chapter 1 · chapter 1 8. words: your recorded balance − forgotten check = statement balance...
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Copyright © Big Ideas Learning, LLC Algebra 1 1All rights reserved. Worked-Out Solutions
Chapter 1 Maintaining Mathematical Profi ciency (p. 1)
1. −5 + (−2) = −7 2. 0 + (−13) = −13
3. −6 + 14 = 8
4. 19 − (−13) = 19 + 13 = 32
5. −1 − 6 = −1 + (−6) = −7
6. −5 − (−7) = −5 + 7 = 2
7. 17 + 5 = 22 8. 8 + (−3) = 5
9. 11 − 15 = 11 + (−15) = −4
10. −3(8) = −24 11. −7 ⋅ (−9) = 63
12. 4 ⋅ (−7) = −28 13. −24 ÷ (−6) = 4
14. −16 ÷ 2 = −8 15. 12 ÷ (−3) = −4
16. 6 ⋅ 8 = 48 17. 36 ÷ 6 = 6
18. −3(−4) = 12
19. a. Sample answer: To add integers with the same sign, add
their absolute values, and the sum has the same sign
as both addends. To add integers with different signs,
subtract their absolute values, and the difference has the
same sign as the addend with the greatest absolute value;
−6 + 2 = −4
b. Sample answer: To subtract integers, change the
subtraction sign to an addition sign, and change the
integer following the sign to its opposite. Then, follow the
rules for adding integers; 5 − (−3) = 5 + 3 = 8
c. Sample answer: To multiply integers, multiply their
absolute values. If the integers have the same sign, then
the product is positive. If the integers have different signs,
then the product is negative; (−6)(−4) = 24
d. Sample answer: To divide integers, divide their absolute
values. If the integers have the same sign, then the
quotient is positive. If the integers have different signs,
then the quotient is negative; −15 ÷ 3 = −5
Chapter 1 Mathematical Practices (p. 2)
1. Population change = 310 million − 280 million
= 30 million people
Time change = 2010 − 2000
= 10 years
Rate of change = 30 million people
—— 10 years
= 3 million people per year
2. Gas mileage = 240 mi
— 8 gal
= 30 mi/gal
3. 18 in. = 1.5 ft
Volume = ℓwh
= (5 ft) × (3 ft) × (1.5 ft)
= 22.5 ft3
Amount of water = 3 —
4 (22.5 ft3)
= 16.875 ft3
Drain time = 16.875 ft3
— 1 ft3/min
= 16.875 min
It takes about 17 minutes for the water to drain.
1.1 Explorations (p. 3)
1.
Quadri-lateral
m∠ A (deg-rees)
m∠ B(deg-rees)
m∠ C(deg-rees)
m∠ D(deg-rees)
m∠ A + m∠ B+ m∠ C + m∠ D
a. 110 90 92 68 360°
b. 65 147 58 90 360°
c. 91 79 75 115 360°
Sample answer: Because a protractor is used, the measurements
are precise.
2. Conjecture: The sum of the angle measures of a quadrilateral
is 360°.
Sample answer:
A B
D C
100° + 40° + 140° + 80° = 360°
A
DC
B
90° + 60° + 115° + 95° = 360°
AB
D
C
150° + 30° + 75° + 105° = 360°
Divide the quadrilateral into two triangles.
The sum of the angle measures of a triangle is 180°,
so the sum of the angle measures of a quadrilateral is
2(180°) = 360°.
Chapter 1
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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.
Copyright © Big Ideas Learning, LLC Algebra 1 3All rights reserved. Worked-Out Solutions
Chapter 1
8. Words: Your recorded
balance −Forgotten
check =Statement
balance
Variable: Let c be the amount of the forgotten check.
Equation: 68 − c = 26
68 − c = 26
− 68 − 68
−c = − 42
c = 42
The check you forgot to record was for $42.
1.1 Exercises (pp. 8–10)
Vocabulary and Core Concept Check
1. Addition, +, and subtraction, −, are inverses of each other.
Multiplication, ×, and division, ÷, are inverses of each other.
2. −2x = 10 −5x = 25
−2x — −2
= 10
— −2
−5x
— −5
= 25
— −5
x = −5 x = −5
yes; The equations are equivalent because they have the same
solution, x = −5.
3. Division Property of Equality; In order to write an equivalent
equation that has x by itself on one side, you must undo
multiplying by 14. So, you would divide each side by 14.
4. The equation x − 6 = 5 does not belong. Sample answer: You would use the Addition Property of Equality to solve
it, whereas you would use the Multiplication Property of
Equality to solve the other three equations.
Monitoring Progress and Modeling with Mathematics
5. x + 5 = 8 Write the equation.
− 5 − 5 Subtract 5 from each side.
x = 3 Simplify.
Check: x + 5 = 8
3 + 5 =?
8
8 = 8 ✓
The solution is x = 3.
6. m + 9 = 2 Write the equation.
− 9 − 9 Subtract 9 from each side.
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.
+ 2 + 2 Add 2 to each side.
s = 3 Simplify.
Check: s − 2 = 1
3 − 2 =?
1
1 = 1 ✓
The solution is s = 3.
9. w + 3 = −4 Write the equation.
− 3 − 3 Subtract 3 from each side.
w = −7 Simplify.
Check: w + 3 = −4
−7 + 3 =?
−4
−4 = −4 ✓
The solution is w = −7.
10. n − 6 = −7 Write the equation.
+ 6 + 6 Add 6 to each side.
n = −1 Simplify.
Check: n − 6 = −7
−1 − 6 =?
−7
−7 = −7 ✓
The solution is n = −1.
11. −14 = p − 11 Write the equation.
+ 11 + 11 Add 11 to each side.
−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 − 4 Subtract 4 from each side.
−4 = q Simplify.
Check: 0 = 4 + q 0 =
? 4 + (−4)
0 = 0 ✓
The solution is q = −4.
4 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.
− 5 − 5 Subtract 5 from each side.
t = 4 Simplify.
Check: t − (−5) = 9 4 − (−5) =
? 9
4 + 5 =?
9
9 = 9 ✓
The solution is t = 4.
15. Words: Discounted
ticket price =Original
price − 12.95
Variable: Let p be the original price.
Equation: 44 = p − 12.95
44 = p − 12.95
+ 12.95 + 12.95
56.95 = p
The original price of an amusement park ticket is $56.95.
16. Words: Your fi nal
score + 12 =Friend’s
fi nal score
Variable: Let x be your fi nal score.
Equation: x + 12 = 195
x + 12 = 195
−12 −12
x = 183
Your fi nal score is 183 points.
17. x + 100 + 120 + 100 = 360
x + 320 = 360
− 320 − 320
x = 40
So, x = 40.
18. x + 48 + 77 + 150 = 360
x + 275 = 360
− 275 − 275
x = 85
So, x = 85.
19. x + 122 + 92 + 76 = 360
x + 290 = 360
− 290 − 290
x = 70
So, x = 70.
20. x + 60 + 115 + 85 = 360
x + 260 = 360
− 260 − 260
x = 100
So, x = 100.
21. 5g = 20 Write the equation.
5g — 5 = 20
— 5 Divide each side by 5.
g = 4 Simplify.
Check: 5g = 20
5(4) =?
20
20 = 20 ✓
The solution is g = 4.
22. 4q = 52 Write the equation.
4q — 4 = 52
— 4 Divide each side by 4.
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 ✓
The solution is y = 7.
Copyright © Big Ideas Learning, LLC Algebra 1 5All rights reserved. Worked-Out Solutions
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 ✓
The solution is x = 48.
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 ✓
The solution is w = −18.
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 ✓
The solution is t = −1.
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.
6 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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
The solution is m = 12.
41. C; Because each carton contains 18 eggs, the total number
of eggs is equal to the product of 18 and the number of
cartons, x.
18x = 162
18x
— 18
= 162 —
18
x = 9
The baker orders 9 cartons of eggs.
42. Words: Temperature
at 5 p.m. −Change in
temperature =Temperature
at 10 p.m.
Variable: Let T be the change in temperature.
Equation: 20 − T = −5
20 − T = −5
− 20 − 20
−T = −25
T = 25
The temperature fell 25°F from 5 p.m. to 10 p.m.
43. Words: Length = 1.9 ⋅ Width
Variable: Let w be the width.
Equation: 9.5 = 1.9 ⋅ w
9.5 = 1.9w
9.5 =
1.9w 1.9 1.9
5 = w
The American fl ag is 5 feet wide.
Copyright © Big Ideas Learning, LLC Algebra 1 7All rights reserved. Worked-Out Solutions
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
8 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 1
49. a. 5x = 10 − 5
5x = 5
5x
— 5 =
5 —
5
x = 1
When a = 5 and b = 10, x is a positive integer.
b. −2x = 9 − 5
−2x = 4
−2x —
−2 =
4 —
−2
x = − 2
When a = −2 and b = 9, x is a negative integer.
50. a. The entire circle represents 100%.
b. The percent of the partitions of a circle graph should sum
to 100 percent.
You can solve the equation to fi nd x, the percent of cats.
7 + 9 + 5 + 48 + x = 100
69 + x = 100
− 69 − 69
x = 31
At a local pet store, 31% of the animals sold are cats.
51. Let g be the number of girls and b be the number of boys in
the marching band.
1 —
6 g = 6
2 —
7 b = 10
6 ⋅ 1 —
6 g = 6 ⋅ 6
7 —
2 ⋅
2 —
7 b =
7 —
2 ⋅ 10
g = 36 b = 35
36 girls + 35 boys = 71 students
The marching band has 71 students.
52. Sample answer: A game has 30 pieces, each player should
get the same number of pieces, and all 30 pieces should be
used. If you and 4 of your friends are playing, how many
pieces should each player get? Let x be how many pieces
each player should get.
5x = 30
5x
— 5 =
30 —
5
x = 6
You and your 4 friends should each get 6 pieces.
53. V = Bh
84π = B(7)
84π —
7 =
7B —
7
12π = B
The area of the base of the cylinder is 12π square inches.
54. V = Bh
1323 = 147h
1323
— 147
= 147h
— 147
9 = h
The height of the rectangular prism is 9 centimeters.
55. V = 1 —
3 Bh
15π = 1 —
3 B(5)
3 —
5 (15π) =
3 —
5 ( 5 —
3 B )
9π = B
The area of the base of the cone is 9π square meters.
56. V = 1 — 3 Bh
35 = 1 — 3 ⋅ 30 ⋅ h
35 = 10h
35
— 10
= 10h —
10
3.5 = h The height of the square pyramid is 3.5 feet.
5 7. a. Batting
average = Number of hits
—— Number of at-bats
.296 = h —
446
446 ⋅ (.296) = 446 ⋅ h —
446
132 ≈ h Player A had 132 hits in the 2011 regular season.
b. no; If player B had fewer hits, then he must have had
fewer at-bats in order to have a greater batting average.
Maintaining Mathematical Profi ciency
58. 8(y + 3) = 8 ⋅ y + 8 ⋅ 3
= 8y + 24
59. 5 — 6 ( x + 1 —
2 + 4 ) = 5 —
6 ⋅ x +
5 —
6 ⋅
1 —
2 + 5 —
6 ⋅ 4
= 5 — 6 x +
5 —
12 +
20 —
6
= 5 — 6 x +
5 —
12 + 40
— 12
= 5 — 6 x +
45 —
12
= 5 — 6 x + 15
— 4
Copyright © Big Ideas Learning, LLC Algebra 1 9All rights reserved. Worked-Out Solutions
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.
− 60 − 60 Subtract 60 from each side.
10x = 120 Simplify.
10x
— 10
= 120 —
10 Divide each side by 10.
x = 12 Simplify.
So, x = 12 and the measures of the angles of the triangle
are 30°, 9x° = (9 ⋅ 12)° = 108°, and (30 + x)° = (30 + 12)° = 42°.
b. (x + 10) + (x + 20) + 50 = 180 Write the equation.
x + 10 + x + 20 + 50 = 180 Associative Property
of Addition
x + x + 10 + 20 + 50 = 180 Commutative Property
of Addition
2x + 80 = 180 Combine like terms.
− 80 − 80 Subtract 80 from each
side.
2x = 100 Simplify.
2x —
2 = 100
— 2 Divide each side by 2.
x = 50 Simplify,
So, x = 50 and the measures of the angles of the
triangle are 50°, (x + 20)° = (50 + 20)° = 70°, and
(x + 10)° = (50 + 10)° = 60°.
c.
50 + (2x + 30) + (2x + 20) + x = 360 Write the
equation.
50 + 2x + 30 + 2x + 20 + x = 360 Associative
Property of
Addition
2x + 2x + x + 50 + 30 + 20 = 360 Commutative
Property of
Addition
5x + 100 = 360 Combine like
terms.
−100 −100 Subtract 100
from each side.
5x = 260 Simplify.
5x —
5 =
260 —
5 Divide each
side by 5.
x = 52 Simplify.
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 the
equation.
x − 17 + x + 35 + x + 42 + x = 360 Associative
Property of
Addition
x + x + x + x − 17 + 35 + 42 = 360 Commutative
Property of
Addition
4x + 60 = 360 Combine like
terms.
− 60 − 60 Subtract 60
from each side.
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°.
10 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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 + 40 = 540 Combine like
terms.
− 40 − 40 Subtract 40
from each side.
25x = 500 Simplify.
25x
— 25
= 500 —
25 Divide each side
by 25.
x = 20 Simplify.
So, x = 20 and the measures of the angles of the pentagon
are (5x + 2)° = (5 ⋅ 20 + 2)° = 102°,
(3x + 5)° = (3 ⋅ 20 + 5)° = 65°,
(8x + 8)° = (8 ⋅ 20 + 8)° = 168°,
(5x + 10)° = (5 ⋅ 20 + 10)° = 110°, and
(4x + 15)° = (4 ⋅ 20 + 15)° = 95°.
f.
2(3x + 16) + (2x + 8) + (4x − 18) Write the
+ (3x − 7) + (2x + 25) = 720 equation.
6x + 32 + (2x + 8) + (4x − 18) Distributive
+ (3x − 7) + (2x + 25) = 720 Property
6x + 32 + 2x + 8 + 4x − 18 Associative
+ 3x − 7 + 2x + 25 = 720 Property of
Addition
6x + 2x + 4x + 3x + 2x + 32 Commutative
+ 8 − 18 − 7 + 25 = 720 Property of
Addition
17x + 40 = 720 Combine like
terms.
− 40 − 40 Subtract 40
from each side.
17x = 680 Simplify.
17x
— 17
= 680
— 17
Divide each side
by 17.
x = 40 Simplify.
So, x = 40 and the measures of the angles of the hexagon
are (3x + 16)° = (3 ⋅ 40 + 16)° = 136°, 136°,
(2x + 8)° = (2 ⋅ 40 + 8)° = 88°,
(4x − 18)° = (4 ⋅ 40 − 18)° = 142°,
(3x − 7)° = (3 ⋅ 40 − 7)° = 113°, and
(2x + 25)° = (2 ⋅ 40 + 25)° = 105°.
Sample answer: You can check the measures of the angles
using a protractor. You can also check to make sure the
angle measures add up to the sum given by the formula.
2. a. Answer should include, but is not limited to: Check that
polygons have 3 or more straight sides of varying lengths
and that all are closed fi gures. Suggest that students draw
large polygons, because it will be easier to measure the
angles.
b. Answer should include, but is not limited to: The sum of
the angle measures of each polygon should satisfy the
formula S = 180(n − 2). Some might be a little off due
to rounding. Have students round to the nearest whole
number of degrees.
c. Answer should include, but is not limited to: The same
value for x should satisfy each expression for the angle
measures of the polygon.
d. and e. Answer should include, but is not limited to: Partners should confi rm that the calculated measures
of the angles are the same as (or at least close to) the
measures obtained with a protractor. In addition, the sum
of the calculated measures of the polygon should satisfy
the formula S = 180(n − 2).
3. Sample answer: If you notice a pattern, you can use inductive
reasoning to write a rule. Then you can test your rule using
several examples. You can use the rule to write an equation
that can be used to solve a problem.
4. Connecting a vertex with each of the other vertices in a
polygon creates n − 2 triangles, each of which has a total
angle measure of 180°.
5. S = 180(n − 2)
1080 = 180(n − 2)
1080
— 180
= 180(n − 2) —
180
6 = n − 2 + 2 + 2 8 = n 8 sides; Use the formula S = 180(n − 2) and inverse
operations to work backward and fi nd that a polygon, whose
angle measures sum to 1080°, is an octagon.
1.2 Monitoring Progress (pp. 12–15)
1. −2n + 3 = 9 −3 −3
− 2n = 6
−2n — −2 = 6
— −2
n = −3
Check: −2n + 3 = 9 −2(−3) + 3 =
? 9
6 + 3 =?
9 9 = 9 ✓
The solution is n = −3.
Copyright © Big Ideas Learning, LLC Algebra 1 11All rights reserved. Worked-Out Solutions
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 ✓
The solution is y = 5.
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 ✓
The solution is x = 5.
12 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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 ✓
The solution is m = −5.
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 ✓
The solution is x = 1.
10. d = 1 — 2 n + 26
50 = 1 — 2 n + 26
− 26 − 26
24 = 1 — 2 n
2 ⋅ 24 = 2 ⋅ 1 —
2 n
48 = n A fi re hose needs 48 pounds per square inch of water
pressure to reach a fi re 50 feet away.
11. Words: Perimeter = 2 ⋅ Width + 2 ⋅ Three times
the width
Variable: Let w be the width of the rectangular pen.
Equation: 96 = 2 ⋅ w + 2 ⋅ (3w)
96 = 2w + 2(3w)
96 = 2w + 6w
96 = 8w
96
— 8 = 8w
— 8
12 = w ℓ = 3w = 3(12) = 36
The pen should be 36 feet by 12 feet.
1.2 Exercises (pp. 16–18)
Vocabulary and Core Concept Check
1. To solve the equation 2x + 3x = 20, fi rst combine 2x and 3x
because they are like terms.
2. Sample answer: One way to solve the equation
2(4x − 11) = 10 is to fi rst use the Distributive Property to
eliminate the parentheses. Then undo subtraction to isolate
the x-term. Finally, undo the multiplication to solve for
x. Another method is to fi rst undo multiplication and use
the Division Property of Equality to divide each side by 2.
Then undo subtraction to isolate the x-term. Finally, undo
multiplication again to solve for x.
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.
Copyright © Big Ideas Learning, LLC Algebra 1 13All rights reserved. Worked-Out Solutions
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 ✓
The solution is g = 8.
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 ✓
The solution is m = −3.
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.
14 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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 ✓
The solution is y = 4.
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.
Copyright © Big Ideas Learning, LLC Algebra 1 15All rights reserved. Worked-Out Solutions
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 ✓
The solution is m = 3.
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.
16 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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 ✓
The solution is y = −19.
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 ✓
The solution is x = 29.
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 ✓
The solution is r = −6.
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°.
Copyright © Big Ideas Learning, LLC Algebra 1 17All rights reserved. Worked-Out Solutions
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.
18 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.
Copyright © Big Ideas Learning, LLC Algebra 1 19All rights reserved. Worked-Out Solutions
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
The solution is x = 34.
42. P = 2ℓ+ 2w
228 = 2(2w + 6) + 2w
228 = 2(2w) + 2(6) + 2w
228 = 4w + 12 + 2w
228 = 6w + 12
− 12 − 12
216 = 6w
216
— 6 =
6w —
6
36 = w
2w + 6 = 2 ⋅ 36 + 6 = 78
The court is 78 feet by 36 feet.
43. P = 2ℓ+ 2w
190 = 2 ( 11 —
8 y ) + 2(y)
190 = 11
— 4 y + 2y
190 = ( 11 —
4 +
8 —
4 ) y
190 = 19
— 4 y
4 —
19 ⋅ 190 =
4 —
19 ⋅ 19
— 4 y
40 = y
11
— 8 y =
11 —
8 ⋅ 40 = 55
The Norwegian fl ag is 55 inches by 40 inches.
44. P = s + (s + 6) + (s + 6) + s + 2s
102 = 6s + 12
− 12 − 12
90 = 6s
90
— 6 =
6s —
6
15 = s
s + 6 = 15 + 6 = 21
2s = 2 ⋅ 15 = 30
The school crossing sign has two sides that are each
15 inches, two sides that are each 21 inches, and one side
that is 30 inches.
45. a. 2(4 − 8x) + 6 = −1
2(4) − 2(8x) + 6 = −1
8 − 16x + 6 = −1
14 − 16x = −1
− 14 − 14
−16x = −15
−16x
— −16
= −15
— −16
x = 15
— 16
b. 2(4 − 8x) + 6 = −1
− 6 − 6 2(4 − 8x) = −7
2(4 − 8x)
— 2 =
−7 —
2
4 − 8x = − 7 — 2
− 4 − 4
−8x = − 15 —
2
− 1 — 8 ⋅ −8x = − 1 —
8 ⋅ ( − 15
— 2 )
x = 15
— 16
The solution is x = 15
— 16
.
Sample answer: Method 1 is preferred because it requires
fewer operations with fractions and was therefore less
complicated, making mistakes less likely.
20 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 1
46. Words:
Total
cost = ( Ticket
price +
Convenience
charge ) ⋅ Number
of tickets + Processing
charge
Variable: Let t be the number of tickets purchased.
Equation: 220.70 = (32.50 + 3.30) ⋅ t + 5.90
220.7 = (32.5 + 3.3)t + 5.9
220.7 = 35.8t + 5.9
− 5.9 − 5.9
214.8 = 35.8t
214.8
— 35.8
= 35.8t
— 35.8
6 = t
For an order that costs $220.70, 6 tickets are purchased.
47. Words
Total
value
=
Value
per
dime⋅
Number
of
dimes+
Value
per
quarter⋅
Number
of
quarters
Variable: Let d be the number of dimes.
Equation: 2.80 = 0.10 ⋅ d + 0.25 ⋅ (d + 8)
2.8 = 0.1d + 0.25(d + 8)
2.8 = 0.1d + 0.25d + 2
2.8 = 0.35d + 2
− 2 − 2
0.8 = 0.35d
0.8
— 0.35
= 0.35d
— 0.35
2.29 ≈ d
no; Sample answer: Because it is not possible to have a
decimal number of dimes, it is not possible for the number
of quarters to be 8 more than the number of dimes when the
total of the dimes and quarters is $2.80.
48. Sample answer:
Component
Student’s score Weight Score × weight
Class
Participation92% 0.20 92% × 0.20 = 18.4%
Homework 95% 0.20 95% × 0.20 = 19%
Midterm
Exam88% 0.25 88% × 0.25 = 22%
Final Exam f 0.35 f × 0.35 = 0.35f
Total 1 18.4 + 19 + 22 + 0.35f
18.4 + 19 + 22 + 0.35f = 90
59.4 + 0.35f = 90
− 59.4 − 59.4
0.35f = 30.6
0.35f — 0.35
= 30.6
— 0.35
f ≈ 87.4
The student must earn at least an 88% on the fi nal exam in
order to earn an A in the class.
49. Let n be an integer. Then 2n is an even integer. The next even
integer is 2 more than 2n: 2n + 2.
The third even integer is 2 more than 2n + 2: 2n + 2 + 2 = 2n + 4.
So, the total is 2n + (2n + 2) + (2n + 4).
2n + (2n + 2) + (2n + 4) = 54
2n + 2n + 2 + 2n + 4 = 54
6n + 6 = 54
− 6 − 6 6n = 48
6n — 6 =
48 —
6
n = 8
2n = 2 ⋅ 8 = 16
2n + 2 = 2 ⋅ 8 + 2 = 18
2n + 4 = 2 ⋅ 8 + 4 = 20
So, the consecutive even integers are 16, 18, and 20.
Copyright © Big Ideas Learning, LLC Algebra 1 21All rights reserved. Worked-Out Solutions
Chapter 1
50. a. greater than; Only one of the numbers is greater than 20,
and two are less than 20. In order for the average to be 20,
the fourth meeting must have had more than 20 students
in attendance.
b. Sample answer: about 25 students
c. Sample answer: You can write and solve an equation
to fi nd how many students were in attendance at the
fourth meeting. If you let x be the number of students in
attendance at the fourth meeting, the equation is
18 + 21 + 17 + x
—— 4 = 20.
18 + 21 + 17 + x ——
4 = 20
4 ⋅ 18 + 21 + 17 + x ——
4 = 4 ⋅ 20
18 + 21 + 17 + x = 80
56 + x = 80
− 56 − 56
x = 24
There were 24 students at the fourth meeting. This is close
to the estimate in part (b).
51. bx = −7
bx
— b =
−7 —
b
x = − 7 — b
52. x + a = 3
— 4
− a − a
x = 3 —
4 − a
53. ax − b = 12.5
+ b + b ax = 12.5 + b
ax
— a =
12.5 + b —
a
x = 12.5 + b
— a
54. ax + b = c
− b − b ax = c − b
ax
— a =
c − b —
a
x = c − b
— a
55. 2bx − bx = −8
x(2b − b) = −8
x(b) = −8
x(b)
— b =
−8 —
b
x = − 8 — b
56. cx − 4b = 5b
+ 4b + 4b
cx = 9b
cx
— c =
9b —
c
x = 9b
— c
Maintaining Mathematical Profi 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 ✓
x = −1 is not a solution. x = 2 is a solution.
22 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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
x = −1 is a solution. x = 2 is not a solution.
64. a. x + 4 = 3x b. x + 4 = 3x
−1 + 4 =?
3(−1) 2 + 4 =?
3(2)
3 ≠ −3 6 = 6 ✓
x = −1 is not a solution. x = 2 is a solution.
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
x = −1 is a solution. x = 2 is not a solution.
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 fi rst polygon is
4 + 5 + 2 + 4 + 2 + 5 = 22.
The perimeter of the second polygon is also
3 + 3 — 2 (4) + 5 + 4 + 2 + ( 3 —
2 ⋅ 4 − 4 )
= 3 + 6 + 5 + 4 + 2 + 2 = 22.
2. a. 5 + 5 + 4 + x + 4 = 1 — 2 ⋅ x ⋅ 3 + 4x
x + 18 = 3 — 2 x + 4x
x + 18 = 11 —
2 x
− x − x
18 = 9 — 2 x
2 —
9 ⋅ 18 = 2 —
9 ⋅
9 —
2 x
4 = x The solution is x = 4.
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
4 = x The solution is x = 4.
Sample answer: Add the side lengths to get the perimeter.
Subtract the area of the small rectangle from the area
of the large rectangle to get the total area. Then set the
perimeter equal to the area and solve for x.
The perimeter of the fi gure is
6 + 4 + 6 + 4 + 1 + 1 = 22 feet.
The area of the fi gure is
6(4) − 1(2) = 24 − 2 = 22 square feet.
c. 1 — 2 ⋅ 2 ⋅ π ⋅ 2 + x + 4 + x = 1 —
2 ⋅ π ⋅ 2
2 + x ⋅ 4
2x + 2π + 4 = 2π + 4x
− 2x − 2x
2π + 4 = 2π + 2x
− 2π − 2π
4 = 2x
4 —
2 = 2x
— 2
2 = x The solution is x = 2.
Sample answer: Add the circumference of the semicircle
to the remaining three side lengths to fi nd the perimeter.
Add the area of the semicircle to the area of the rectangle
to fi nd the total area. Then set the perimeter equal to the
area and solve for x.
The perimeter of the fi gure is
1 —
2 ⋅ 2 ⋅ π ⋅ 2 + 2 + 4 + 2 = 2π + 8 feet.
The area of the fi gure is
1 —
2 ⋅ π ⋅ 22 + 2 ⋅ 4 = 2π + 8 square feet.
Copyright © Big Ideas Learning, LLC Algebra 1 23All rights reserved. Worked-Out Solutions
Chapter 1
3. To solve an equation that has variables on both sides, collect
the variable terms on one side of the equation and the
constant terms on the other side of the equation, then solve.
4. Sample answer: Some sample equations are 3x = 4x + 4
{x = −4}, 5x − 7 = 7x − 1{x = −3}, and 2(3x − 4) + 3 = 4x + 5{x = 5}. Note that students may write equations with
no solution or infi nitely many solutions.
1.3 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 ✓
The solution is x = −2.
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 ✓
The solution is n = 0.
4. 4(1 − p) = −4p + 4
4(1) − 4(p) = −4p + 4
4 − 4p = −4p + 4
+ 4p + 4p
4 = 4
The statement 4 = 4 is always true. So, the equation is an
identity and has infi nitely many solutions.
5. 6m − m = 5 —
6 (6m − 10)
5m = 5 —
6 (6m) −
5 —
6 (10)
5m = 5m − 25
— 3
− 5m − 5m
0 = − 25 —
3
The statement 0 = − 25 —
3 is never true. So, the equation has
no solution.
6. 10k + 7 = −3 − 10k
+ 10k + 10k
20k + 7 = −3
− 7 − 7 20k = −10
20k — 20
= −10
— 20
k = − 1 — 2
The solution is k = − 1 — 2 .
24 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 1
7. 3(2a − 2) = 2(3a − 3)
3(2a) − 3(2) = 2(3a) − 2(3)
6a − 6 = 6a − 6
− 6a − 6a
−6 = −6
The statement −6 = −6 is always true. So, the equation is
an identity and has infi nitely many solutions.
8. Words: Distance upstream = Distance downstream
Variable: Let x be the speed (in miles per hour) of the boat
upstream.
Equation: x mi
— 1 h
⋅ 3.5 h = (x + 2) mi
— 1 h
⋅ 2.5 h
3.5x = 2.5(x + 2)
3.5x = 2.5x + 5
− 2.5x − 2.5x
1x = 5
x = 5
3.5x = 3.5(5) = 17.5
The boat travels 17.5 miles upstream.
1.3 Exercises (pp. 23–24)
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 fi rst step
is to use the Distributive Property on the left side of the
equal sign to remove the parentheses and then simplify so
that the equation becomes 9x − 24 = 4x + 6. In order to
eliminate the x-term from the right side, subtract 4x from
each side. Then, in order to isolate the remaining x-term,
undo subtraction by adding 24 to each side. Finally, you can
solve for x by undoing multiplication, so divide each side
by 5. The solution is x = 6.
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 ✓
The solution is x = 3.
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 ✓
The solution is s = 2.
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 ✓
The solution is p = 7.
Copyright © Big Ideas Learning, LLC Algebra 1 25All rights reserved. Worked-Out Solutions
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 ✓
The solution is g = 5.
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 ✓
The solution is t = −1.
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 ✓
The solution is w = −4.
26 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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 ✓
The solution is t = 6.
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 ✓
The solution is x = −3.
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 ✓
The solution is t = 1.
Copyright © Big Ideas Learning, LLC Algebra 1 27All rights reserved. Worked-Out Solutions
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 ✓
The solution is y = −12.
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 ✓
The solution is x = 1 —
2 .
17. 50h = 190 − 45h
+ 45h + 45h
95h = 190
95h — 95
= 190
— 95
h = 2 You and your friend will meet after you have been driving
toward each other for 2 hours.
18. 1.5r +15 = 2.25r
− 1.5r − 1.5r
15 = 0.75r
15
— 0.75
= 0.75r
— 0.75
20 = r
You must rent 20 movies to spend the same amount at each
store.
19. 3t + 4 = 12 + 3t
− 3t − 3t
4 = 12
The statement 4 = 12 is never true. So, the equation has no
solution.
20. 6d + 8 = 14 + 3d
− 3d − 3d
3d + 8 = 14
− 8 − 8 3d = 6
3d — 3 =
6 —
3
d = 2
The equation has one solution: d = 2.
21. 2(h + 1) = 5h − 7
2(h) + 2(1) = 5h − 7
2h + 2 = 5h − 7
− 2h − 2h
2 = 3h − 7
+ 7 + 7 9 = 3h
9 —
3 =
3h —
3
3 = h
The equation has one solution: h = 3.
28 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 1
22. 12y + 6 = 6(2y + 1)
12y + 6 = 6(2y) + 6(1)
12y + 6 = 12y + 6
− 12y − 12y
6 = 6
The statement 6 = 6 is always true. So, the equation is an
identity and has infi nitely many solutions.
23. 3(4g + 6) = 2(6g + 9)
3(4g) + 3(6) = 2(6g) + 2(9)
12g + 18 = 12g + 18
− 12g − 12g
18 = 18
The statement 18 = 18 is always true. So, the equation is an
identity and has infi nitely many solutions.
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
The statement 5 = 4 is never true. So, the equation has no
solution.
25. In order to undo subtraction, 3c should have been added to
each side.
5c − 6 = 4 − 3c
+ 3c + 3c
8c − 6 = 4
+ 6 + 6 8c = 10
8c — 8 =
10 —
8
c = 5 —
4
The solution is c = 5 —
4 .
26. Because the statement 0 = 0 is always true, the equation has
infi nitely many solutions. It is better to subtract a variable
term from each side before subtracting a constant from
each side.
6(2y + 6) = 4(9 + 3y)
6(2y) + 6(6) = 4(9) + 4(3y)
12y + 36 = 36 + 12y
− 12y − 12y
36 = 36
Because the statement 36 = 36 is always true, the equation
has infi nitely many solutions.
27. Words: Total cost of
Company A’s
Internet service
=
Total cost of
Company B’s
Internet service
Variable: Let m be the number of months you have Internet
service.
Equation: 60.00 + 42.95m = 25.00 + 49.95m
60 + 42.95m = 25 + 49.95m
−42.95m −42.95m
60 = 25 + 7m
− 25 − 25
35 = 7m
35
— 7 = 7m
— 7
5 = m
After 5 months, you would pay the same total amount for
each Internet service.
28. Words: 4% of total
protein needed
daily
+ 48 = Total protein
needed daily
Variable: Let p be the total amount (in grams) of protein you
need daily.
Equation: 0.04 ⋅ p + 48 = p
0.04p + 48 = p
− 0.04p − 0.04p
48 = 0.96p
48
— 0.96
= 0.96p —
0.96
50 = p
You need 50 grams of protein daily.
29. 8(x + 6) − 10 + r = 3(x + 12) + 5x
8(x) + 8(6) − 10 + r = 3(x) + 3(12) + 5x
8x + 48 − 10 + r = 3x + 36 + 5x
8x + 38 + r = 8x + 36
− 8x − 8x
38 + r = 36
− 38 − 38
r = −2
So, r = −2.
Copyright © Big Ideas Learning, LLC Algebra 1 29All rights reserved. Worked-Out Solutions
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.
30 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 1
34. 90 ft —
1 sec ⋅ t sec = 650 ft +
60 ft —
1 sec ⋅ t sec
90t = 650 + 60t
− 60t − 60t
30t = 650
30t — 30
= 650
— 30
t = 21 2 —
3 sec
no; In order to catch the antelope, the cheetah would have to
be running at top speed for over 20 seconds, so the antelope
is probably safe.
35. a(2x+3) = 9x + 15 + x
a(2x) + a(3) = 10x + 15
2ax + 3a = 10x + 15
Set the coeffi cients of x equal to each other and set the
constant terms equal to each other.
2a = 10 3a = 15
2a — 2 =
10 —
2
3a —
3 =
15 —
3
a = 5 a = 5
If a = 5, then 2ax + 3a = 2(5) x + 3(5) = 10x + 15.
Because the other side of the equal sign is also 10x + 15, the
equation is an identity when a = 5.
36. 8x − 8 + 3ax = 5ax − 2a
8x + 3ax − 8 = 5ax − 2a
(8 + 3a)x − 8 = 5ax − 2a
Set the coeffi cients of x equal to each other and set the
constant terms equal to each other.
8 + 3a = 5a −8 = −2a
− 3a − 3a −8
— −2
= −2a
— −2
8 = 2a 4 = a
8 —
2 =
2a —
2
4 = a If a = 4, then 8x − 8 + 3ax = 8x − 8 + 3(4)x = 8x − 8 +
12x = 20x − 8, and 5ax − 2a = 5(4)x − 2(4) = 20x − 8.
Because the expressions on each side of the equation are the
same, the equation is an identity when a = 4.
37. Words: 2 ⋅ Greater
consecutive
integer
= 3 ⋅ Lesser
consecutive
integer
− 9
Variable: Let n be the lesser consecutive integer. Then n + 1
is the greater consecutive integer.
Equation: 2 ⋅ (n + 1) = 3 ⋅ n − 9
2(n + 1) = 3n − 9
2(n) + 2(1) = 3n − 9
2n + 2 = 3n − 9
− 2n − 2n
2 = n − 9 + 9 + 9 11 = n
n + 1 = 11 + 1 = 12
The integers are 11 and 12.
38. a. After 6 years, there will be equal enrollment in Spanish
and French classes because the graphs meet at this point.
b. The left side, 355 − 9x, represents the predicted Spanish
class enrollment, and the right side, 229 + 12x, represents
the predicted French class enrollment. So, the equation
represents when there will be equal enrollment in Spanish
and French classes. The solution should give the same
result as the graph.
355 − 9x = 229 + 12x
+ 9x + 9x
355 = 229 + 21x
− 229 − 229
126 = 21x
126
— 21
= 21x
— 21
6 = x
The equation confi rms that there will be equal enrollment
in Spanish and French classes after 6 years if the trend
continues as predicted.
Copyright © Big Ideas Learning, LLC Algebra 1 31All rights reserved. Worked-Out Solutions
Chapter 1
39. a. Sample answer: 2(x + 5) = 2x − 7
2(x) + 2(5) = 2x − 7
2x + 10 = 2x − 7 − 2x − 2x
10 = −7
The statement 10 = −7 is never true. So, the equation
2(x + 5) = 2x − 7 has no solution.
b. Sample answer: 2(3x + 6) = 3(2x + 4)
2(3x)+2(6) = 3(2x) + 3(4)
6x + 12 = 6x + 12
− 6x − 6x
12 = 12
The statement 12=12 is always true. So, the equation
2(3x + 6) = 3(2x + 4) is an identity and has infi nitely
many solutions.
40. Sample answer: The perimeter of the given triangle is
P = (x + 3) + (2x + 1) + 3x = 6x + 4.
Another fi gure with the same perimeter is shown.
22
3x
3x
P = 3x + 2 + 3x + 2 = 6x + 4
Maintaining Mathematical Profi 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 defi nition of an identity, which has
infi nitely many solutions
1.1–1.3 Quiz (p. 26)
1. x + 9 = 7 Write the equation.
− 9 − 9 Subtract 9 from each side.
x = −2 Simplify.
Check: x + 9 = 7
−2 + 9 =?
7
7 = 7 ✓
The solution is x = −2.
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 ✓
The solution is r = −5.
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 ✓
The solution is p = 24.
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 ✓
The solution is m = 8.
32 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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 ✓
The solution is x = 14.
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.
Copyright © Big Ideas Learning, LLC Algebra 1 33All rights reserved. Worked-Out Solutions
Chapter 1
12. −8 − 5n = 64 + 3n
+ 5n + 5n
−8 = 64 + 8n
− 64 − 64
−72 = 8n
−72
— 8 =
8n —
8
−9 = n
The solution is n = −9.
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
The solution is y = −2.
15. 4(g + 8) = 7 + 4g
4(g) + 4(8) = 7 + 4g
4g + 32 = 7 + 4g
− 4g − 4g
32 = 7
The statement 32 = 7 is never true. So, the equation has no
solution.
16. −4(−5h − 4) = 2(10h + 8)
−4(−5h) − 4(−4) = 2(10h) + 2(8)
20h + 16 = 20h + 16
− 20h − 20h
16 = 16
The statement 16 = 16 is always true. So, the equation is an
identity and has infi nitely many solutions.
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.
34 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 1
19. a. Words: Total cost
for Studio A= Total cost for
Studio B
Variable: Let h be the time (in hours) spent painting
at the studio.
Equation: 10 + 8h = 16 + 6h
10 + 8h = 16 + 6h
−6h −6h
10 + 2h = 16
− 10 − 10
2h = 6
2h — 2 =
6 —
2
h = 3
After 3 hours of painting, the total costs will be the same
at both studios.
b. If Studio B increases their studio fee by $2, then both
studios have the same studio fee of $8.
10 + 8h = 16 + 8h
− 8h − 8h
10 = 16
The statement 10 = 16 is never true. So, the equation has
no solution. In other words, because Studio B charges
more for the vase, if their studio fee is the same, their total
costs will never be the same. More specifi cally, Studio B
will always charge more.
1.4 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
fi nd the ones that give the expected solution.
4. Sample answer: You can solve an absolute value equation
algebraically, graphically, or numerically. For the algebraic
method, write and solve two linear equations, one that has
the expression equal to the constant and one that has the
expression equal to the opposite of the constant. For the
graphical method, fi rst identify the point on the number
line that makes the absolute value expression equal to 0.
Using the constant that the absolute value expression is
equal to, fi nd both of the points that are this many units
from the original point in either direction. These two values
are the solutions. For the numerical method, you can use
a spreadsheet to calculate the values of the absolute value
expression for many values of the variable until you identify
the value or values that make the absolute value expression
equal to the given constant.
Chapter 1
Copyright © Big Ideas Learning, LLC Algebra 1 35All rights reserved. Worked-Out Solutions
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
The solutions are x = −3 and x = 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 ✓
The solutions are x = −2 and x = 6.
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.
36 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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 ✓
The solutions are x = −3 and x = 7.
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.
Copyright © Big Ideas Learning, LLC Algebra 1 37All rights reserved. Worked-Out Solutions
Chapter 1
12. ∣ 2 + x ∣ = ∣ x − 8 ∣ 2 + x = x − 8 or 2 + x = −(x − 8)
− x − x 2 + x = −x + 8
2 = −8 + x + x 2 + 2x = 8
− 2 − 2 2x = 6
2x
— 2 = 6 —
2
x = 3
Check: ∣ 2 + x ∣ = ∣ x − 8 ∣ ∣ 2 + 3 ∣ =? ∣ 3 − 8 ∣ ∣ 5 ∣ =? ∣ −5 ∣ 5 = 5 ✓
The solution is x =3.
13. ∣ 5x − 2 ∣ = ∣ 5x + 4 ∣ 5x − 2 = 5x + 4 or 5x − 2 = −(5x + 4)
− 5x − 5x 5x − 2 = −5x − 4
−2 = 4 + 5x + 5x
10x − 2 = −4
+2 +2
10x = −2
10x
— 10
= −2
— 10
x = − 1 — 5
Check: ∣ 5x − 2 ∣ = ∣ 5x + 4 ∣
∣ 5 ( − 1 — 5 ) − 2 ∣ =? ∣ 5 ( − 1 —
5 ) + 4 ∣
∣ −1 − 2 ∣ =? ∣ −1 + 4 ∣ ∣ −3 ∣ =? ∣ 3 ∣ 3 = 3 ✓
The solution is x = − 1 — 5 .
1.4 Exercises (pp. 32–34)
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 − 3
m = 4 m = −10
2 40−2−4−6−8−10
The solutions are m = −10 and m = 4.
16. ∣ q − 8 ∣ = 14
q − 8 = 14 or q − 8 = −14
+ 8 + 8 + 8 + 8
q = 22 q = −6
−6
8 16 240−8
22
The solutions are q = −6 and q = 22.
The statement
2 = –8 is false.
So, the original
equation has only
one solution.
The statement –2 = 4
is false. So, the original
equation has only one
solution.
38 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 1
17. ∣ −3d ∣ = 15
−3d = 15 or −3d = −15
−3d —
−3 =
15 —
−3 −3d
— −3
= −15
— −3
d = −5 d = 5
−5
2 4 60−2−4−6
5
The solutions are d = −5 and d = 5.
18. ∣ t — 2 ∣ = 6
t —
2 = 6 or
t —
2 = −6
2⋅ t —
2 = 2⋅ 6 2⋅
t —
2 = 2(−6)
t = 12 t = −12
4 8 120−4−8−12
The solutions are t = −12 and t = 12.
19. ∣ 4b − 5 ∣ = 19
4b − 5 = 19 or 4b − 5 = −19
+ 5 + 5 + 5 + 5
4b = 24 4b = −14
4b
— 4 =
24 —
4
4b —
4 =
−14 —
4
b = 6 b = − 7 — 2
−3.5
2 4 60−2−4
The solutions are b = − 7 — 2 and b = 6.
20. ∣ x − 1 ∣ + 5 = 2
− 5 − 5∣ x − 1 ∣ = −3
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
The solution is s = 20.
Copyright © Big Ideas Learning, LLC Algebra 1 39All rights reserved. Worked-Out Solutions
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
The equation is ∣ x − 13 ∣ = 5.
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
The equation is ∣ x − 2 ∣ = 8.
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 ✓
40 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.
Copyright © Big Ideas Learning, LLC Algebra 1 41All rights reserved. Worked-Out Solutions
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.
42 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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 ✓
The solution is x = 3 —
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.
Copyright © Big Ideas Learning, LLC Algebra 1 43All rights reserved. Worked-Out 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.
44 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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 ✓
The solutions are x = −5 and x = 1.
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 Profi 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.
Copyright © Big Ideas Learning, LLC Algebra 1 45All rights reserved. Worked-Out Solutions
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) —
5 Divide each side by 5.
6 = b Simplify.
So, b = 6, and the base of the parallelogram is 6 inches.
c. A = bh Write the formula.
A — h =
bh —
h Divide each side by h.
A — h = b Simplify.
d. Sample answer: In both cases, you use the Division
Property of Equality to undo multiplication. The second
process involved all variables.
2. a. A = 1 —
2 h(b1 + b2)
2 ⋅ A = 2 ⋅ 1 —
2 h(b1 + b2)
2A = h(b1 + b2)
2A —
(b1 + b2) =
h(b1 + b2) — (b1 + b2)
2A —
b1 + b2
= h
h = 2A —
b1 + b2
= 2(63)
— 8 + 10
= 126
— 18
= 7
The height of the trapezoid is 7 centimeters.
b. C = 2πr
C — 2π
= 2πr
— 2π
C — 2π
= r
r = C
— 2π
= 24π — 2π
= 12
The radius of the circle is 12 feet.
c. V = Bh
V — B
= Bh
— B
V — B
= h
h = V
— B
= 75
— 15
= 5
The height of the rectangular prism is 5 yards.
d. V = 1 —
3 Bh
3 ⋅ V = 3 ⋅ 1 —
3 Bh
3V = Bh
3V — B
= Bh
— B
3V
— B
= h
h = 3V
— B
= 3(24π)
— 12π
= 72π — 12π
= 6
The height of the cone is 6 meters.
3. Sample answer: You can solve a given formula for a different
variable to form a new formula that can be used to solve for
the variable. For instance, suppose a rectangle has a length
of 15 inches and an area of 60 square inches. You can solve
the formula for the area of a rectangle for w, and use the new
formula to fi nd the width.
A = ℓw
A
— ℓ = ℓw
— ℓ
A
— ℓ = w
w = A
— ℓ = 60
— 15
= 4
The width of the rectangle is 4 inches.
1.5 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.
46 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 1
2. 2x − 2y = 5
2x − 2x − 2y = 5 − 2x
−2y = 5 − 2x
−2y — −2
= 5 − 2x
— −2
y = − 5 — 2 + x
The rewritten literal equation is y = − 5 —
2 + x.
3. 20 = 8x + 4y
20 − 8x = 8x − 8x + 4y
20 − 8x = 4y
20 − 8x —
4 =
4y —
4
5 − 2x = y
The rewritten literal equation is y = 5 − 2x.
4. y = 5x − 4x
y = x(5 − 4)
y = x(1)
y = x
The rewritten literal equation is x = y.
5. 2x + kx = m
x(2 + k) = m
x(2 + k)
— (2 + k)
= m —
(2 + k)
x = m —
2 + k
The rewritten literal equation is x = m —
2 + k .
6. 3 + 5x − kx = y
3 − 3 + 5x − kx = y − 3
5x − kx = y − 3
x(5 − k) = y − 3
x(5 − k)
— (5 − k)
= y − 3
— (5 − k)
x = y − 3
— 5 − k
The rewritten literal equation is x = y − 3
— 5 − k
.
7. A = 1 —
2 bh
2 ⋅ A = 2 ⋅ 1 —
2 bh
2A = bh
2A — b =
bh —
b
2A — b = h
When you solve for h, you obtain h = 2A
— b .
8. S = πr2 + πrℓ S − πr2 = πr2 − πr2 + πrℓ S − πr2 = πrℓ
S − πr2
— πr
= πrℓ — πr
S − πr2
— πr
= ℓ
When you solve for ℓ, you obtain ℓ = S − πr2
— πr
.
9. F = 9 —
5 C + 32 =
9 —
5 (37) + 32 = 66.6 + 32 = 98.6
Because 98.6°F is less than 100°F, your friend does not have
a fever.
10. I = Prt
I — rt
= Prt
— rt
I — rt
= P
P = I —
rt =
500 —
(0.04)(5) =
500 —
0.2 = 2500
You must deposit $2500.
11. d = rt
d —
r =
rt —
r
d —
r = t
So, d —
60 is how long (in hours) it takes for the truck driver to
deliver the freight, and d —
45 is the driving time (in hours) of
the return trip.
d —
60 +
d —
45 = 7
3 ⋅ d
— 3 ⋅ 60
+ 4 ⋅ d
— 4 ⋅ 45
= 7
3d + 4d —
180 = 7
7d — 180
= 7
180
— 7 ⋅
7d —
180 =
180 —
7 ⋅ 7
d = 180
So, it takes d — 60
= 180
— 60
= 3 hours to deliver the freight, and
the return trip takes d — 45
= 180
— 45
= 4 hours.
1.5 Exercises (pp. 40–42)
Vocabulary and Core Concept Check
1. no; The equation 9r + 16 = π
— 5 is not a literal equation
because it has only one variable.
Copyright © Big Ideas Learning, LLC Algebra 1 47All rights reserved. Worked-Out Solutions
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
The rewritten literal equation is 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
The rewritten literal equation is y = 3 − 4x.
7. 9x − y = 45
9x − 9x − y = 45 − 9x
−y = 45 − 9x
−y — −1 =
45 − 9x —
−1
y = − 45 + 9x
The rewritten literal equation is y = −45 + 9x.
8. 6x − 3y = −6
6x − 6x − 3y = −6 − 6x
− 3y = −6 − 6x
−3y — −3 =
−6 − 6x —
−3
y = 2 + 2x
The rewritten literal equation is y = 2 + 2x.
9. 4x − 5 = 7 + 4y
4x − 5 − 7 = 7 − 7 + 4y
4x − 12 = 4y
4x − 12
— 4 =
4y —
4
x − 3 = y
The rewritten literal equation is y = x − 3.
10. 16x + 9 = 9y − 2x
16x + 2x + 9 = 9y − 2x + 2x
18x + 9 = 9y
18x + 9
— 9 =
9y —
9
2x + 1 = y
The rewritten literal equation is y = 2x + 1.
11. 2 + 1 —
6 y = 3x + 4
2 − 2 + 1 —
6 y = 3x + 4 − 2
1 —
6 y = 3x + 2
6⋅ 1
— 6 y = 6 ⋅ (3x + 2)
y = 6(3x) + 6(2)
y = 18x + 12
The rewritten literal equation is y = 18x + 12.
12. 11 − 1 —
2 y = 3 + 6x
11 − 11 − 1 —
2 y = 3 − 11 + 6x
− 1 —
2 y = −8 + 6x
−2 ⋅ ( − 1 —
2 y ) = −2 ⋅ (−8 + 6x)
y = −2(−8) − 2(6x)
y = 16 − 12x
The rewritten literal equation is y = 16 − 2x.
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 .
48 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.
Copyright © Big Ideas Learning, LLC Algebra 1 49All rights reserved. Worked-Out Solutions
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 ) .
50 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 1
32. F = G ( m1m2 — d2
)
F — G
=
G ( m1m2 — d2
) —
G
F — G
= m1m2 —
d2
d2 ⋅ F
— G
= d2 ⋅ m1m2 —
d2
d2F — G
= m1m2
d2F — G
⋅ 1 —
m2
= m1m2 ⋅ 1 —
m2
d2F — Gm2
= m1
When you solve the formula for m1, you obtain m1 = d2F
— Gm2
.
33. a. S = L − rL
S − L = L − L − rL
S − L = −rL
S − L — −L
= −rL
— −L
− S — L
+ 1 = r
1 − S — L
= r
When you solve the formula for r, you obtain r = 1 − S — L
.
b. r = 1 − S — L
= 1 − 18 —
30 = 1 − 0.6 = 0.4
The discount rate is 0.4, or 40%.
34. a. d = m
— V
V ⋅ d = V ⋅ m
— V
dV = m
When you solve the formula for m, you obtain m = dV.
b. m = dV = (5.01)(1.2) = 6.012.
The mass of the pyrite sample is 6.012 grams.
35. I = Prt
I —
Pr =
Prt —
Pr
I —
Pr = t
t = I —
Pr =
500 ——
(2000)(0.04) = 6.25 = 6
1 —
4
You must leave the money in the account for 6.25 years, or
6 years and 3 months.
36. d = rt
d —
r =
rt —
r
d —
r = t
So, d —
460 is how long the fi rst fl ight takes, and
d —
500 is how long
the return fl ight takes. Add these expressions and solve for
the one-way distance:
d —
460 +
d —
500 = 4.8
25 ⋅ d —
25 ⋅ 460 +
23 ⋅ d —
23 ⋅ 500 = 4.8
25d + 23d
— 11,500
= 4.8
48d
— 11,500
= 4.8
11,500
— 48
⋅ 48d —
11,500 =
11,500 —
48 ⋅ 4.8
d = 1150
Substitute this distance into each of the original expressions.
So, the fi rst fl ight takes d —
460 =
1150 —
460 = 2.5 hours, and the
return fl ight takes d —
500 =
1150 —
500 = 2.3 hours.
37. a. P = 2x + 2 ( 1 — 2 ⋅ C )
P = 2x + C
P = 2x + 2πr
b. P = 2x + 2πr
P − 2πr = 2x
P − 2πr —
2 = x
c. x = P − 2πr
— 2 =
660 − 2π(50) ——
2 =
660 − 100π — 2 ≈ 173
So, the length of the rectangular portion of the track is
about 173 feet.
38. a. Because d = 55t and d = 20g,
55t = 20g.
b. 55t = 20g
55t — 20
= 20g
— 20
2.75t = g
c. The amount of gasoline used can be found using the
formula from part (b). Either of the original formulas can
be used to fi nd the distance. If you travel for 6 hours, you
will use g = 2.75t = 2.75(6) = 16.5 gallons of gas, and
you will go d = 55t = 55(6) = 330 miles.
Copyright © Big Ideas Learning, LLC Algebra 1 51All rights reserved. Worked-Out Solutions
Chapter 1
39. a. C = 2πr
C
— 2π
= 2πr
— 2π
C
— 2π
= r
So, the radius of a column is given by the rewritten
formula r = C
— 2π
.
b. r = C
— 2π
= 7 —
2π ≈ 1.1 ft
r = C
— 2π
= 8 —
2π ≈ 1.3 ft
r = C
— 2π
= 9 —
2π ≈ 1.4 ft
So, the radius of a column that has a circumference of
7 feet, 8 feet, or 9 feet is 1.1 feet, 1.3 feet, or 1.4 feet,
respectively.
c. Sample answer: First you can use the formula r = C
— 2π
to
fi nd the radius of the cross section. Then you can use the
value of the radius and the formula A = πr2 to calculate
the area of the cross section.
40. a. The rectangular prism has two square bases with side
length b and four rectangular sides that are b units wide
andℓ units long.
S = 2b2 + 4bℓ b. Sample answer: Choose sideℓbecause this variable is only
in the formula one time. So, it will be easier to isolate.
41. F = 9 —
5 C + 32 =
9 —
5 (20) + 32 = 36 + 32 = 68
no; Because 68°F is less than 70°F, Thermometer A displays
a lesser temperature than Thermometer B.
42. Sample answer: One possible value for h is 8. Then, using
the formula from Exercise 29, the missing base is
b2 = 2A
— b − b1
= 2(40)
— 8 − 8
= 80
— 8 − 8
= 10 − 8
= 2 centimeters.
h = 8 cm
8 cm
2 cm
A = 40 cm2
43.
h b
A = 5 ( 1 — 2 bh )
A = 5bh
— 2
2A = 2 ⋅ 5bh —
2
2A = 5bh
2A
— 5b
= 5bh
— 5b
2A
— 5b
= h
So, the height, h, is given by h = 2A
— 5b
.
44.
h
b
A = 8 ( 1 — 2 bh )
A = 4bh
A
— 4b
= 4bh
— 4b
A
— 4b
= h
So the height, h, is given by h = A
— 4b
.
45. x = a + b + c
— ab
ab ⋅ x = ab ⋅ a + b + c —
ab
abx = a + b + c
abx − a = a − a + b + c
abx − a = b + c
a(bx − 1) = b + c
a(bx − 1)
— (bx − 1)
= b + c
— (bx − 1)
a = b + c
— bx − 1
So, the rewritten literal equation is a = b + c
— bx − 1
.
52 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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 Profi 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
The solutions are x = −2 and x = 8.
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 fi rst step is to add 9 to each side in order
to isolate the absolute value expression. After this step, the
absolute value expression is equal to a positive value. So,
you know that the equation has two solutions.
2. Sample answer: The absolute value expressions were the
same on each side. So, use another variable to represent the
absolute value expression and solve the new equation for the
value of this variable. Because the absolute value expression
is equal to this value, write two linear equations, one of
which has the expression equal to the value and one of which
has the expression equal to its opposite. These equations give
two solutions for the original equation.
3. Sample answer: First, partition each shape into congruent
triangles by connecting the center to each of the vertices.
Then multiply the number of triangles by the formula for the
area of a triangle. This gives the area of the original shape.
Copyright © Big Ideas Learning, LLC Algebra 1 53All rights reserved. Worked-Out Solutions
Chapter 1
Chapter 1 Review (pp. 44–46)
1. z + 3 = −6 Write the equation.
− 3 − 3 Subtract 3 from each side.
z = −9 Simplify.
Check: z + 3 = −6
−9 + 3 =?
−6
−6 = −6 ✓
The solution is z = −9.
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 ✓
The solution is n = 10.
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 ✓
The solution is y = −9.
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 ✓
The solution is n = 6.
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 ✓
The solution is z = −5.
54 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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 ✓
The solution is x = 18.
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
The solution is n = −4.
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 infi 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.
Copyright © Big Ideas Learning, LLC Algebra 1 55All rights reserved. Worked-Out Solutions
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 ✓
The solution is x = −1.
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.
56 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
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.
+ 7 + 7 Add 7 to each side.
x = 22 Simplify.
Check: x − 7 = 15
22 − 7 =?
15
15 = 15 ✓
The solution is x = 22.
2. 2 —
3 x + 5 = 3 Write the equation.
− 5 − 5 Subtract 5 from each side.
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 ✓
The solution is x = −3.
3. 11x + 1 = −1 + x Write the equation.
− x − x Subtract x from each side.
10x + 1 = −1 Simplify.
− 1 − 1 Subtract 1 from each side.
10x = −2 Simplify.
10x
— 10
= −2
— 10
Divide each side by 10.
x = − 1 — 5
Check: 11x + 1 = −1 + x
11 ( − 1 — 5 ) + 1 =
? −1 + ( − 1 —
5 )
− 11 —
5 + 1 =
? −
5 —
5 −
1 —
5
− 11 —
5 +
5 —
5 =
? −
6 —
5
− 6 — 5 = − 6 —
5 ✓
The solution is x = − 1 —
5 .
4. 2 ∣ x − 3 ∣ − 5 = 7
+ 5 + 5 2 ∣ x − 3 ∣ = 12
2 ∣ x − 3 ∣
— 2 =
12 —
2
∣ x − 3 ∣ = 6
x − 3 = 6 or x − 3 = −6
+ 3 + 3 + 3 + 3 x = 9 x = −3
The solutions are x = −3 and x = 9.
Copyright © Big Ideas Learning, LLC Algebra 1 57All rights reserved. Worked-Out Solutions
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 infi 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
The solution is 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
The statement 18 = 19 is never true. So, the equation has no
solution.
10. 3x − 5 = 3x − c
− 3x − 3x
−5 = −c
− 5 = − c −1 = −1
5 = c
If c = 5, the equation is an identity. So, for c ≠ 5, the
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 ✓
58 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 1
13. a. P = 2ℓ+ 2w
P − 2ℓ = 2ℓ − 2ℓ + 2w
P − 2ℓ = 2w
P − 2ℓ —
2 =
2w —
2
P − 2ℓ —
2 = w
When you solve the formula for w, you obtain
w = P − 2ℓ —
2 .
b. w = P − 2ℓ —
2
= 330 − 2(100)
—— 2
= 330 − 200
— 2
= 130
— 2
= 65 yd
The fi eld is 65 yards wide.
c. A =ℓw A = πr2 314.16
— 6500
≈ 0.048, or 4.8%
A = (100)(65) A = π(10)2
A = 6500 A ≈ 314.16
About 4.8% of the fi eld is inside the circle.
14. a. Words: Dealership
total cost =Local mechanic
total cost
Variable: Let h be the time (in hours) it takes to do the
repairs.
Equation: 24 + 99 ⋅ h = 45 + 89 ⋅ h
24 + 99h = 45 + 89h
− 89h − 89h
24 + 10h = 45
− 24 − 24
10h = 21
10h
— 10
= 21
— 10
h = 2.1, or 2 hours and 0.1(60) = 6 minutes
After 2.1 hours, or 2 hours and 6 minutes, of work the
total costs are the same at both places.
b. For less than 2.1 hours of work, the repairs cost less at the
dealership because the parts cost less there, and for more
than 2.1 hours of work, the repairs cost less at the local
mechanic.
15. Sample answer: If x = −2, then the right side of the
equation, 6x, is negative, and the absolute value expression
cannot equal a negative value. So, x = −2 cannot make the
equation true and must be an extraneous solution.
16. Sample answer: The variables cancelled out of the equation,
and the resulting statement, −8 = −8, is always true. So, the
equation is an identity and has infi nitely many solutions.
Chapter 1 Standards Assessment (pp. 48–49)
1. B; 37.5% of 48 = 0.375 ⋅ 48 = 18 beginner trails
48 − 18
— 2 =
30 —
2 = 15 each of intermediate and expert trails
2. cx − a + b = 2b, x = a + b
— c , b + a = cx;
cx − a + b = 2b
− b − bcx − a = b ✓
0 = cx − a + b
− b − b−b = cx − a ✗
2cx − 2a = b —
2
2(cx − a) = b —
2
1 —
2 ⋅ 2(cx − a) =
1 —
2 ⋅
b —
2
cx − a = b —
4 ✗
x − a = b —
c x = a + b —
c b + a = cx
c ⋅ (x − a) = c⋅ b — c c⋅ x = c⋅
a + b —
c cx = b + a
c(x) − c(a) = b cx = a + b cx − a = b + a − a
cx − ca = b ✗ cx − a = a − a + b cx − a = b ✓
cx − a = b ✓
3. 3(x − a) = 3x − 6
3(x) − 3(a) = 3x − 6
3x − 3a = 3x − 6
− 3x − 3x
−3a = −6
a. −3a = −6 b. −3a = −6 c. −3a = −6
−3(3) =?
−6 −3(−3) =?
−6 −3(2) =?
−6
−9 = −6 ✗ 9 = −6 ✗ −6 = −6 ✓
When a = 3, N < 1. When a = −3, N < 1. When a = 2, N > 1.
d. −3a = −6 e. −3a = −6 f. −3a = −6
−3(−2) =?
−6 −3(x) = −6 −3(−x) = −6
6 = −6 ✗ −3x = −6 3x = −6
When a = −2, N < 1. −3x —
−3 =
−6 —
−3
3x — 3 =
−6 —
3
x = 2 x = −2
When a = x, N = 1. When a = −x, N = 1.
Copyright © Big Ideas Learning, LLC Algebra 1 59All rights reserved. Worked-Out Solutions
Chapter 1
4. a.
Words:Total
cost =
Cost per
can of
white
paint
⋅Number
of cans
of white
paint
+
Cost
per can
of blue
paint
⋅Number
of cans
of blue
paint
Variable: Let x be the number of cans of white paint. Then
(5 − x) is the number of cans of blue paint.
Equation: 132 = 24 ⋅ x + 28 ⋅ (5 − x)
132 = 24x + 28(5 − x)
b. 132 = 24x + 28(5 − x)
132 = 24x + 28(5) − 28(x)
132 = 24x + 140 − 28x
132 = −4x + 140
− 140 − 140
− 8 = −4x
−8
— −4
= −4x
— −4
2 = x and 5 − x = 5 − 2 = 3
So, you bought 2 cans of white paint and 3 cans of blue
paint. However, if you switched the colors and bought 3 cans
of white paint and 2 cans of blue paint, you would have spent
24(3) + 28(2) = $128, which is $132 − $128 = $4 less.
5. The equations 8x + 6 = −2x − 14 and 5x + 3 = −7 are
equivalent because they have the same solution, x = −2.
6x + 6 = −14
− 6 − 66x = −20
6x
— 6 =
−20 —
6
x = − 10 —
3
8x + 6 = −2x − 14
+ 2x + 2x
10x + 6 = −14
−6 −6
10x = −20
10x
— 10
= −20
— 10
x = −2
7x + 3 = 2x − 13 5x + 3 = −7
− 2x − 2x − 3 − 3 5x + 3 = 13 5x = −10
− 3 − 3 5x
— 5 =
−10 —
5
5x = 10 x = −2
5x — 5 =
10 —
5
x = 2
6. B; (x − 5) + x —
2 + 6 = 13
x + 1 —
2 x − 5 + 6 = 13
3 —
2 x + 1 = 13
− 1 − 1
3 —
2 x = 12
2 —
3 ⋅
3 —
2 x =
2 —
3 ⋅ 12
x = 8
So, the sides are 6 inches, x − 5 = 8 − 5 = 3 inches,
and x —
2 =
8 —
2 = 4 inches, and the shortest side is 3 inches.
7. a.
Words:
Cable
TV
monthly
cost
⋅Number
of
months=
Cost of
satellite
TV
receiver
+
Satelite
TV
monthly
cost
⋅Number
of
months
Variable: Let m be the number of months.
Equation: 45 ⋅ m = 99 + 36 ⋅ m
45m = 99 + 36m
− 36m − 36m
9m = 99
9m — 9 =
99 —
9
m = 11
After 11 months, you and your friend will have paid the
same amount for TV services.
b. yes; After 11 months, you will have paid the same
amount, and because your friend’s monthly cost is less, he
will pay less in the 12th month and therefore less overall
for the year.
60 Algebra 1 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 1
8. ∣ 8x + 3 ∣ = 0
8x + 3 = 0
− 3 − 38x = −3
8x
— 8 =
−3 —
8
x = − 3 — 8
−6 = 5x − 9
+ 9 + 93 = 5x
3 —
5 =
5x —
5
3 —
5 = x
3x − 12 = 3(x − 4) + 1
3x − 12 = 3(x) − 3(4) + 1
3x − 12 = 3x − 12 + 1
3x − 12 = 3x − 11
− 3x − 3x
−12 = −11 ✗
−2x + 4 = 2x + 4
+ 2x + 2x
4 = 4x + 4
− 4 − 40 = 4x
0 —
4 =
4x —
4
0 = x
0 = ∣ x + 13∣ + 2− 2 − 2
−2 = ∣ x + 13 ∣ ✗
−4(x + 4) = −4x − 16
−4(x) − 4(4) = −4x − 16
−4x − 16 = −4x − 16
+ 4x + 4x
−16 = −16
12x − 2x = 10x − 8
10x = 10x − 8
− 10x − 10x
0 = 8 ✗
9 = 3 ∣ 2x − 11 ∣
9 —
3 =
3 ∣ 2x − 11 ∣ —
3
3 = ∣ 2x − 11 ∣ 3 = 2x − 11 or −3 = 2x − 11
+ 11 + 11 + 11 + 11
14 = 2x 8 = 2x
14
— 2 =
2x —
2
8 —
2 =
2x —
2
7 = x 4 = x
7 − 2x = 3 − 2(x − 2)
7 − 2x = 3 − 2(x) − 2(−2)
7 − 2x = 3 − 2x + 4
7 − 2x = −2x + 7
+ 2x + 2x
7 = 7
No solution One solutionTwo
solutionsInfi nitely
many solutions
3x − 12 =
3(x − 4) + 1
∣ 8x + 3 ∣ = 0 9 = 3
∣ 2x − 11 ∣ −4(x + 4) =
−4x − 16
0 = ∣ x + 13 ∣ + 2 −6 = 5x − 9 7 − 2x =
3 − 2(x − 2)
12x − 2x = 10x − 8 −2x + 4 =
2x + 4
9. 1000 ft —
12.5 sec = 80
feet —
second =
80 feet —
second
So, the expressions that do not represent the average speed of
the car are 80 second
— feet
and second
— 80 feet
.