Appendix A — Answer Key, Chapter 5: Multistep Linear Equations

SOL 8.PFA.4 · Covers textbook Chapter 5 and the companion workbook. Item numbers match the textbook; workbook items are the same problems, so this key serves both. Item numbers run continuously from 1 to 124 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Every solution below was substituted back into the original equation, and the check is shown wherever a bare answer would not teach. Modeling convention: a long rectangle is an $x$ tile, a small square is a unit tile, a circle is a chip; blue pieces are positive and red pieces are negative.

Lesson 5.1 — Simplify Each Side First

Workbook table (page 2). $5x + 2x + (-3)$, so $7x - 3$. · $4x + 9 + (-x)$, so $3x + 9$. · $8n + (-3n) + 5$, so $5n + 5$. · $10 + (-4k) + k$, so $-3k + 10$. The blank in $5x + 2x - 3 = 18$ is $7$.

Guided practice

1. $7x - 5 = 16$, so $7x = 21$ (addition property of equality) and $x = 3$ (division property of equality). Check: $3(3) + 4(3) - 5 = 9 + 12 - 5 = 16$. True.
2. $4a + 7 = 27$, so $4a = 20$ and $a = 5$. Check: $6(5) - 2(5) + 7 = 30 - 10 + 7 = 27$. True.
3. $23 = 5m + 3$, so $20 = 5m$ and $m = 4$. Check: $9(4) + 3 - 4(4) = 36 + 3 - 16 = 23$. True.
4. $2x + 3 = 11$, so $2x = 8$ and $x = 4$. Check: $1.2(4) + 0.8(4) + 3 = 4.8 + 3.2 + 3 = 11$. True.

Independent practice

5. a) $7x + 4 = 25$, $x = 3$; check $2(3) + 5(3) + 4 = 6 + 15 + 4 = 25$. b) $5y - 6 = 14$, $y = 4$; check $8(4) - 3(4) - 6 = 32 - 12 - 6 = 14$. c) $-3k + 10 = 1$, so $-3k = -9$ and $k = 3$; check $10 - 4(3) + 3 = 10 - 12 + 3 = 1$. d) $7 = 7n - 21$, so $28 = 7n$ and $n = 4$; check $2(4) + 5(4) - 21 = 8 + 20 - 21 = 7$.
6. a) $\tfrac{1}{3} + \tfrac{2}{3} = 1$, so $x + 5 = 12$ and $x = 7$; check $\tfrac{1}{3}(7) + \tfrac{2}{3}(7) + 5 = 7 + 5 = 12$. b) $0.25 + 0.75 = 1$, so $p - 2.5 = 4.5$ and $p = 7$; check $0.25(7) + 0.75(7) - 2.5 = 7 - 2.5 = 4.5$.
7. $5x - 11 = 19$, so $5x = 30$ and $x = 6$. Check: $-3(6) + 8(6) - 11 = -18 + 48 - 11 = 19$. True.
8. $14 = 3w + 6$, so $8 = 3w$ and $w = \tfrac{8}{3}$. Check: $6 - 2\left(\tfrac{8}{3}\right) + 5\left(\tfrac{8}{3}\right) = 6 + 3\left(\tfrac{8}{3}\right) = 6 + 8 = 14$. True. Leave the answer as the exact fraction $\tfrac{8}{3}$; rounding to $2.67$ makes the check fail.
9. Answers vary. One equation is $3x + 2x - 5 = 15$: it simplifies to $5x - 5 = 15$, so $5x = 20$ and $x = 4$. Check: $3(4) + 2(4) - 5 = 12 + 8 - 5 = 15$. True.
10. Let $x$ be the shortest side in inches. $x + 2x + (3x + 4) = 34$, which simplifies to $6x + 4 = 34$, so $6x = 30$ and $x = 5$. The sides are 5 in, 10 in, and 19 in. Check: $5 + 10 + 19 = 34$ in. True.
11. Combining like terms rewrites one side without changing what it is worth — $5x + 2x$ and $7x$ name the same number for every value of $x$, by the distributive property read backwards. Since neither side's value moved, the equation is still true and no balancing move is needed. A property of equality is required only when you change a side's value, and then the other side must change identically.
12. The student subtracted the $2$ from the coefficient instead of leaving it as a separate constant term. Only $4x$ and $3x$ are like terms, so the left side is $7x - 2$, not $5x$. Correctly: $7x - 2 = 19$, so $7x = 21$ and $x = 3$. Check: $4(3) + 3(3) - 2 = 12 + 9 - 2 = 19$. True.

Exit ticket 5.1

13. $8x - 7 = 17$, so $8x = 24$ and $x = 3$. Check: $5(3) + 3(3) - 7 = 15 + 9 - 7 = 17$. True.
14. $9 = 3t + 3$, so $6 = 3t$ and $t = 2$. Check: $4(2) - 2 + 3 = 8 - 2 + 3 = 9$. True.
15. $3m - 1 = 11$, so $3m = 12$ and $m = 4$. Check: $0.5(4) + 2.5(4) - 1 = 2 + 10 - 1 = 11$. True.
16. Combining like terms changes only how a side is written, not what it is worth, so the set of values making the sentence true is exactly the same before and after. Nothing was added to or removed from either side.

Lesson 5.2 — Modeling Equations with the Variable on Both Sides

Workbook fill-ins (pages 6–7). Three blocks and two chips on the left; one block and eight chips on the right; the equation is $3x + 2 = x + 8$. The rule: do the same thing to the other pan. Panel 1 → 2: take one $x$ block off each pan. Panel 2 → 3: take 2 chips off each pan. Panel 3 → 4: split each pan into 2 equal groups. Blocks come off first because that is what clears the variable from one side and turns the problem into a two-step equation. Sliding the $x$ tiles together is combining like terms. A zero pair is worth $0$.

Guided practice

17. $4x + 3 = x + 12$
18. Remove one $x$ block from each pan: $3x + 3 = 12$. Remove 3 chips from each pan: $3x = 9$. Split each pan into 3 equal groups: $x = 3$. Rebuilt check: the left pan holds four blocks and 3 chips, worth $4(3) + 3 = 15$, and the right pan holds one block and 12 chips, worth $3 + 12 = 15$. Both pans are worth 15, so the scale is level and $x = 3$ is confirmed.
19. Left mat: two $x$ tiles and six $1$ tiles. Right mat: four $x$ tiles. Remove two $x$ tiles from each mat, leaving $6 = 2x$. Split both mats into 2 equal groups: $x = 3$. Check: $2(3) + 6 = 12$ and $4(3) = 12$. True.
20. Left mat: three $x$ tiles and four $-1$ tiles. Right mat: one $x$ tile and two $1$ tiles. Add four $1$ tiles to each mat; four zero pairs form on the left and come off, leaving $3x = x + 6$. Remove one $x$ tile from each mat: $2x = 6$. Two equal groups: $x = 3$. Model check: left $3(3) - 4 = 5$, right $3 + 2 = 5$. Level.

Independent practice

21. a) $5x + 2 = 3x + 10$. Remove three $x$ blocks from each pan: $2x + 2 = 10$. Remove 2 chips from each: $2x = 8$. Split into 2 groups: $x = 4$. Check: $5(4) + 2 = 22$ and $3(4) + 10 = 22$. b) $2x + 3x + 1 = 16$. Slide the $x$ tiles together: $5x + 1 = 16$. Remove one unit tile from each mat: $5x = 15$. Five equal groups: $x = 3$. Check: $2(3) + 3(3) + 1 = 6 + 9 + 1 = 16$.
22. Remove two $x$ tiles from each mat, leaving $4x = 12$. Split both mats into 4 equal groups: $x = 3$. Check: $6(3) = 18$ and $2(3) + 12 = 18$. True.
23. Left mat: four $x$ tiles and three $-1$ tiles. Right mat: two $x$ tiles and five $1$ tiles. Add three $1$ tiles to each mat; three zero pairs form on the left and come off, leaving $4x = 2x + 8$. Remove two $x$ tiles from each mat: $2x = 8$. Two equal groups: $x = 4$. Model check: left $4(4) - 3 = 13$, right $2(4) + 5 = 13$. Level.
24. Left mat: twelve $1$ tiles and one $-x$ tile. Add one $x$ tile to each mat, forming a zero pair on the left: $12 = 3x + 3$. Remove three unit tiles from each mat: $9 = 3x$. Three equal groups: $x = 3$. Check: $12 - 3 = 9$ and $2(3) + 3 = 9$. True.
25. Remove four unit tiles from each mat: $5x = 3x$. Remove three $x$ tiles from each mat: $2x = 0$. Two equal groups: $x = 0$. Zero is a perfectly good solution — it is a number, and it makes the sentence true: $5(0) + 4 = 4$ and $3(0) + 4 = 4$. "No solution" would mean no number works, which is a different situation.
26. A move must be made on both mats or the two sides stop being equal. Removing one $x$ tile from the left only makes the left smaller than the right, so the new line is false. Removing one $x$ tile from each mat gives $2x + 5 = 11$, then $2x = 6$ and $x = 3$. Check: $3(3) + 5 = 14$ and $3 + 11 = 14$. True.
27. Let $b$ be the mass of one box in pounds: $2b + 5 = b + 9$. Remove one box from each pan: $b + 5 = 9$. Remove 5 pounds from each pan: $b = 4$. Scale check: left holds $4 + 4 + 5 = 13$ lb, right holds $4 + 9 = 13$ lb. Level, so one box weighs 4 pounds.
28. You do not need to know what one block weighs; you only need to know that all the blocks are identical. Taking one block off each pan removes the same mass from both sides, so the beam stays level whatever that mass is. In symbols, this is the subtraction property of equality applied to the quantity $x$.

Exit ticket 5.2

29. $3x + 7 = x + 15$. Remove one $x$ block from each pan: $2x + 7 = 15$. Remove 7 chips: $2x = 8$. Split into 2 groups: $x = 4$. Check: $3(4) + 7 = 19$ and $4 + 15 = 19$. True.
30. Left mat: four $x$ tiles and one $-1$ tile. Add one $1$ tile to each mat, forming a zero pair on the left: $4x = 2x + 8$. Remove two $x$ tiles from each mat: $2x = 8$. Two equal groups: $x = 4$. Check: $4(4) - 1 = 15$ and $2(4) + 7 = 15$. True.
31. Replace every $x$ piece with 2 unit pieces. Left: $2 + 2 + 2 + 2 + 2 + 1 = 11$. Right: $2 + 2 + 2 + 5 = 11$. The counts match, so the beam is level and $x = 2$ is the solution.
32. Because the mats stand for the two sides of an equation, and the equation is only true while the two sides are worth the same. A move that changes one mat and not the other produces a false statement, and every line after it is untrustworthy.

Lesson 5.3 — Expanding with the Distributive Property

Workbook fill-ins (page 11). Route 1: $3x + 12 = 21$, $x = 3$. Route 2: $x + 4 = 7$, $x = 3$. Route 2 fails when the outside factor is not a factor of the whole side — for instance when a term sits outside the parentheses, as in $5(x + 2) - 3x$. Sign table: $-4x + 12$; $-x + 7$; $3x + 4$; $6x - 4$; $-6x + 10$.

Guided practice

33. Dividing first: $x + 3 = 7$, so $x = 4$. Expanding first: $4x + 12 = 28$, so $4x = 16$ and $x = 4$. Check: $4(4 + 3) = 4(7) = 28$. True.
34. $6x - 3 = 21$, so $6x = 24$ and $x = 4$. Check: $3(2 \cdot 4 - 1) = 3(7) = 21$. True.
35. $2x + 10 + 3x = 30$, so $5x + 10 = 30$, $5x = 20$, and $x = 4$. Check: $2(4 + 5) + 3(4) = 18 + 12 = 30$. True.
36. $-3x + 6 = 18$, so $-3x = 12$ and $x = -4$. Check: $-3(-4 - 2) = -3(-6) = 18$. True.

Independent practice

37. a) $5x - 10 = 25$, so $x = 7$; check $5(7 - 2) = 5(5) = 25$. b) $12x + 6 = 42$, so $12x = 36$ and $x = 3$; check $6(2 \cdot 3 + 1) = 6(7) = 42$. c) $-2x - 14 = 4$, so $-2x = 18$ and $x = -9$; check $-2(-9 + 7) = -2(-2) = 4$. d) $6x - 4 = 20$, so $6x = 24$ and $x = 4$; check $\tfrac{2}{3}(9 \cdot 4 - 6) = \tfrac{2}{3}(30) = 20$.
38. a) $4x + 4 + 2x = 34$, so $6x + 4 = 34$, $6x = 30$, and $x = 5$; check $4(5 + 1) + 2(5) = 24 + 10 = 34$. b) $3x - 12 - x = 6$, so $2x - 12 = 6$, $2x = 18$, and $x = 9$; check $3(9 - 4) - 9 = 15 - 9 = 6$.
39. $2x + 6 + 3x - 3 = 23$, so $5x + 3 = 23$, $5x = 20$, and $x = 4$. Check: $2(4 + 3) + 3(4 - 1) = 14 + 9 = 23$. True.
40. $2x + 3 = 11$, so $2x = 8$ and $x = 4$. Check: $0.5(4 \cdot 4 + 6) = 0.5(22) = 11$. True.
41. Dividing first: $x + 4 = 7$, so $x = 3$. Expanding first: $3x + 12 = 21$, so $3x = 9$ and $x = 3$. They must agree because the distributive property guarantees $3(x + 4)$ and $3x + 12$ are the same number for every $x$, so the two routes are operating on the same equation. Different legal paths through an equation cannot reach different solutions.
42. Answers vary. One equation is $5(x + 6) = 20$: dividing by 5 gives $x + 6 = 4$, so $x = -2$. Check: $5(-2 + 6) = 5(4) = 20$. True.
43. Let $x$ be the number of stickers in one bag: $4(x + 3) = 40$. Dividing by 4 gives $x + 3 = 10$, so $x = 7$. Check: $4(7 + 3) = 40$. True. Each bag holds 7 stickers, and since $4 \times 3 = 12$ of the 40 items are pencils, the remaining 28 are stickers, 7 per bag.
44. The factor $-2$ must multiply the $-5$ as well, and $(-2)(-5) = +10$, not $-10$. Correctly: $-2x + 10 = 8$, so $-2x = -2$ and $x = 1$. Check: $-2(1 - 5) = -2(-4) = 8$. True. The student's answer fails the check: $-2(-9 - 5) = -2(-14) = 28 \neq 8$.

Exit ticket 5.3

45. $6x - 12 = 18$, so $6x = 30$ and $x = 5$. Check: $6(5 - 2) = 6(3) = 18$. True.
46. $6x + 8 - 4x = 20$, so $2x + 8 = 20$, $2x = 12$, and $x = 6$. Check: $2(3 \cdot 6 + 4) - 4(6) = 44 - 24 = 20$. True.
47. $-5x - 5 = 15$, so $-5x = 20$ and $x = -4$. Check: $-5(-4 + 1) = -5(-3) = 15$. True.
48. Because the parentheses say the whole sum was multiplied. $3(x + 4)$ means three copies of the quantity $x + 4$, which is three $x$'s and three $4$'s. Multiplying only the first term would leave out the other copies and change the value.

Lesson 5.4 — Solving with the Variable on Both Sides

Workbook fill-ins (page 15). The four steps: expand, combine, gather, divide. Move the smaller variable term. Property table: subtraction property of equality, subtraction property of equality, division property of equality.

Guided practice

49. Subtract $4x$ from both sides (subtraction property of equality): $2x + 1 = 9$. Subtract 1 from both sides (subtraction property of equality): $2x = 8$. Divide both sides by 2 (division property of equality): $x = 4$. Check: $6(4) + 1 = 25$ and $4(4) + 9 = 25$. True.
50. Subtract $3x$: $5x - 5 = 20$. Add 5: $5x = 25$. Divide by 5: $x = 5$. Check: $8(5) - 5 = 35$ and $3(5) + 20 = 35$. True.
51. Expand: $3x - 6 = x + 4$. Subtract $x$: $2x - 6 = 4$. Add 6: $2x = 10$. Divide by 2: $x = 5$. Check: $3(5 - 2) = 9$ and $5 + 4 = 9$. True.
52. Add $2x$ to both sides: $9 = 6x - 15$. Add 15: $24 = 6x$. Divide by 6: $x = 4$. Check: $9 - 2(4) = 1$ and $4(4) - 15 = 1$. True.

Independent practice

53. a) Subtract $5x$: $2x + 2 = 14$, so $2x = 12$ and $x = 6$; check $7(6) + 2 = 44$ and $5(6) + 14 = 44$. b) Subtract $x$: $3x - 9 = 3$, so $3x = 12$ and $x = 4$; check $4(4) - 9 = 7$ and $4 + 3 = 7$. c) Add $3x$: $10 = 2 + 2x$, so $8 = 2x$ and $x = 4$; check $10 - 3(4) = -2$ and $2 - 4 = -2$. d) Subtract $2x$: $7 = 3x - 8$, so $15 = 3x$ and $x = 5$; check $2(5) + 7 = 17$ and $5(5) - 8 = 17$.
54. a) Expand: $5x - 5 = 3x + 7$. Subtract $3x$: $2x - 5 = 7$, so $2x = 12$ and $x = 6$; check $5(6 - 1) = 25$ and $3(6) + 7 = 25$. b) Expand both sides: $4x + 6 = 3x + 15$. Subtract $3x$: $x + 6 = 15$, so $x = 9$; check $2(2 \cdot 9 + 3) = 2(21) = 42$ and $3(9 + 5) = 3(14) = 42$.
55. Combine on the left: $4x + 4 = x + 19$. Subtract $x$: $3x + 4 = 19$, so $3x = 15$ and $x = 5$. Check: $6(5) + 4 - 2(5) = 30 + 4 - 10 = 24$ and $5 + 19 = 24$. True.
56. Subtract $0.2x$: $0.2x + 1.2 = 2$, so $0.2x = 0.8$ and $x = 4$. Check: $0.4(4) + 1.2 = 2.8$ and $0.2(4) + 2 = 2.8$. True.
57. Subtract $\tfrac{1}{3}x$: $\tfrac{1}{3}x + 1 = 4$, so $\tfrac{1}{3}x = 3$ and $x = 9$. Check: $\tfrac{2}{3}(9) + 1 = 7$ and $\tfrac{1}{3}(9) + 4 = 7$. True. Multiplying both sides by 3 at the start gives $2x + 3 = x + 12$ and the same answer.
58. Subtracting $4x$: $6 = 2x - 2$, so $8 = 2x$ and $x = 4$. Subtracting $6x$: $-2x + 6 = -2$, so $-2x = -8$ and $x = 4$. Both are correct. Moving the smaller variable term leaves a positive coefficient, so the final division is by a positive number and there is one less negative sign to lose.
59. Let $c$ be the number of classes. $4c + 30 = 6c$, so $30 = 2c$ and $c = 15$. Check: $4(15) + 30 = 90$ and $6(15) = 90$. True. The studios cost the same, $90, at 15 classes.
60. A property of equality must be applied to both sides. Subtracting $2x$ from the left only makes the two sides unequal, so the new line is false. Subtracting $2x$ from both sides gives $3x + 2 = 14$, so $3x = 12$ and $x = 4$. Check: $5(4) + 2 = 22$ and $2(4) + 14 = 22$. True.

Exit ticket 5.4

61. Subtract $6x$: $3x - 4 = 11$, so $3x = 15$ and $x = 5$. Check: $9(5) - 4 = 41$ and $6(5) + 11 = 41$. True.
62. Expand: $2x + 10 = 4x - 2$. Subtract $2x$: $10 = 2x - 2$, so $12 = 2x$ and $x = 6$. Check: $2(6 + 5) = 22$ and $4(6) - 2 = 22$. True.
63. Add $x$: $12 = 4x + 4$, so $8 = 4x$ and $x = 2$. Check: $12 - 2 = 10$ and $3(2) + 4 = 10$. True.
64. (1) Expand parentheses, so every term stands on its own. (2) Combine like terms within each side, so each side is as short as possible. (3) Gather the variable terms on one side and the constants on the other, using the addition and subtraction properties of equality. (4) Divide by the coefficient, using the division property of equality, to leave the variable alone.

Lesson 5.5 — Writing Equations and Writing Situations

Workbook fill-ins (page 19). A $30 fee plus $12 per month is $12m + 30$; $18 per month with no fee is $18m$; "the two totals are equal" is the equal sign, giving $12m + 30 = 18m$. "Five less than three times a number" is $3n - 5$.

Guided practice

65. Let $h$ be the number of hours. $3h + 8 = 5h$, so $8 = 2h$ and $h = 4$. Check: $3(4) + 8 = 20$ and $5(4) = 20$. True. The rinks cost the same, $20, at 4 hours.
66. $2n + 4 = n + 10$, so $n + 4 = 10$ and $n = 6$. Check: $2(6) + 4 = 16$ and $6 + 10 = 16$. True.
67. $x + (x + 2) + (2x + 1) = 27$, which simplifies to $4x + 3 = 27$, so $4x = 24$ and $x = 6$. The sides are 6 in, 8 in, and 13 in. Check: $6 + 8 + 13 = 27$ in. True.
68. Answers vary. A club charges a $15 registration fee plus $5 per event; a rival club charges $8 per event with no fee. For how many events do the two cost the same? Solving: $15 = 3x$, so $x = 5$ events. Check: $5(5) + 15 = 40$ and $8(5) = 40$. True.

Independent practice

69. $5n - 7 = 3n + 5$, so $2n - 7 = 5$, $2n = 12$, and $n = 6$. Check: $5(6) - 7 = 23$ and $3(6) + 5 = 23$. True.
70. Let $g$ be the number of gigabytes. $2g + 40 = 6g$, so $40 = 4g$ and $g = 10$. Check: $2(10) + 40 = 60$ and $6(10) = 60$. True. The plans cost the same, $60, at 10 gigabytes.
71. Let $c$ be the mass of one crate in kilograms. $3c + 12 = 5c + 2$, so $12 = 2c + 2$, $10 = 2c$, and $c = 5$. Check: $3(5) + 12 = 27$ and $5(5) + 2 = 27$. True. One crate has a mass of 5 kg.
72. Let $w$ be the width in centimeters, so the length is $w + 4$. $2(w + 4) + 2w = 36$, which expands to $2w + 8 + 2w = 36$ and simplifies to $4w + 8 = 36$, so $4w = 28$ and $w = 7$. The width is 7 cm and the length is 11 cm. Check: $2(11) + 2(7) = 22 + 14 = 36$ cm. True.
73. Answers vary. Printer A charges $6 per poster plus a $20 design fee. Printer B charges $10 per poster with no fee. For how many posters are the bills equal? Solving: $20 = 4x$, so $x = 5$ posters. Check: $6(5) + 20 = 50$ and $10(5) = 50$. True.
74. Answers vary. Four identical lunch boxes each hold $x$ crackers and 3 grapes, and the boxes hold 32 items in all. How many crackers are in one box? Solving: $x + 3 = 8$, so $x = 5$ crackers. Check: $4(5 + 3) = 32$. True.
75. Answers vary. A book costs $x$ dollars. Sam buys 2 copies, Ana buys 4 copies, and together with a $5 gift wrap charge the order comes to $41. What does one book cost? Solving: $6x + 5 = 41$, so $6x = 36$ and $x = 6$ dollars. Check: $2(6) + 4(6) + 5 = 12 + 24 + 5 = 41$. True.
76. The second store charges nothing for shipping, so there is no $20 on its side; writing $+20$ on both sides describes a situation where both stores charge shipping, and it makes the shipping charge cancel out. The correct equation is $12s + 20 = 16s$, so $20 = 4s$ and $s = 5$ shirts. Check: $12(5) + 20 = 80$ and $16(5) = 80$. True.

Exit ticket 5.5

77. Let $v$ be the number of visits. $4v + 18 = 6v$, so $18 = 2v$ and $v = 9$. Check: $4(9) + 18 = 54$ and $6(9) = 54$. True. The clubs cost the same, $54, at 9 visits.
78. $4n + 3 = 2n + 15$, so $2n + 3 = 15$, $2n = 12$, and $n = 6$. Check: $4(6) + 3 = 27$ and $2(6) + 15 = 27$. True.
79. Answers vary. Five identical party bags each hold $x$ stickers and 2 pencils, and the bags hold 40 items in all. How many stickers are in one bag? Solving: $x + 2 = 8$, so $x = 6$ stickers. Check: $5(6 + 2) = 40$. True.
80. Each side of the equation is one complete quantity being compared. Build the total for one option on the left and the total for the other option on the right; the equal sign is the claim that the two totals match. If only one quantity is being described, the known total goes on the other side by itself.

Lesson 5.6 — Problems in Context: Solving, Confirming, Interpreting

Workbook fill-ins (page 23). Confirming answers: does this number make the original equation true? Interpreting answers: what does the number mean in the situation, and can it be used? The lines cross at $c = 22.5$, where each total is $112.50. For 22 classes or fewer, Plan B is cheaper; for 23 or more, Plan A is cheaper. Trimming the four equal parts leaves $60 = 6s$, so $s = 10$.

Guided practice

81. Let $m$ be the number of months. $10m + 50 = 20m$, so $50 = 10m$ and $m = 5$. Check: $10(5) + 50 = 100$ and $20(5) = 100$. True. Interpretation: the two gyms cost the same, $100, after 5 months; after that, Gym A is cheaper because its monthly rate is lower.
82. Let $w$ be the width in centimeters, so the length is $3w$. $2(3w) + 2w = 48$, which simplifies to $8w = 48$, so $w = 6$. The width is 6 cm and the length is 18 cm. Perimeter check: $2(18) + 2(6) = 36 + 12 = 48$ cm. True.
83. Let $t$ be the number of minutes. $80 - 6t = 20 + 4t$, so $80 = 20 + 10t$, $60 = 10t$, and $t = 6$. Check: $80 - 6(6) = 44$ and $20 + 4(6) = 44$. True. After 6 minutes both tanks hold 44 liters.
84. Let $t$ be the price of one ticket in dollars. $5t + 4 = 46.50$, so $5t = 42.50$ and $t = 8.50$. Check: $5(8.50) + 4 = 42.50 + 4 = 46.50$. True. Yes, the answer is usable: money comes in halves of a dollar, so $8.50 is an ordinary ticket price.

Independent practice

85. Let $s$ be the number of shirts. $4s + 60 = 10s$, so $60 = 6s$ and $s = 10$. Check: $4(10) + 60 = 100$ and $10(10) = 100$. True. The bills are equal at 10 shirts, and each total is $100.
86. Let $w$ be the width in centimeters, so the length is $2w + 5$. $2(2w + 5) + 2w = 64$, which expands to $4w + 10 + 2w = 64$ and simplifies to $6w + 10 = 64$, so $6w = 54$ and $w = 9$. The width is 9 cm and the length is 23 cm. Check: $2(23) + 2(9) = 46 + 18 = 64$ cm. True.
87. Let $w$ be the number of weeks. $240 - 15w = 120 + 15w$, so $240 = 120 + 30w$, $120 = 30w$, and $w = 4$. Check: $240 - 15(4) = 180$ and $120 + 15(4) = 180$. True. After 4 weeks each has $180.
88. Let $p$ be one monthly payment in dollars. $3p + 25 = 187$, so $3p = 162$ and $p = 54$. Check: $3(54) + 25 = 162 + 25 = 187$. True. Each payment is $54.
89. Let $x$ be one friend's share in dollars. $6x - 9 = 51$, so $6x = 60$ and $x = 10$. Check: $6(10) - 9 = 51$. True. Interpretation: each friend's share of the bill before the coupon was $10, so the bill was $60 and the coupon brought it down to $51. The $10 is a share of the original bill, not the amount each friend actually handed over.
90. Let $m$ be the number of miles. $1.25m + 3.50 = 0.75m + 8.00$, so $0.5m + 3.50 = 8.00$, $0.5m = 4.50$, and $m = 9$. Check: $1.25(9) + 3.50 = 11.25 + 3.50 = 14.75$ and $0.75(9) + 8.00 = 6.75 + 8.00 = 14.75$. True. At 9 miles both services charge $14.75.
91. $6v = 4v + 15$, so $2v = 15$ and $v = 7.5$. Check: $6(7.5) = 45$ and $4(7.5) + 15 = 45$. True, and both sides equal 45 students. Interpretation: the exact solution says the two arrangements carry the same number of students at 7.5 vans, which no one can actually run — you cannot send half a van. It locates the break-even point for 45 students. If 45 students must travel in vans holding 6, you need $45 \div 6 = 7.5$, so 8 vans, with the last one not full.
92. The solution $c = 10$ is the point where the two plans cost the same, $90 each, not a point where one plan wins. Below 10 classes Plan B is cheaper (at 5 classes, $45 against $65), and above 10 classes Plan A is cheaper (at 20 classes, $140 against $180). Correct conclusion: the plans cost the same at 10 classes; Plan A is cheaper only for more than 10 classes.

Exit ticket 5.6

93. Let $v$ be the number of visits. $3v + 24 = 7v$, so $24 = 4v$ and $v = 6$. Check: $3(6) + 24 = 42$ and $7(6) = 42$. True. Interpretation: the clubs cost the same, $42, at 6 visits; Club A is cheaper for more than 6 visits.
94. Let $w$ be the width in meters, so the length is $w + 2$. $2(w + 2) + 2w = 28$, which simplifies to $4w + 4 = 28$, so $4w = 24$ and $w = 6$. The width is 6 m and the length is 8 m. Check: $2(8) + 2(6) = 16 + 12 = 28$ m. True.
95. Left side: $4(7) - 5 = 28 - 5 = 23$. Right side: $2(7) + 9 = 14 + 9 = 23$. The sides match, so $x = 7$ is the solution.
96. Confirming is arithmetic: substitute the value into the original equation, simplify each side separately, and see whether the sides match. Interpreting is meaning: say what the number counts or measures, check that its size and type make sense in the situation, and state the result in a sentence with units. A solution can confirm perfectly and still be useless in context — half a bus, or a negative number of hours.

Chapter 5 Review

Part A — Modeling with concrete and pictorial representations (8.PFA.4a)

97. $5x + 2 = 2x + 11$. Remove two $x$ blocks from each pan: $3x + 2 = 11$. Remove 2 chips from each pan: $3x = 9$. Split each pan into 3 equal groups: $x = 3$. Scale check: left holds $3 + 3 + 3 + 3 + 3 + 2 = 17$, right holds $3 + 3 + 11 = 17$. Level.
98. Left mat: three $x$ tiles and four $1$ tiles. Right mat: one $x$ tile and ten $1$ tiles. Remove one $x$ tile from each mat: $2x + 4 = 10$. Remove four unit tiles from each mat: $2x = 6$. Split both mats into 2 equal groups: $x = 3$. Check: $3(3) + 4 = 13$ and $3 + 10 = 13$. True.
99. Left mat: four $x$ tiles and three $-1$ tiles. Right mat: two $x$ tiles and one $1$ tile. Add three $1$ tiles to each mat; three zero pairs form on the left and come off, leaving $4x = 2x + 4$. Remove two $x$ tiles from each mat: $2x = 4$. Two equal groups: $x = 2$. Check: $4(2) - 3 = 5$ and $2(2) + 1 = 5$. True.
100. Build two identical groups on the left mat, each holding one $x$ tile and three $1$ tiles, and ten $1$ tiles on the right mat. Because the two groups are identical, split both mats into 2 equal parts: one group, $x + 3$, matches 5 unit tiles. Remove three unit tiles from each side: $x = 2$. Check: $2(2 + 3) = 2(5) = 10$. True.

Part B — Solving with properties of equality (8.PFA.4b)

101. a) $7x - 8 = 20$, so $7x = 28$ and $x = 4$; check $4(4) + 3(4) - 8 = 16 + 12 - 8 = 20$. b) $5x - 15 = 20$, so $5x = 35$ and $x = 7$; check $5(7 - 3) = 5(4) = 20$. c) Subtract $2x$: $4x + 5 = 21$, so $4x = 16$ and $x = 4$; check $6(4) + 5 = 29$ and $2(4) + 21 = 29$. d) Expand: $3x + 3 = x + 11$. Subtract $x$: $2x + 3 = 11$, so $2x = 8$ and $x = 4$; check $3(4 + 1) = 15$ and $4 + 11 = 15$.
102. a) $0.6 + 0.4 = 1$, so $x - 2 = 5$ and $x = 7$; check $0.6(7) + 0.4(7) - 2 = 7 - 2 = 5$. b) $\tfrac{3}{4} + \tfrac{1}{4} = 1$, so $x + 6 = 15$ and $x = 9$; check $\tfrac{3}{4}(9) + \tfrac{1}{4}(9) + 6 = 9 + 6 = 15$.
103. Expand: $6x - 8 = 4x + 6$. Subtract $4x$: $2x - 8 = 6$, so $2x = 14$ and $x = 7$. Check: $2(3 \cdot 7 - 4) = 2(17) = 34$ and $4(7) + 6 = 34$. True.
104. Add $4x$: $9 = 6x - 15$. Add 15: $24 = 6x$, so $x = 4$. Check: $9 - 4(4) = -7$ and $2(4) - 15 = -7$. True.
105. Subtract $3x$: $2x + 2 = 9$. Subtract 2: $2x = 7$, so $x = \tfrac{7}{2}$. Check: $5\left(\tfrac{7}{2}\right) + 2 = \tfrac{35}{2} + \tfrac{4}{2} = \tfrac{39}{2}$ and $3\left(\tfrac{7}{2}\right) + 9 = \tfrac{21}{2} + \tfrac{18}{2} = \tfrac{39}{2}$. True. Keep the exact value $\tfrac{7}{2}$, or $3.5$.
106. $4(x - 2) + 3x = 27$ — given. $4x - 8 + 3x = 27$ — distributive property. $7x - 8 = 27$ — combining like terms, which is the distributive property used in reverse. $7x = 35$ — addition property of equality. $x = 5$ — division property of equality, together with the multiplicative identity property, since $\tfrac{7x}{7} = 1 \cdot x = x$. Check: $4(5 - 2) + 3(5) = 12 + 15 = 27$. True.

Part C — Writing an equation from a situation (8.PFA.4c)

107. Let $h$ be the number of hours. $4h + 16 = 8h$, so $16 = 4h$ and $h = 4$. Check: $4(4) + 16 = 32$ and $8(4) = 32$. True. The shops cost the same, $32, at 4 hours.
108. $6n - 8 = 4n + 2$, so $2n - 8 = 2$, $2n = 10$, and $n = 5$. Check: $6(5) - 8 = 22$ and $4(5) + 2 = 22$. True.
109. Let $s$ be the length of one leg in inches, so the base is $s - 3$. $2s + (s - 3) = 27$, which simplifies to $3s - 3 = 27$, so $3s = 30$ and $s = 10$. The sides are 10 in, 10 in, and 7 in. Check: $10 + 10 + 7 = 27$ in. True.

Part D — Creating a situation from an equation (8.PFA.4d)

110. Answers vary. Camp A charges $3 per day plus a $12 registration fee. Camp B charges $5 per day with no fee. For how many days do the two cost the same? Solving: $12 = 2x$, so $x = 6$ days. Check: $3(6) + 12 = 30$ and $5(6) = 30$. True.
111. Answers vary. Six identical crates each hold $x$ apples and 2 pears, and the crates hold 42 pieces of fruit in all. How many apples are in one crate? Solving: $x + 2 = 7$, so $x = 5$ apples. Check: $6(5 + 2) = 42$. True.
112. Answers vary. Tickets cost $x$ dollars each. A family buys 2 tickets, their neighbors buy 4, and with a $5 booking fee the order comes to $41. What does one ticket cost? Solving: $6x + 5 = 41$, so $6x = 36$ and $x = 6$ dollars. Check: $2(6) + 4(6) + 5 = 41$. True.