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

Chapter 1 Maintaining Mathematical Pro! 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 −

Forgottencheck =

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 ! nal score + 12 =

Friend’s ! nal score

Variable: Let x be your ! nal score.

Equation: x + 12 = 195

x + 12 = 195

−12 −12

x = 183

Your ! 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 ✓

The solution is y = −10.

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 $ 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 ! 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 Pro! 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 = 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


−100 −100 Subtract 100 from each side.

5x = 260 Simplify.

5x —

5 =

260 —

5 Divide each

side by 5.

x = 52 Simplify.

So, x = 52 and the measures of the angles of the 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 theequation.

x − 17 + x + 35 + x + 42 + x = 360 Associative Property of Addition

x + x + x + x − 17 + 35 + 42 = 360 Commutative Property of Addition



4x = 300 Simplify.

4x —

4 = 300

— 4 Divide each

side by 4.

x = 75 Simplify.

So, x = 75 and the measures of the angles of the 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



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



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 ! 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 con! 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 ! 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 ✓

The solution is x = −6.

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 ! re hose needs 48 pounds per square inch of water

pressure to reach a ! 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.


Vocabulary and Core Concept Check 1. To solve the equation 2x + 3x = 20, ! 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 ! 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 ! 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.

Monitoring Progress and Modeling with Mathematics 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 $ 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 Participation

92% 0.20 92% × 0.20 = 18.4%

Homework 95% 0.20 95% × 0.20 = 19%

Midterm Exam

88% 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 ! 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 ! 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

Maintaining Mathematical Pro! ciency

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


1.3 Explorations (p. 19) 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 ! 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 ! gure is 6 + 4 + 6 + 4 + 1 + 1 = 22 feet.

The area of the ! 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 ! nd the perimeter. Add the area of the semicircle to the area of the rectangle to ! nd the total area. Then set the perimeter equal to the area and solve for x.

The perimeter of the ! gure is

1 — 2 ⋅ 2 ⋅ π ⋅ 2 + 2 + 4 + 2 = 2π + 8 feet.

The area of the ! 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 in! nitely many solutions.

1.3 Monitoring Progress (pp. 20–22) 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 in! 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 in! 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.


Vocabulary and Core Concept Check 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 ! 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.

Monitoring Progress and Modeling with Mathematics 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 ✓


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


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 in! 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 in! 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 coef! 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 coef! 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 con! 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 in! nitely many solutions.

40. Sample answer: The perimeter of the given triangle is P = (x + 3) + (2x + 1) + 3x = 6x + 4.

Another ! gure with the same perimeter is shown.

22

3x

3x

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

Maintaining Mathematical Pro! ciency 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 de! nition of an identity, which has in! 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


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 speci! cally, Studio B will always charge more.

1.4 Explorations (p. 27) 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 ! 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, ! 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, ! 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.

1.4 Monitoring Progress (pp. 28–31) 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 ✓



Vocabulary and Core Concept Check 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.

Monitoring Progress and Modeling with Mathematics 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 − 3m = 4 m = −10

2 40−2−4−6−8−10

The solutions are m = −10 and m = 4.

16. ∣ q − 8 ∣ = 14q − 8 = 14 or q − 8 = −14

+ 8 + 8 + 8 + 8q = 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 ∣ = 194b − 5 = 19 or 4b − 5 = −19

+ 5 + 5 + 5 + 54b = 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

The absolute value of an expression must be greater than or equal to 0. The expression ∣ x − 1 ∣ cannot equal −3. So, the equation has no solution.

21. −4 ∣ 8 − 5n ∣ = 13

−4 ∣ 8 − 5n ∣ — −4

= 13 — −4

∣ 8 − 5n ∣ = − 13 —

4

The absolute value of an expression must be greater than or 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 point to the minimum and maximum respectively is 2.

29. C; The halfway point is 2, and the distance from the halfway point to the minimum and maximum respectively is 4.

30. A; The halfway point is −4, and the distance from the halfway point to the minimum and maximum respectively is 2.

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 ✓


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 equation has no solution.

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.

Maintaining Mathematical Pro! ciency 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.

1.5 Explorations (p. 35) 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 ! 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.

1.5 Monitoring Progress (pp. 36–39) 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.


Vocabulary and Core Concept Check

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

Monitoring Progress and Modeling with Mathematics 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. 6 − 3y = −6

6 − 6 − 3y = −6 − 6

− 3y = −12

−3y — −3 = −12 —

−3

y = 4

The rewritten literal equation is y = 4.

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 ! rst $ ight takes, and d — 500

is how long

the return $ 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 ! rst $ ight takes d — 460

= 1150 — 460

= 2.5 hours, and the

return $ 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 ! 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

! 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

.

Maintaining Mathematical Pro! ciency 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 equation has no solution.

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 ! 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 an identity and has in! nitely many solutions.

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 ✓


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

an identity and has in! nitely many solutions.

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 ✓ The solutions are x = −4 and x = 8.

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


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 equation has no solution.

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 ! 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 ! 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 in! 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 ⋅ m45m = 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

solutionsIn! 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

.

hscc alg1 wsk 01 - schoolwires...2 2 = 2 the solution is m = −7. 7. y − 4 = 3 write the...

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