Appendix A — Answer Key, Chapter 10: Two-Step Linear Equations

SOL 7.PFA.3 · Covers textbook Chapter 10 and the companion workbook. Item numbers match the textbook; workbook items are the same problems, so this key serves both. Every solution below has been substituted back into the original equation. Reasoning answers show an acceptable response, not the only wording.

Lesson 10.1 — What It Means to Solve an Equation

Guided practice

1. $4(3) - 5 = 12 - 5 = 7$, and the right side is $7$. Yes, $x = 3$ is a solution.
2. $k = 4$. Tests: $2(-2) - 3 = -7$; $2(1) - 3 = -1$; $2(4) - 3 = 5$. Only $4$ gives $5$.
3. Coefficient $7$; constant term $-9$. Undo the subtraction of 9 first, by adding 9 to both sides.
4. $-2(-5) + 1 = 10 + 1 = 11$, and the right side is $11$. Confirmed.

Independent practice

5. a) $3(6) - 8 = 18 - 8 = 10$ → yes b) $4(-3) + 9 = -12 + 9 = -3$ → yes c) $5(2) - 4 = 10 - 4 = 6$, and $6 \neq 8$ → no d) $8\left(\tfrac{1}{2}\right) + 2 = 4 + 2 = 6$ → yes
6. $x = -3$, since $9 - 2(-3) = 9 + 6 = 15$. ($9 - 2(3) = 3$ and $9 - 2(12) = -15$ both fail.)
7. a) coefficient $5$, constant $12$ b) coefficient $\tfrac{1}{4}$, constant $-6$ c) coefficient $-3$, constant $8$
8. Applied to the variable: multiply by 2, then subtract 7. To undo: add 7, then divide by 2. Check: $2(4) - 7 = 8 - 7 = 1$. Confirmed.
9. Many answers. One: $5x + 2 = 17$. Check: $5(3) + 2 = 15 + 2 = 17$. True.
10. $2(5) + 3 = 10 + 3 = 13$, so $m = 5$ is a solution. It means the $13.00 fare corresponds to a 5-mile ride.
11. Because a check is supposed to test the answer against what the problem actually said. If you substitute into a line you wrote yourself and that line contains an arithmetic error, the check will agree with your error and report success. Only the original equation is known to be correct.
12. Jordan multiplied the signs incorrectly: $-3(-2) = 6$, not $-6$, because a negative times a negative is positive. The correct check is $-3(-2) + 5 = 6 + 5 = 11$, which matches the right side. So $x = -2$ is a solution.

Exit ticket 10.1

13. $2(7) - 5 = 14 - 5 = 9$, and the right side is $9$. Yes.
14. $x = -4$, since $3(-4) + 10 = -12 + 10 = -2$.
15. Coefficient $-6$; constant term $1$. Undo the addition of 1 first, by subtracting 1 from both sides.
16. A number is a solution when substituting it for the variable makes the two sides of the equation name the same number — that is, makes the sentence true.

Lesson 10.2 — Modeling Two-Step Equations with Tiles and a Balance

Guided practice

17. $2x + 5 = 13$
18. Move 1: remove 5 chips from each pan, leaving $2x = 8$. Move 2: split each pan into 2 equal groups, leaving one block matched with 4 chips, so $x = 4$. Confirmation: replacing each block with 4 chips gives $4 + 4 + 5 = 13$ chips on the left and $13$ on the right, so the beam is level. Check: $2(4) + 5 = 13$. True.
19. Left mat: four $x$ tiles and three $1$ tiles; right mat: fifteen $1$ tiles. Move 1: remove three $1$ tiles from both mats, leaving $4x = 12$. Move 2: split both mats into 4 equal groups, giving $x = 3$. Check: $4(3) + 3 = 12 + 3 = 15$. True.
20. A zero pair is a $+1$ tile together with a $-1$ tile; the pair is worth $0$, so adding or removing one does not change a mat's value. Adding three $1$ tiles to each mat is legal because the same amount goes on both sides, and on the left the three new tiles form zero pairs with the three $-1$ tiles and clear away, leaving $2x = 10$. So $x = 5$. Check: $2(5) - 3 = 10 - 3 = 7$. True.

Independent practice

21. a) $3x + 2 = 17$. Remove 2 chips from each pan to get $3x = 15$, then split into 3 equal groups: $x = 5$. Check: $3(5) + 2 = 17$. True. b) $2x - 4 = 6$. Add four $1$ tiles to each mat; four zero pairs clear the left mat, leaving $2x = 10$; split into 2 equal groups: $x = 5$. Check: $2(5) - 4 = 6$. True.
22. Left mat: five $x$ tiles and six $1$ tiles; right mat: twenty-one $1$ tiles. Move 1: remove six $1$ tiles from both mats, leaving $5x = 15$. Move 2: split both mats into 5 equal groups, giving $x = 3$. Check: $5(3) + 6 = 15 + 6 = 21$. True.
23. Left mat: three $x$ tiles and five $-1$ tiles; right mat: seven $1$ tiles. Move 1: add five $1$ tiles to each mat; the five zero pairs on the left clear, leaving $3x = 12$. Move 2: split both mats into 3 equal groups, giving $x = 4$. Confirmation: replace each $x$ tile with 4 unit tiles — the left mat holds $12$ positive tiles and $5$ negative tiles, which is $7$ after zero pairs are removed, matching the right mat.
24. Left mat: two $-x$ tiles and nine $1$ tiles; right mat: three $1$ tiles. Move 1: add nine $-1$ tiles to each mat. On the left, nine zero pairs form and clear, leaving $-2x$; on the right, three zero pairs clear and six $-1$ tiles remain, so the right mat is $-6$. That gives $-2x = -6$. Move 2: split both mats into 2 equal groups, so one $-x$ tile matches three $-1$ tiles: $-x = -3$, and therefore $x = 3$. Check: $-2(3) + 9 = -6 + 9 = 3$. True.
25. Left mat: six $1$ tiles; right mat: two $x$ tiles and ten $1$ tiles. Move 1: add ten $-1$ tiles to each mat. On the right, ten zero pairs clear, leaving $2x$; on the left, six zero pairs clear and four $-1$ tiles remain, so the left mat is $-4$. That gives $-4 = 2x$. Move 2: split both mats into 2 equal groups: $x = -2$. Check: $2(-2) + 10 = -4 + 10 = 6$. True.
26. Removing tiles from one mat only tips the balance, so the equation Devon wrote after that move was no longer true. He should remove 5 unit tiles from both mats, leaving $2x = 8$, and then split both mats into 2 equal groups. The correct solution is $x = 4$. Check: $2(4) + 5 = 13$. True.
27. $3b + 4 = 22$. Remove the 4-pound weight from the left pan and 4 pounds from the right pan, leaving $3b = 18$. Split each pan into 3 equal groups: one box balances 6 pounds, so $b = 6$. Confirmation: three 6-pound boxes plus 4 pounds is $18 + 4 = 22$ pounds, which balances the 22-pound weight. Check: $3(6) + 4 = 22$. True.
28. Splitting into equal groups is the division property of equality, and it only preserves the balance if both sides are divided by the same number. Splitting the left mat into 3 groups and the right into 2 would divide the left by 3 and the right by 2, which changes the two sides by different amounts. The mats would no longer hold equal values, so the equation you read off would be false.

Exit ticket 10.2

29. $2x + 7 = 15$. Remove 7 chips from each pan to get $2x = 8$, then split each pan into 2 equal groups: $x = 4$. Check: $2(4) + 7 = 15$. True.
30. Left mat: three $x$ tiles and two $-1$ tiles; right mat: ten $1$ tiles. Add two $1$ tiles to each mat; two zero pairs clear the left, leaving $3x = 12$. Split both mats into 3 equal groups: $x = 4$. Check: $3(4) - 2 = 12 - 2 = 10$. True.
31. Replace each $x$ tile with 5 unit tiles: the left mat holds $5 + 5 + 3 = 13$ tiles and the right mat holds $13$. They match, so yes, $x = 5$ is the solution. Check: $2(5) + 3 = 13$. True.
32. Removing the same number of tiles from both mats subtracts the same amount from each side, which is the subtraction property of equality. Both sides change by an equal amount, so they stay equal and the model stays level.

Lesson 10.3 — Solving Two-Step Equations

Guided practice

33. $2x + 9 = 21$; subtract 9 from both sides (subtraction property of equality) to get $2x = 12$; divide both sides by 2 (division property of equality) to get $x = 6$. Check: $2(6) + 9 = 12 + 9 = 21$. True.
34. Add 8 to both sides: $5x = 15$. Divide both sides by 5: $x = 3$. Check: $5(3) - 8 = 15 - 8 = 7$. True.
35. Subtract 4 from both sides: $\tfrac{x}{3} = 5$. Multiply both sides by 3: $x = 15$. Check: $\tfrac{15}{3} + 4 = 5 + 4 = 9$. True.
36. Subtract 5 from both sides: $-6x = 18$. Divide both sides by $-6$: $x = -3$. Check: $-6(-3) + 5 = 18 + 5 = 23$. True.

Independent practice

37. a) $4x = 28$, so $x = 7$. Check: $4(7) + 11 = 28 + 11 = 39$. True. b) $7x = -28$, so $x = -4$. Check: $7(-4) - 6 = -28 - 6 = -34$. True. c) $-3x = 15$, so $x = -5$. Check: $-3(-5) + 2 = 15 + 2 = 17$. True. d) $\tfrac{x}{4} = 4$, so $x = 16$. Check: $\tfrac{16}{4} - 5 = 4 - 5 = -1$. True.
38. a) $\tfrac{3}{4}x = 6$; multiply both sides by $\tfrac{4}{3}$: $x = 8$. Check: $\tfrac{3}{4}(8) + 5 = 6 + 5 = 11$. True. b) $0.5x = 5.5$, so $x = 11$. Check: $0.5(11) - 3 = 5.5 - 3 = 2.5$. True. c) $2.4x = 12$, so $x = 5$. Check: $2.4(5) + 1.2 = 12 + 1.2 = 13.2$. True. d) $-\tfrac{2}{5}x = -6$; multiply both sides by $-\tfrac{5}{2}$: $x = 15$. Check: $-\tfrac{2}{5}(15) + 7 = -6 + 7 = 1$. True.
39. a) $20 = 5x$, so $x = 4$. Check: $5(4) + 6 = 26$. True. b) $-10 = -2x$, so $x = 5$. Check: $3 - 2(5) = 3 - 10 = -7$. True.
40. $6x = 7$, so $x = \tfrac{7}{6}$. Check: $6\left(\tfrac{7}{6}\right) + 5 = 7 + 5 = 12$. True. (Do not round; $1.17$ would fail the check.)
41. $8 - 5x = 43$. Subtract 8 from both sides (subtraction property of equality): $-5x = 35$. Divide both sides by $-5$ (division property of equality): $x = -7$. Check: $8 - 5(-7) = 8 + 35 = 43$. True. Reverse order works because the equation records operations done to $x$ in a fixed order — multiply first, then add or subtract. To retrace a path you take the last step first, so the constant comes off before the coefficient is undone. Undoing the multiplication first would require dividing the whole side, constant included, which is legal but creates extra fractions.
42. Many answers. One: $-4x + 1 = 9$. Check: $-4(-2) + 1 = 8 + 1 = 9$. True.
43. $18m = 126$, so $m = 7$. Check: $18(7) + 25 = 126 + 25 = 151$. True. The member has been paying for 7 months.
44. Dividing by 3 has to be applied to the entire side, not just the first term: $\tfrac{3x + 12}{3} = x + 4$, not $x + 12$. Correctly: subtract 12 from both sides to get $3x = 18$, then divide by 3 to get $x = 6$. Check: $3(6) + 12 = 18 + 12 = 30$. True. Priya's answer fails its own check, since $3(-2) + 12 = -6 + 12 = 6 \neq 30$.

Exit ticket 10.3

45. $5x = 35$, so $x = 7$. Check: $5(7) - 7 = 35 - 7 = 28$. True.
46. $-2x = -8$, so $x = 4$. Check: $-2(4) + 11 = -8 + 11 = 3$. True.
47. $\tfrac{x}{6} = 3$, so $x = 18$. Check: $\tfrac{18}{6} + 2 = 3 + 2 = 5$. True.
48. Because solving reverses the order of operations. Going forward the variable is multiplied by the coefficient first and the constant is added last, so going backward the constant comes off first. Undoing the multiplication first would force you to divide the constant term as well, which usually produces fractions for no benefit.

Lesson 10.4 — Writing an Equation for a Situation

Guided practice

49. Let $h$ = the number of hours. $2h + 5 = 15$; $2h = 10$; $h = 5$ hours. Check: $2(5) + 5 = 15$. True.
50. Let $n$ = the number. $2n + 3 = 17$; $2n = 14$; $n = 7$. Check: $2(7) + 3 = 17$. True.
51. Let $h$ = the number of hours worked. $45h + 60 = 285$; $45h = 225$; $h = 5$ hours. Check: $45(5) + 60 = 225 + 60 = 285$. True.
52. Many answers. One: A bookstore charges $3 per used paperback plus an $8 membership fee. Marisol's total was $29. How many paperbacks did she buy? Solution: $3x = 21$, so $x = 7$ paperbacks. Check: $3(7) + 8 = 21 + 8 = 29$. True.

Independent practice

53. Let $n$ = the number. $5n - 8 = 32$; $5n = 40$; $n = 8$. Check: $5(8) - 8 = 40 - 8 = 32$. True.
54. Let $n$ = the number. $\tfrac{n}{4} + 6 = 10$; $\tfrac{n}{4} = 4$; $n = 16$. Check: $\tfrac{16}{4} + 6 = 4 + 6 = 10$. True.
55. Let $t$ = the number of tacos. $2.50t + 1 = 11$; $2.50t = 10$; $t = 4$ tacos. Check: $2.50(4) + 1 = 10 + 1 = 11$. True.
56. $-4h + 30 = 10$; $-4h = -20$; $h = 5$ hours. Check: $-4(5) + 30 = -20 + 30 = 10$. True.
57. Many answers. One: A camp charges $7 per day plus a $12 registration fee. Deshawn's family paid $61. How many days is the camp? Solution: $7x = 49$, so $x = 7$ days. Check: $7(7) + 12 = 49 + 12 = 61$. True.
58. Many answers. One: A length of rope is cut exactly in half, then 3 feet are trimmed from one half, leaving 6 feet. How long was the rope? Solution: $\tfrac{1}{2}x = 9$, so $x = 18$ feet. Check: $\tfrac{1}{2}(18) - 3 = 9 - 3 = 6$. True. The story must halve first because the equation multiplies $x$ by $\tfrac{1}{2}$ before subtracting 3. A story that subtracted 3 first and then halved would be $\tfrac{1}{2}(x - 3) = 6$, a different equation with the solution $x = 15$.
59. Many answers. One: A 25-gallon barrel loses 3 gallons of water each day. After $x$ days, 4 gallons are left. How many days have passed? Solution: $-3x = -21$, so $x = 7$ days. Check: $-3(7) + 25 = -21 + 25 = 4$. True.
60. Sam's equation multiplies the $4 fee by the number of tickets, but the fee is charged once for the whole order, not once per ticket. His equation says $6t + 24 = 34$, which is not the situation. The correct equation is $6t + 4 = 34$; then $6t = 30$ and $t = 5$ tickets. Check: $6(5) + 4 = 30 + 4 = 34$. True.

Exit ticket 10.4

61. Let $r$ = the number of rides. $3r + 7 = 28$; $3r = 21$; $r = 7$ rides. Check: $3(7) + 7 = 28$. True.
62. Let $n$ = the number. $3n - 6 = 9$; $3n = 15$; $n = 5$. Check: $3(5) - 6 = 15 - 6 = 9$. True.
63. Many answers. One: A craft kit costs $4 per bracelet plus a $9 shipping charge. The order came to $33. How many bracelets were ordered? Solution: $4x = 24$, so $x = 6$ bracelets. Check: $4(6) + 9 = 24 + 9 = 33$. True.
64. The variable stands for the quantity the problem does not tell you and is asking about. Read the question sentence, name that quantity together with its units, and write the definition down before writing the equation.

Lesson 10.5 — Solving Problems in Context

Guided practice

65. Let $w$ = the width in centimeters. Two lengths contribute $2(12) = 24$, so $2w + 24 = 40$; $2w = 16$; $w = 8$ cm. Check: $2(8) + 24 = 16 + 24 = 40$. True, and $12 + 12 + 8 + 8 = 40$ cm.
66. Let $n$ = the number of lawns. $12n + 50 = 146$; $12n = 96$; $n = 8$ lawns. Check: $12(8) + 50 = 96 + 50 = 146$. True.
67. Let $h$ = the number of hours. $-7h + 60 = 18$; $-7h = -42$; $h = 6$ hours. Check: $-7(6) + 60 = -42 + 60 = 18$. True.
68. Let $s$ = the price of one sandwich in dollars. $5s + 2.50 = 32.50$; $5s = 30$; $s = 6$, so $6.00. Check: $5(6) + 2.50 = 30 + 2.50 = 32.50$. True.

Independent practice

69. Let $c$ = the number of classes. $8.50c + 40 = 91$; $8.50c = 51$; $c = 6$. Check: $8.50(6) + 40 = 51 + 40 = 91$. True. Kara took 6 classes.
70. Let $m$ = the number of miles. $1.75m + 3.25 = 17.25$; $1.75m = 14$; $m = 8$ miles. Check: $1.75(8) + 3.25 = 14 + 3.25 = 17.25$. True.
71. $2s + 7 = 31$; $2s = 24$; $s = 12$ inches. Check: $2(12) + 7 = 24 + 7 = 31$. True, and $12 + 12 + 7 = 31$ in.
72. Let $t$ = the number of seconds. $1.5t - 6 = 3$; $1.5t = 9$; $t = 6$ seconds. Check: $1.5(6) - 6 = 9 - 6 = 3$. True.
73. Let $n$ = the number of dimes. Three quarters are worth $0.75, so $0.10n + 0.75 = 1.85$; $0.10n = 1.10$; $n = 11$ dimes. Check: $0.10(11) + 0.75 = 1.10 + 0.75 = 1.85$. True.
74. Let $p$ = the length of one piece in feet. $4p + 1 = 10$; $4p = 9$; $p = \tfrac{9}{4}$ feet, or $2\tfrac{1}{4}$ feet. Check: $4\left(\tfrac{9}{4}\right) + 1 = 9 + 1 = 10$. True.
75. Check: $6(10) + 45 = 60 + 45 = 105$. True. Beyond the substitution, Ana still has to interpret the result: say what $x$ counts and in what units, confirm the size is sensible for the story (10 weeks of saving $6 on top of $45 already saved reaches $105), and state the answer in a sentence, such as "it takes 10 weeks."
76. Leo swapped the roles of the two numbers. The $5 is the amount charged for each hour, so it belongs with $h$; the $12 happens once. The correct equation is $5h + 12 = 47$; then $5h = 35$ and $h = 7$ hours. Check: $5(7) + 12 = 35 + 12 = 47$. True. Against the story: 7 hours at $5 is $35, plus the $12 fee is $47.

Exit ticket 10.5

77. Let $w$ = the width in meters. $2w + 30 = 54$; $2w = 24$; $w = 12$ m. Check: $2(12) + 30 = 24 + 30 = 54$. True.
78. Let $n$ = the number of songs. $-1.25n + 20 = 8.75$; $-1.25n = -11.25$; $n = 9$ songs. Check: $-1.25(9) + 20 = -11.25 + 20 = 8.75$. True.
79. Let $w$ = the number of weeks. $9w + 35 = 116$; $9w = 81$; $w = 9$ weeks. Check: $9(9) + 35 = 81 + 35 = 116$. True.
80. Substitution only proves the arithmetic is consistent with the equation you wrote. Units and a check against the story test whether the equation itself models the situation, which is where most real mistakes live. A bare number also does not answer the question that was asked — "12" could be weeks, dollars, or meters.

Chapter 10 Review

Part A — Modeling with concrete and pictorial representations (7.PFA.3a)

81. $3x + 4 = 19$. Remove 4 chips from each pan, leaving $3x = 15$; split each pan into 3 equal groups, so one block matches 5 chips: $x = 5$. Check: $3(5) + 4 = 15 + 4 = 19$. True.
82. Left mat: two $x$ tiles and six $1$ tiles; right mat: fourteen $1$ tiles. Remove six $1$ tiles from both mats, leaving $2x = 8$; split both mats into 2 equal groups: $x = 4$. Check: $2(4) + 6 = 8 + 6 = 14$. True.
83. Left mat: three $x$ tiles and four $-1$ tiles; right mat: eight $1$ tiles. Add four $1$ tiles to each mat; on the left, four zero pairs form and clear, leaving $3x$; on the right, $8 + 4 = 12$. Split both mats into 3 equal groups: $x = 4$. Check: $3(4) - 4 = 12 - 4 = 8$. True.
84. Left mat: two $-x$ tiles and seven $1$ tiles; right mat: one $1$ tile. Add seven $-1$ tiles to each mat. On the left, seven zero pairs clear, leaving $-2x$; on the right, one zero pair clears and six $-1$ tiles remain, so the right mat is $-6$. That gives $-2x = -6$. Split both mats into 2 equal groups: $-x = -3$, so $x = 3$. Check: $-2(3) + 7 = -6 + 7 = 1$. True.

Part B — Solving with properties of equality (7.PFA.3b)

85. a) $4x = 16$, so $x = 4$. Check: $4(4) + 9 = 16 + 9 = 25$. True. b) $6x = 24$, so $x = 4$. Check: $6(4) - 11 = 24 - 11 = 13$. True. c) $-5x = 25$, so $x = -5$. Check: $-5(-5) + 3 = 25 + 3 = 28$. True. d) $\tfrac{x}{7} = 3$, so $x = 21$. Check: $\tfrac{21}{7} + 2 = 3 + 2 = 5$. True.
86. a) $\tfrac{2}{3}x = 6$; multiply both sides by $\tfrac{3}{2}$: $x = 9$. Check: $\tfrac{2}{3}(9) - 5 = 6 - 5 = 1$. True. b) $1.5x = 9$, so $x = 6$. Check: $1.5(6) + 2.5 = 9 + 2.5 = 11.5$. True.
87. Subtract 8 from both sides: $22 = -2x$. Divide both sides by $-2$: $x = -11$. Check: $8 - 2(-11) = 8 + 22 = 30$. True.
88. $10x = 5$, so $x = \tfrac{1}{2}$. Check: $10\left(\tfrac{1}{2}\right) + 3 = 5 + 3 = 8$. True.
89. $-3x + 14 = 2$. Subtract 14 from both sides (subtraction property of equality): $-3x = -12$. Divide both sides by $-3$ (division property of equality): $x = 4$. The multiplicative identity property is what lets $\tfrac{-3x}{-3}$ be written as $x$. Check: $-3(4) + 14 = -12 + 14 = 2$. True.

Part C — Confirming algebraic solutions (7.PFA.3c)

90. $4(6) - 5 = 24 - 5 = 19$, which matches the right side. Yes, $x = 6$ is the solution.
91. $5(3) - 4 = 15 - 4 = 11$, and $11 \neq 6$, so no. Solving: $5x = 10$, so $x = 2$. Check: $5(2) - 4 = 10 - 4 = 6$. True.
92. Marcus multiplied two negatives and kept a negative result. Correctly, $-2(-3) = 6$, so the check is $-2(-3) + 5 = 6 + 5 = 11$, which matches the right side. Yes, $x = -3$ is a solution.

Part D — Writing an equation from a situation (7.PFA.3d)

93. Let $t$ = the number of tickets. $9t + 5 = 68$; $9t = 63$; $t = 7$ tickets. Check: $9(7) + 5 = 63 + 5 = 68$. True.
94. Let $n$ = the number. $6n - 11 = 31$; $6n = 42$; $n = 7$. Check: $6(7) - 11 = 42 - 11 = 31$. True.
95. $-2h + 24 = 9$; $-2h = -15$; $h = \tfrac{15}{2}$ hours, or $7\tfrac{1}{2}$ hours. Check: $-2\left(\tfrac{15}{2}\right) + 24 = -15 + 24 = 9$. True.

Part E — Creating a situation from an equation (7.PFA.3e)

96. Many answers. One: A soccer club charges $5 per practice plus a $12 uniform fee. Elena's total was $47. How many practices did she pay for? Solution: $5x = 35$, so $x = 7$ practices. Check: $5(7) + 12 = 35 + 12 = 47$. True.
97. Many answers. One: A 30-page notebook loses 4 pages each week as sheets are torn out. After $x$ weeks, 6 pages remain. How many weeks have passed? Solution: $-4x = -24$, so $x = 6$ weeks. Check: $-4(6) + 30 = -24 + 30 = 6$. True.

Part F — Solving problems in context (7.PFA.3f)

98. Let $w$ = the width in feet. Two lengths contribute $2(11) = 22$, so $2w + 22 = 38$; $2w = 16$; $w = 8$ ft. Check: $2(8) + 22 = 16 + 22 = 38$. True.
99. Let $w$ = the number of weeks. $6.50w + 18 = 63.50$; $6.50w = 45.50$; $w = 7$ weeks. Check: $6.50(7) + 18 = 45.50 + 18 = 63.50$. True.
100. Let $g$ = the number of gigabytes. $0.25g + 12 = 19.50$; $0.25g = 7.50$; $g = 30$ gigabytes. Check: $0.25(30) + 12 = 7.50 + 12 = 19.50$. True.

Part G — Mixed application and reasoning

101. Route 1: subtract 5 from both sides to get $3x = 15$, then divide both sides by 3 to get $x = 5$. Route 2: divide both entire sides by 3 first, giving $x + \tfrac{5}{3} = \tfrac{20}{3}$; subtract $\tfrac{5}{3}$ from both sides: $x = \tfrac{20}{3} - \tfrac{5}{3} = \tfrac{15}{3} = 5$. They agree because both routes apply properties of equality to the whole of each side, and any sequence of legal moves preserves the solution. Route 1 is preferred only because it avoids fractions. Check: $3(5) + 5 = 15 + 5 = 20$. True.
102. Substituting: $-2(-4) + 9 = 8 + 9 = 17$, and $17 \neq 1$, so the student is incorrect. Solving: $-2x = -8$, so $x = 4$. Check: $-2(4) + 9 = -8 + 9 = 1$. True.
103. Many answers. Negative coefficient: $-3x + 20 = 2$. Check: $-3(6) + 20 = -18 + 20 = 2$. True. Fractional coefficient: $\tfrac{1}{2}x + 4 = 7$. Check: $\tfrac{1}{2}(6) + 4 = 3 + 4 = 7$. True.
104. Plan A: $4c + 30 = 70$; $4c = 40$; $c = 10$ classes. Check: $4(10) + 30 = 40 + 30 = 70$. True. Plan B: $8c + 10 = 70$; $8c = 60$; $c = \tfrac{15}{2}$, or $7.5$ classes. Check: $8\left(\tfrac{15}{2}\right) + 10 = 60 + 10 = 70$. True. Since classes cannot be split, no whole number of Plan B classes costs exactly $70. The value $7.5$ is the exact break-even point: 7 classes cost $8(7) + 10 = \$66$ and 8 classes cost $8(8) + 10 = \$74$. So $70 buys 7 full classes with $4 left over.

Workbook-only items

Page 2, fill in the blanks. $a$ is the coefficient and $b$ is the constant term. Going forward, the variable is multiplied by $a$ and then $b$ is added. Going backward, you undo in the reverse order.

Page 2, table.

Page 7, balance. The figure shows $2x + 3 = 11$. The rule: whatever you do to one pan, you must do to the other. Panel 1: $2x + 3 = 11$. Panel 2: $2x = 8$. Panel 3: $x = 4$. Panel 1 to panel 2: remove 3 chips from each pan. Panel 2 to panel 3: split each pan into 2 equal groups.

Page 8, tiles. The top row models $3x + 2 = 11$. Removing 2 unit tiles from both mats gives $3x = 9$. Splitting each mat into 3 equal groups gives $x = 3$. A zero pair is a $+1$ tile together with a $-1$ tile, worth $0$ in total. You may add three $1$ tiles to each mat because the same amount is added to both sides, so the sides stay equal.

Page 13, properties table. Addition property: add the same number to both sides. Subtraction property: subtract the same number from both sides. Multiplication property: multiply both sides by the same number. Division property: divide both sides by the same nonzero number.

Page 13, procedure. 1. Name the coefficient and the constant term. 2. Undo the addition or subtraction first. 3. Undo the multiplication or division second. 4. Check by substituting into the original equation.

Page 13, coefficient table. $4x + 3 = 10$: divide both sides by 4, because division undoes multiplication. $-5x + 3 = 28$: divide both sides by $-5$, because the sign belongs to the coefficient. $\tfrac{x}{6} + 2 = 5$: multiply both sides by 6, because the coefficient is $\tfrac{1}{6}$. $\tfrac{2}{3}x + 4 = 10$: multiply both sides by $\tfrac{3}{2}$, the reciprocal of $\tfrac{2}{3}$. $1.5x - 2 = 7$: divide both sides by $1.5$.

Page 14, item 33 frame. $2x + 9 - \mathbf{9} = 21 - \mathbf{9}$ (subtraction property of equality); $2x = \mathbf{12}$; $\tfrac{2x}{\mathbf{2}} = \tfrac{\mathbf{12}}{\mathbf{2}}$ (division property of equality); $x = \mathbf{6}$.

Page 18, labeling $ax + b = c$. $x$ is the unknown number of units; $a$ is the amount per unit; $b$ is the fixed one-time amount; $c$ is the total.

Page 22, six-step finish. All six boxes should be checked for every context problem in Lesson 10.5.