0 —

4 =

4x —

4

0 = x

0 = ∣ x + 13∣ + 2− 2 − 2

−2 = ∣ x + 13 ∣ ✗

−4(x + 4) = −4x − 16

−4(x) − 4(4) = −4x − 16

−4x − 16 = −4x − 16

+ 4x + 4x

−16 = −16

12x − 2x = 10x − 8

10x = 10x − 8

− 10x − 10x

0 = 8 ✗

9 = 3 ∣ 2x − 11 ∣

9 —

3 =

3 ∣ 2x − 11 ∣ —

3

3 = ∣ 2x − 11 ∣ 3 = 2x − 11 or −3 = 2x − 11

+ 11 + 11 + 11 + 11

14 = 2x 8 = 2x

14

— 2 =

2x —

2

8 —

2 =

2x —

2

7 = x 4 = x

7 − 2x = 3 − 2(x − 2)

7 − 2x = 3 − 2(x) − 2(−2)

7 − 2x = 3 − 2x + 4

7 − 2x = −2x + 7

+ 2x + 2x

7 = 7

No solution One solutionTwo

solutionsInfi nitely

many solutions

3x − 12 =

3(x − 4) + 1

∣ 8x + 3 ∣ = 0 9 = 3

∣ 2x − 11 ∣ −4(x + 4) =

−4x − 16

0 = ∣ x + 13 ∣ + 2 −6 = 5x − 9 7 − 2x =

3 − 2(x − 2)

12x − 2x = 10x − 8 −2x + 4 =

2x + 4

9. 1000 ft —

12.5 sec = 80

feet —

second =

80 feet —

second

So, the expressions that do not represent the average speed of

the car are 80 second

— feet

and second

— 80 feet

.

Chapter 1 · Chapter 1 8. Words: Your recorded balance − Forgotten check = Statement balance...

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Transcript of Chapter 1 · Chapter 1 8. Words: Your recorded balance − Forgotten check = Statement balance...