Appendix A — Answer Key, Chapter 6: Multistep Linear Inequalities

SOL 8.PFA.5 · Covers textbook Chapter 6 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 134 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Graphing convention used throughout: an open circle marks a boundary excluded by a strict symbol ($<$, $>$), a closed circle marks a boundary included by an inclusive symbol ($\le$, $\ge$), and the shaded ray runs right for $>$ or $\ge$ and left for $<$ or $\le$. Every solution below has been checked with one value inside the solution set and one outside.

Lesson 6.1 — What Carries Over from One- and Two-Step Inequalities

Guided practice

1. Yes. Left: $3(6) - 4 = 14$. Right: $2(6) + 3 = 15$. The sentence $14 \le 15$ is true.
2. $-1$, $0$, and $2$. Solving: $4x - 3 < 2x + 5$, so $2x - 3 < 5$, so $2x < 8$, so $x < 4$. Check the two nearest values in the original: $x = 2$ gives $5 < 9$, true; $x = 5$ gives $17 < 15$, false.
3. $x \le 3$. Distributive property: $3x + 6 - 4 \le 11$. Combining like terms: $3x + 2 \le 11$. Subtraction property of inequality: $3x \le 9$. Division property of inequality (positive divisor, no reversal): $x \le 3$. Check: $3(5) - 4 = 11 \le 11$ is true; $x = 4$ gives $14 \le 11$, false.
4. $x > 4$. Combining like terms gives $3x + 7 > 19$, then $3x > 12$, then $x > 4$. Inside, $x = 5$: $25 - 10 + 7 = 22 > 19$, true. Outside, $x = 4$: $20 - 8 + 7 = 19 > 19$, false.

Independent practice

5. a) $7x - 6 \le 15$, so $7x \le 21$, so $x \le 3$. Check $x = 3$: $21 - 6 = 15 \le 15$, true; $x = 4$: $22 \le 15$, false. b) $4x - 4 > 12$, so $4x > 16$, so $x > 4$. Check $x = 5$: $16 > 12$, true; $x = 4$: $12 > 12$, false. c) $\tfrac{3}{4} + \tfrac{1}{4} = 1$, so $x - 2 \ge 6$ and $x \ge 8$. Check $x = 8$: $6 + 2 - 2 = 6 \ge 6$, true; $x = 4$: $2 \ge 6$, false. d) $14 - 3x \ge -1$, so $-3x \ge -15$, and dividing by $-3$ reverses: $x \le 5$. Check $x = 5$: $9 - 15 + 5 = -1 \ge -1$, true; $x = 6$: $-4 \ge -1$, false.
6. $4$ and $7$. Solving: $2x - 2 + 3 \ge x + 4$, so $2x + 1 \ge x + 4$, so $x \ge 3$. The values $-3$, $0$, and $\tfrac{1}{2}$ are all below 3. Check $x = 4$: $2(3) + 3 = 9$ and $4 + 4 = 8$, and $9 \ge 8$ is true; check $x = \tfrac{1}{2}$: $2(-\tfrac{1}{2}) + 3 = 2$ and $4.5$, and $2 \ge 4.5$ is false.
7. Subtract $2x$ from both sides, add 4 to both sides, divide both sides by 4 — three steps. $6x - 4 \le 2x + 8$ gives $4x - 4 \le 8$, then $4x \le 12$, then $x \le 3$. Check $x = 3$: $14 \le 14$, true; $x = 4$: $20 \le 16$, false.
8. $x < 4$. Distributing: $2x + 3 < 11$, so $2x < 8$. Check $x = 3$: $\tfrac{1}{2}(18) = 9 < 11$, true; $x = 4$: $\tfrac{1}{2}(22) = 11 < 11$, false.
9. $x \le 3$. Subtracting $2x$: $-1.5x + 1.5 \ge -3$, so $-1.5x \ge -4.5$, and dividing by $-1.5$ reverses to $x \le 3$. Check $x = 3$: $3 \ge 3$, true; $x = 4$: $3.5 \ge 5$, false.
10. Solving gives $x > -3$. Any three values above $-3$ work, for example $-2.5$, $0$, and $5$. Check $-2.5$: $2(-2.5) + 9 = 4$, and $4 > 3$ is true. Check $-3$ to see the boundary excluded: $3 > 3$ is false.
11. Let $m$ = the number of miles. $1.25m + 3.50 \le 21$, so $1.25m \le 17.50$, so $m \le 14$. Check $m = 14$: $17.50 + 3.50 = 21 \le 21$, true; $m = 15$: $22.25 \le 21$, false. Jordan can travel at most 14 miles.
12. Rosa subtracted 5 from the right side instead of adding 5. Since the left side reads $5x - 5$, the additive inverse of $-5$ is $+5$, so both sides gain 5: $5x \le 25$, giving $x \le 5$. Test $x = 4$ in the original: $12 + 8 - 5 = 15 \le 20$, true, so 4 is a solution — but Rosa's answer $x \le 3$ excludes every value from 3 to 5, which really are solutions.

Exit ticket 6.1

13. $3x + 6 \le 18$, so $3x \le 12$, so $x \le 4$. Check $x = 4$: $16 - 4 + 6 = 18 \le 18$, true; $x = 5$: $21 \le 18$, false.
14. No. Left: $5(-2) + 1 = -9$. Right: $3(-2) - 3 = -9$. The sentence $-9 > -9$ is false, because a number is not strictly greater than itself.
15. $2x + 8 \ge 3x + 1$, so $-x \ge -7$, and dividing by $-1$ reverses: $x \le 7$. Check $x = 7$: $22 \ge 22$, true; $x = 8$: $24 \ge 25$, false.
16. What changes: there are more moves, one or both sides may need expanding or combining first, and the variable may sit on both sides. What stays the same: the meaning of the symbols, the fact that the solution set is infinite, the properties that authorize each move, the single rule about reversing, and the habit of checking with one value inside and one outside.

Lesson 6.2 — The Reversal Rule, Revisited and Deepened

Guided practice

17. $4(-2) = -8$ and $9(-2) = -18$. On a number line $-8$ is right of $-18$, so the true statement is $-8 > -18$. The direction reversed.
18. $4 + (-12) = -8$ and $9 + (-12) = -3$. The true statement is $-8 < -3$. The direction did not change, because both points slid 12 units left together.
19. $x < -4$. Subtracting 9 gives $-4x > 16$; the reversal happens at the next step, dividing both sides by $-4$. Check $x = -5$: $20 + 9 = 29 > 25$, true; $x = -4$: $25 > 25$, false; $x = 0$: $9 > 25$, false.
20. $x \ge -9$. Adding 5 gives $-\tfrac{2}{3}x \le 6$; multiplying both sides by $-\tfrac{3}{2}$ reverses to $x \ge -9$. Boundary check: $-\tfrac{2}{3}(-9) - 5 = 6 - 5 = 1 \le 1$, true. Outside, $x = -12$: $8 - 5 = 3 \le 1$, false.

Independent practice

21. a) $-5x \ge 15$, and dividing by $-5$ reverses: $x \le -3$. Check $x = -3$: $18 \ge 18$, true; $x = -2$: $13 \ge 18$, false. b) $-2x < -6$, reversing: $x > 3$. Check $x = 4$: $-1 < 1$, true; $x = 3$: $1 < 1$, false. c) $-\tfrac{x}{4} \ge 3$, and multiplying by $-4$ reverses: $x \le -12$. Check $x = -12$: $3 + 2 = 5 \ge 5$, true; $x = -8$: $4 \ge 5$, false. d) $-0.5x > 3.5$, and dividing by $-0.5$ reverses: $x < -7$. Check $x = -8$: $4 - 1.5 = 2.5 > 2$, true; $x = -7$: $2 > 2$, false.
22. a) $x \ge 4$, by adding 6 to both sides — a slide, so no reversal. Check $x = 4$: $-2 \ge -2$, true; $x = 3$: $-3 \ge -2$, false. b) $x \le \tfrac{1}{3}$, by dividing both sides by $-6$ — a reflection, so the symbol reverses. Check $x = 0$: $0 \ge -2$, true; $x = 1$: $-6 \ge -2$, false. Only the second multiplies or divides both sides by a negative number. In the first, the negative number is merely being added and subtracted.
23. Distributing: $8 - 3x - 3 > 14$, so $5 - 3x > 14$, so $-3x > 9$, and dividing by $-3$ reverses to $x < -3$. Check $x = -4$: $8 + 9 = 17 > 14$, true; $x = -3$: $14 > 14$, false.
24. Subtracting $2x$: $12 - 7x \le -9$, so $-7x \le -21$, and dividing by $-7$ reverses to $x \ge 3$. Check $x = 3$: $-3 \le -3$, true; $x = 2$: $2 \le -5$, false.
25. a) No — subtracting a term from both sides is a slide. b) Yes — dividing both sides by a negative number reflects both sides across zero. c) No — adding a negative number is still a slide. d) No — $\tfrac{2}{5}$ is positive, so no point crosses zero.
26. Let $t$ = minutes. $1{,}200 - 75t < 600$, so $-75t < -600$, and dividing by $-75$ reverses to $t > 8$. Check $t = 9$: $1{,}200 - 675 = 525 < 600$, true; $t = 8$: $600 < 600$, false. The balloon is below 600 feet after 8 minutes of descent.
27. $-6 \cdot \left(-\tfrac{1}{2}\right) = 3$ and $2 \cdot \left(-\tfrac{1}{2}\right) = -1$, so the true statement is $3 > -1$. Multiplying by a negative number reflects both points across zero: $-6$ was far to the left and lands far to the right, while $2$ was to the right and lands just left of zero. The reflection swapped which number is larger, so the symbol had to swap too.
28. Test $x = 0$: $-2(0) + 5 = 5$, and $5 \ge 11$ is false — yet Nico's answer $x \ge -3$ includes 0. That single test proves his set contains non-solutions. He divided both sides by $-2$ without reversing. Correct: $-2x \ge 6$ gives $x \le -3$. Check $x = -3$: $6 + 5 = 11 \ge 11$, true; $x = -4$: $13 \ge 11$, true; $x = 0$: false.

Exit ticket 6.2

29. $-3x < -15$, and dividing by $-3$ reverses: $x > 5$. Check $x = 6$: $-11 < -8$, true; $x = 5$: $-8 < -8$, false.
30. $-\tfrac{x}{2} \ge 2$, and multiplying by $-2$ reverses: $x \le -4$. Check $x = -4$: $10 + 2 = 12 \ge 12$, true; $x = 0$: $10 \ge 12$, false.
31. $-5 \div (-1) = 5$ and $3 \div (-1) = -3$, so the true statement is $5 > -3$. The direction reversed.
32. Distributing changes one side of the inequality into an equivalent expression; the two sides are not both being scaled, so nothing about their comparison changes. Dividing both sides by a negative number reflects both sides across zero, which swaps which one is larger, so the symbol must be reversed to keep the sentence true.

Lesson 6.3 — Expanding and Combining Like Terms

Guided practice

33. $x < 7$. Distributive property: $2x - 6 + 5 < 13$. Combining like terms: $2x - 1 < 13$. Addition property: $2x < 14$. Division property (positive divisor): $x < 7$. Check $x = 6$: $2(3) + 5 = 11 < 13$, true; $x = 7$: $13 < 13$, false.
34. $8x + 4 - 3x \ge 24$, so $5x + 4 \ge 24$, so $5x \ge 20$, so $x \ge 4$. Boundary check: $4(9) - 12 = 24 \ge 24$, true. Outside, $x = 3$: $28 - 9 = 19 \ge 24$, false.
35. $-3x + 12 > 18$, so $-3x > 6$; the reversal happens when both sides are divided by $-3$, giving $x < -2$. Check $x = -3$: $21 > 18$, true; $x = -2$: $18 > 18$, false.
36. $3x - 4 + 3 \le 11$, so $3x - 1 \le 11$, so $3x \le 12$, so $x \le 4$. Inside, $x = 4$: $\tfrac{1}{2}(16) + 3 = 11 \le 11$, true. Outside, $x = 5$: $\tfrac{1}{2}(22) + 3 = 14 \le 11$, false.

Independent practice

37. a) $3x + 15 \le 27$, so $3x \le 12$, so $x \le 4$. Check $x = 4$: $27 \le 27$, true; $x = 5$: $30 \le 27$, false. b) $-2x + 14 < 4$, so $-2x < -10$, reversing to $x > 5$. Check $x = 6$: $-2(-1) = 2 < 4$, true; $x = 5$: $4 < 4$, false. c) $5x - 2x - 2 \ge 13$, so $3x - 2 \ge 13$, so $x \ge 5$. Check $x = 5$: $25 - 12 = 13 \ge 13$, true; $x = 4$: $20 - 10 = 10 \ge 13$, false. d) $6x - 4 < 20$, so $6x < 24$, so $x < 4$. Check $x = 3$: $\tfrac{2}{3}(21) = 14 < 20$, true; $x = 4$: $\tfrac{2}{3}(30) = 20 < 20$, false.
38. $7 - 8x + 12 \le 27$, so $19 - 8x \le 27$, so $-8x \le 8$, reversing to $x \ge -1$. Check $x = -1$: $7 - 4(-5) = 27 \le 27$, true; $x = -2$: $35 \le 27$, false.
39. $4x - 8 > 8$, so $4x > 16$, so $x > 4$. Check $x = 5$: $30 + 3 - 10 - 11 = 12 > 8$, true; $x = 4$: $24 + 3 - 8 - 11 = 8 > 8$, false.
40. $2x - 1 \le 9$, so $2x \le 10$, so $x \le 5$. Check $x = 5$: $0.25(36) = 9 \le 9$, true; $x = 6$: $0.25(44) = 11 \le 9$, false.
41. $-2x - 3 > 5$, so $-2x > 8$, reversing to $x < -4$. Check $x = -5$: $-\tfrac{1}{3}(-21) = 7 > 5$, true; $x = -4$: $-\tfrac{1}{3}(-15) = 5 > 5$, false.
42. Let $g$ = the number of bags. Before combining: $4g + 6g + 25 \le 325$. After combining: $10g + 25 \le 325$, so $10g \le 300$, so $g \le 30$. Check $g = 30$: $120 + 180 + 25 = 325 \le 325$, true; $g = 31$: $335 \le 325$, false. The club can assemble at most 30 bags.
43. $-2x + 10 > 8$, so $-2x > -2$, reversing to $x < 1$. Check $x = 0$: $-2(-5) = 10 > 8$, true; $x = 1$: $-2(-4) = 8 > 8$, false. The two sign hazards are distributing $-2$ across $-5$, which must give $+10$ rather than $-10$, and dividing by $-2$ at the end, which must reverse the symbol. Guard against the first by writing the factor over each term before multiplying, and against the second by asking the reversal question explicitly at the division step.
44. Ivy distributed $-3$ to the first term only. The $-3$ multiplies both terms: $-3x - 6 \le 9$, so $-3x \le 15$, and dividing by $-3$ reverses to $x \ge -5$. Test $x = -4$ in the original: $-3(-2) = 6 \le 9$, true, so $-4$ is a solution — but Ivy's answer, $-3x \le 7$ and so $x \ge -\tfrac{7}{3}$, excludes it.

Exit ticket 6.3

45. $5x - 10 + 4 \ge 19$, so $5x - 6 \ge 19$, so $5x \ge 25$, so $x \ge 5$. Check $x = 5$: $15 + 4 = 19 \ge 19$, true; $x = 4$: $14 \ge 19$, false.
46. $9 - 2x - 8 > 3$, so $1 - 2x > 3$, so $-2x > 2$, reversing to $x < -1$. Check $x = -2$: $9 - 2(2) = 5 > 3$, true; $x = -1$: $9 - 2(3) = 3 > 3$, false.
47. $8x - 3x + 3 \le 23$, so $5x + 3 \le 23$, so $5x \le 20$, so $x \le 4$. Check $x = 4$: $32 - 9 = 23 \le 23$, true; $x = 5$: $40 - 12 = 28 \le 23$, false.
48. Both moves rewrite one side as an equivalent expression; they do not scale or shift both sides. The reversal rule is triggered only when both sides are multiplied or divided by a negative number, because only then are both quantities reflected across zero. Distributing a $-3$ across parentheses reflects nothing — it just renames what one side already was.

Lesson 6.4 — The Variable on Both Sides

Guided practice

49. $x > 4$. Subtraction property (subtract $2x$): $3x - 3 > 9$. Addition property: $3x > 12$. Division property (positive divisor): $x > 4$. Check $x = 5$: $22 > 19$, true; $x = 4$: $17 > 17$, false.
50. Both routes give $x \ge 3$. Subtracting $4x$: $7 \le 5x - 8$, then $15 \le 5x$, then $3 \le x$, that is $x \ge 3$, with no reversal. Subtracting $9x$: $-5x + 7 \le -8$, then $-5x \le -15$, then, dividing by $-5$ and reversing, $x \ge 3$. Check $x = 3$: $19 \le 19$, true; $x = 2$: $15 \le 10$, false.
51. $3x - 6 < 5x + 4$, so $-2x - 6 < 4$, so $-2x < 10$, reversing to $x > -5$. Boundary check: $x = -5$ gives $-21$ on both sides, and $-21 < -21$ is false, so $-5$ is correctly excluded. Inside, $x = 0$: $-6 < 4$, true.
52. $6x - 4x - 4 \le 3x + 5$, so $2x - 4 \le 3x + 5$, so $-x \le 9$, reversing to $x \ge -9$. Inside, $x = 0$: $-4 \le 5$, true. Outside, $x = -10$: $-60 + 36 = -24$ and $-25$, and $-24 \le -25$ is false.

Independent practice

53. a) $3x + 2 \ge 17$, so $3x \ge 15$, so $x \ge 5$. Check $x = 5$: $37 \ge 37$, true; $x = 4$: $30 \ge 33$, false. b) Subtracting $6x$: $-4x - 9 < 7$, so $-4x < 16$, reversing to $x > -4$. Check $x = 0$: $-9 < 7$, true; $x = -4$: $-17 < -17$, false. c) Subtracting $3x$ and adding $-10$: $-4x \le -16$, reversing to $x \ge 4$. Check $x = 4$: $6 \le 6$, true; $x = 3$: $7 \le 3$, false. d) Subtracting $\tfrac{3}{2}x$: $-x + 4 > -1$, so $-x > -5$, reversing to $x < 5$. Check $x = 4$: $6 > 5$, true; $x = 5$: $6.5 > 6.5$, false.
54. $8 - 3x \ge 2x + 18$, so $-5x \ge 10$, reversing to $x \le -2$. Check $x = -2$: $8 + 6 = 14$ and $2(7) = 14$, and $14 \ge 14$ is true; $x = 0$: $8 \ge 18$, false.
55. $4x + 4 < 6x - 10$, so $-2x < -14$, reversing to $x > 7$. Check $x = 8$: $4(9) = 36$ and $2(19) = 38$, and $36 < 38$ is true; $x = 7$: $32 < 32$, false.
56. Subtracting $0.5x$: $x - 2 \le 3$, so $x \le 5$. Check $x = 5$: $5.5 \le 5.5$, true; $x = 6$: $7 \le 6$, false.
57. $9x - 5x + 10 \ge 2x + 16$, so $4x + 10 \ge 2x + 16$, so $2x \ge 6$, so $x \ge 3$. Check $x = 3$: $27 - 5 = 22$ and $6 + 16 = 22$, and $22 \ge 22$ is true; $x = 2$: $18$ and $20$, false.
58. Let $c$ = the number of classes. $30 + 5c < 10c$, so $30 < 5c$, so $c > 6$. Check $c = 7$: Gym A costs $30 + 35 = 65$ and Gym B costs $70$, and $65 < 70$ is true; $c = 6$: both cost $60, and $60 < 60$ is false. Gym A costs less starting at 7 classes.
59. Both routes give $x \ge 3$. Subtracting $3x$: $8 \le 4x - 4$, so $12 \le 4x$, so $3 \le x$ — no reversal. Subtracting $7x$: $-4x + 8 \le -4$, so $-4x \le -12$, and dividing by $-4$ reverses to $x \ge 3$. The answers agree because both routes are legal sequences of the same properties applied to the same inequality; reversing at the right moment on the second route is exactly what keeps the two in step. Check $x = 3$: $17 \le 17$, true; $x = 2$: $14 \le 10$, false.
60. Test $x = 3$: the left side is $11$ and the right side is $15$, and $11 > 15$ is false — yet Marco's answer $x > 2$ includes 3. He divided by $-4$ without reversing. Correct: $-4x > -8$ gives $x < 2$. Check $x = 0$: $5 > -3$, true; $x = 2$: $9 > 9$, false.

Exit ticket 6.4

61. $2x - 1 \le 7$, so $2x \le 8$, so $x \le 4$. Check $x = 4$: $23 \le 23$, true; $x = 5$: $29 \le 27$, false.
62. $5 - 6x > 23$, so $-6x > 18$, reversing to $x < -3$. Check $x = -4$: $13 > 7$, true; $x = -3$: $11 > 11$, false.
63. $3x + 12 \ge 5x - 2$, so $-2x \ge -14$, reversing to $x \le 7$. Check $x = 7$: $33 \ge 33$, true; $x = 8$: $36 \ge 38$, false.
64. If the variable ends up with a positive coefficient, the final division is by a positive number and the reversal question answers itself — one fewer chance to make the chapter's most common error. The other route is still correct because dividing by a negative number is a legal move; it simply requires reversing the symbol, and it lands on the same solution set.

Lesson 6.5 — Representing Solutions Algebraically and Graphically

Guided practice

65. Endpoint $-3$; open circle, because $>$ is strict; shade to the right. Check: $-2 > -3$ is true, $-3 > -3$ is false.
66. Endpoint $\tfrac{5}{2} = 2.5$, halfway between 2 and 3; closed circle, because $\le$ is inclusive; shade to the left. Check: $2.5 \le 2.5$ is true, $3 \le 2.5$ is false.
67. $2x - 7 \ge 1$, so $2x \ge 8$, so $x \ge 4$. Closed circle at 4, shaded right. Two values in the set: 4 and 7 — check $x = 4$: $9 \ge 9$, true; $x = 7$: $21 \ge 15$, true. Outside, $x = 3$: $5 \ge 7$, false.
68. $-2x \ge 6$, and dividing by $-2$ reverses to $x \le -3$. Closed circle at $-3$, shaded left. The graph runs left because the direction is read off the solved inequality, not the original; the division by a negative number reversed the symbol along the way. Check $x = -4$: $9 \ge 7$, true; $x = 0$: $1 \ge 7$, false.

Independent practice

69. a) endpoint 1, open, shade left. b) endpoint $-5$, closed, shade right. c) endpoint $\tfrac{1}{2}$, open, shade right. d) endpoint $-2.5$, closed, shade left.
70. a) $2x + 5 \le 11$, so $2x \le 6$, so $x \le 3$; closed circle at 3, shaded left. Check $x = 3$: $14 \le 14$, true; $x = 4$: $17 \le 15$, false. b) $9 - 2x - 2 > 3x + 2$, so $7 - 2x > 3x + 2$, so $-5x > -5$, reversing to $x < 1$; open circle at 1, shaded left. Check $x = 0$: $7 > 2$, true; $x = 1$: $5 > 5$, false.
71. $-5x + 5 \ge 20$, so $-5x \ge 15$, reversing to $x \le -3$; closed circle at $-3$, shaded left. Check $x = -3$: $-5(-4) = 20 \ge 20$, true; $x = -2$: $15 \ge 20$, false.
72. $12 \le 3x$ gives $4 \le x$, written variable-first as $x \ge 4$; closed circle at 4, shaded right. Swapping the sides also swaps the symbol so that the wide end still faces $x$. Check $x = 4$: $12 \le 12$, true; $x = 3$: $12 \le 9$, false.
73. $x < -1$. One multistep inequality with the same solution set is $2x + 5 < x + 4$: subtracting $x$ gives $x + 5 < 4$, so $x < -1$. Check $x = -2$: $1 < 2$, true; $x = -1$: $3 < 3$, false.
74. Let $t$ = minutes. $500 - 20t > 100$, so $-20t > -400$, reversing to $t < 20$; open circle at 20, shaded left. Check $t = 19$: $120 > 100$, true; $t = 20$: $100 > 100$, false. The part of the graph to the left of 0 describes no real moment, since time since the drain opened cannot be negative; in context the answer is $0 \le t < 20$.
75. Solving $-3x > 12$ requires dividing both sides by $-3$, which reverses the symbol and gives $x < -4$. The graph is drawn from the solved form, so it runs left. Check $x = -5$: $15 > 12$, true; $x = -4$: $12 > 12$, false.
76. Two errors: the circle should be closed, not open, because $\le$ is inclusive after solving; and the shading should run right, not left, because the division by $-1$ reverses the symbol. Solving: $-x \le 5$, so $x \ge -5$. Test $x = 0$: $4 - 0 = 4 \le 9$ is true, so 0 is a solution, but Priya's graph excludes it. The correct graph is a closed circle at $-5$ shaded to the right.

Exit ticket 6.5

77. $2x - 4 < 6$, so $2x < 10$, so $x < 5$; open circle at 5, shaded left. Check $x = 4$: $24 < 26$, true; $x = 5$: $31 < 31$, false.
78. $-\tfrac{1}{2}x \ge 2$, and multiplying by $-2$ reverses to $x \le -4$; closed circle at $-4$, shaded left. Check $x = -4$: $2 + 3 = 5 \ge 5$, true; $x = 0$: $3 \ge 5$, false.
79. $x \ge \tfrac{3}{2}$.
80. Use an open circle for $<$ and $>$, and a closed circle for $\le$ and $\ge$. The rule exists because the circle records whether the boundary number itself satisfies the inequality: $4 > 4$ is false, so 4 is not in the set and the circle is hollow, while $4 \ge 4$ is true, so 4 is in the set and the circle is filled.

Lesson 6.6 — Inequalities in Context: Write, Solve, Interpret, Invent

Guided practice

81. Let $h$ = the number of hours. $90 + 40h < 150 + 25h$, so $15h < 60$, so $h < 4$. Check $h = 3$: $210 versus $225, true; $h = 4$: $250 versus $250, and $250 < 250$ is false. The first company is less expensive for any job shorter than 4 hours.
82. Let $n$ = the number. $3n - 5 \ge 2n + 4$, so $n \ge 9$. Check $n = 9$: $22 \ge 22$, true; $n = 8$: $19 \ge 20$, false.
83. Let $w$ = the number of weeks. $120 - 12 - 15w > 30$, so $108 - 15w > 30$, so $-15w > -78$, reversing to $w < \tfrac{78}{15} = 5.2$. Test $w = 5$: $108 - 75 = 33 > 30$, true; $w = 6$: $18 > 30$, false. Devon still has more than $30 for 5 whole weeks.
84. Sample situation: Company A charges a $20 booking fee plus $8 per mile; Company B charges a $60 booking fee plus $4 per mile. For which trip lengths does Company A cost no more than Company B? Solving: $4x \le 40$, so $x \le 10$. Company A costs no more for trips up to and including 10 miles. Check $x = 10$: $100 \le 100$, true; $x = 11$: $108 \le 104$, false.

Independent practice

85. Let $b$ = the number of boxes. $190 + 2(150) + 45b \le 1{,}500$, so $490 + 45b \le 1{,}500$, so $45b \le 1{,}010$, so $b \le \tfrac{202}{9} \approx 22.4$. Test $b = 22$: $490 + 990 = 1{,}480 \le 1{,}500$, true; $b = 23$: $1{,}525 \le 1{,}500$, false. At most 22 boxes may be loaded.
86. Let $g$ = gigabytes. $25 + 8g \le 55 + 4g$, so $4g \le 30$, so $g \le 7.5$. Check $g = 7$: $81 versus $83, true; $g = 8$: $89 versus $87, false. Plan A costs no more as long as data use is at most 7.5 gigabytes; above that, Plan B is cheaper. Since data need not come in whole gigabytes, 7.5 is a usable boundary here.
87. Let $n$ = the number. $2n + 4 < n - 6$, so $n < -10$. Check $n = -11$: $-18 < -17$, true; $n = -10$: $-16 < -16$, false.
88. Sample situation: A tutoring service charges a $45 registration fee plus $12 per session. How many sessions must a student book for the total to be at least $165? Solving: $12x \ge 120$, so $x \ge 10$. The student must book at least 10 sessions. Check $x = 10$: $120 + 45 = 165 \ge 165$, true; $x = 9$: $153 \ge 165$, false.
89. Sample situation: A phone battery starts at 80 percent and drops 6 percentage points each hour. For how many hours does the charge stay above 20 percent? Solving: $-6x > -60$, reversing to $x < 10$. The charge stays above 20 percent for the first 10 hours. Check $x = 9$: $80 - 54 = 26 > 20$, true; $x = 10$: $20 > 20$, false.
90. At most 22 crates, since $c \le 22.4$ and crates come in whole numbers that cannot be negative, so the usable values are $0$ through $22$. The greatest usable value is 22. Testing confirms the edge: 22 satisfies the original inequality and 23 does not, so the boundary sits between them.
91. At $h = 4$ the two companies charge exactly the same amount, $250. The question asked when the first company is less expensive, which is a strict comparison, and equal is not less, so 4 is excluded. Had the question asked "costs no more than," the symbol would have been $\le$ and 4 would be included.
92. Sam multiplied the fixed fee by the per-page cost. The $80 is paid once and the $0.20 is paid for each page, so the two amounts are added, not multiplied: $0.20p + 80 \le 140$, so $0.20p \le 60$, so $p \le 300$. Check $p = 300$: $60 + 80 = 140 \le 140$, true; $p = 301$: $140.20 \le 140$, false. The budget covers at most 300 pages.

Exit ticket 6.6

93. Let $g$ = the number of guests. $18g + 200 \le 920$, so $18g \le 720$, so $g \le 40$. Check $g = 40$: $720 + 200 = 920 \le 920$, true; $g = 41$: $938 \le 920$, false. At most 40 guests.
94. Let $n$ = the number. $4n + 7 \le n + 25$, so $3n \le 18$, so $n \le 6$. Check $n = 6$: $31 \le 31$, true; $n = 7$: $35 \le 32$, false.
95. Sample situation: Renting equipment from the shop costs a $30 delivery charge plus $5 per day. Buying the same use from a rival costs $8 per day with no delivery charge. For how many days is the shop the cheaper choice, or a tie? Solving: $30 \le 3x$, so $x \ge 10$. The shop costs no more once the rental runs 10 days or longer. Check $x = 10$: $80 \le 80$, true; $x = 9$: $75 \le 72$, false.
96. To interpret a solution in context is to say what the numbers mean for the situation — naming the units, deciding whether fractions and negatives are allowed, and giving the answer as a sentence rather than as a symbol string. Example: a ticket problem that solves to $t \le 6\tfrac{1}{3}$ has the practical answer "at most 6 tickets," because tickets are whole and a third of a ticket cannot be purchased.

Chapter 6 Review

Part A — Solving multistep inequalities (8.PFA.5a)

97. a) $7x - 9 \le 19$, so $7x \le 28$, so $x \le 4$. Check $x = 4$: $19 \le 19$, true; $x = 5$: $26 \le 19$, false. b) $5x - 15 > 10$, so $5x > 25$, so $x > 5$. Check $x = 6$: $15 > 10$, true; $x = 5$: $10 > 10$, false. c) $-4x \ge -8$, reversing to $x \le 2$. Check $x = 2$: $3 \ge 3$, true; $x = 3$: $-1 \ge 3$, false. d) $\tfrac{3}{5}x < 6$, so $x < 10$. Check $x = 5$: $-1 < 2$, true; $x = 10$: $2 < 2$, false.
98. $6x - 2 + 4x \le 38$, so $10x - 2 \le 38$, so $10x \le 40$, so $x \le 4$. Check $x = 4$: $22 + 16 = 38 \le 38$, true; $x = 5$: $48 \le 38$, false.
99. $3x - 3 > 12$, so $3x > 15$, so $x > 5$. Check $x = 6$: $45 > 42$, true; $x = 5$: $37 > 37$, false.
100. $7 - 2x + 8 \le 3x$, so $15 - 2x \le 3x$, so $15 \le 5x$, so $x \ge 3$. Check $x = 3$: $9 \le 9$, true; $x = 2$: $11 \le 6$, false.
101. $2x + 3 \ge x + 8$, so $x \ge 5$. Check $x = 5$: $\tfrac{1}{4}(52) = 13 \ge 13$, true; $x = 4$: $11 \ge 12$, false.
102. $-2x + 3 < 9$, so $-2x < 6$, reversing to $x > -3$. Check $x = 0$: $3 < 9$, true; $x = -3$: $-0.5(-18) = 9 < 9$, false.
103. $6x - 5x - 10 \ge 2x - 14$, so $x - 10 \ge 2x - 14$, so $-x \ge -4$, reversing to $x \le 4$. Check $x = 4$: $-6 \ge -6$, true; $x = 5$: $-5 \ge -4$, false.