Appendix A — Answer Key, Chapter 3: Linear Inequalities in One Variable

SOL A.EI.1 (c, f) · Covers textbook Chapter 3 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 88 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 set below was checked with a value inside the set, the boundary value itself, and a value outside the set, substituted into the original inequality.

Lesson 3.1 — Inequalities, Solution Sets, and the Number Line

Guided practice

1. Yes. Left: $4(-2) + 7 = -1$. Right: $3(-2) + 5 = -1$. The sentence $-1 \le -1$ is true, because $\le$ allows equality.
2. $x < 4$; open circle at 4, shaded left. Adding 5 to both sides gives $3x < 12$, and dividing by the positive number 3 gives $x < 4$. Inside, $x = 3$: $4 < 7$, true. Boundary, $x = 4$: $7 < 7$, false. Outside, $x = 5$: $10 < 7$, false.
3. $x \ge 3$; closed circle at 3, shaded right. Subtraction property of inequality: $2x \ge 6$. Additive inverse property: the $+1$ is gone. Division property of inequality with a positive divisor: $x \ge 3$. Inside, $x = 5$: $11 \ge 7$, true. Boundary, $x = 3$: $7 \ge 7$, true. Outside, $x = 2$: $5 \ge 7$, false.
4. $x > 4$. Subtraction property of inequality (subtract $2x$): $3x - 3 > 9$. Addition property of inequality: $3x > 12$. Division property of inequality, positive divisor: $x > 4$. Boundary, $x = 4$: $17 > 17$, false, so 4 is correctly excluded. Inside, $x = 5$: $22 > 19$, true.

Independent practice

5. a) $x \le 4$; closed circle at 4, shaded left. Check $x = 4$: $10 \le 10$, true; $x = 5$: $11 \le 10$, false. b) $x > 3$; open circle at 3, shaded right. Check $x = 4$: $28 > 21$, true; $x = 3$: $21 > 21$, false. c) $4x \ge 20$, so $x \ge 5$; closed circle at 5, shaded right. Check $x = 5$: $11 \ge 11$, true; $x = 4$: $7 \ge 11$, false. d) $\tfrac{1}{3}x < 3$, so $x < 9$; open circle at 9, shaded left. Check $x = 6$: $4 < 5$, true; $x = 9$: $5 < 5$, false.
6. $x \le 6$; closed circle at 6, shaded left. Subtraction property of inequality (subtract $4x$): $2x + 5 \le 17$. Subtraction property of inequality: $2x \le 12$. Division property of inequality, positive divisor: $x \le 6$. Check $x = 6$: $41 \le 41$, true; $x = 7$: $47 \le 45$, false.
7. $15 < 5x$ gives $3 < x$, written variable-first as $x > 3$; open circle at 3, shaded right. Swapping the sides also swaps the symbol so the wide end still faces $x$. Check $x = 4$: $15 < 20$, true; $x = 3$: $15 < 15$, false.
8. $-5$, $0$, $2.5$, and $4$. Solving: $3x - 4 \le x + 6$ gives $2x \le 10$, so $x \le 5$, which excludes 9 only. Check $x = 4$: $8 \le 10$, true; check $x = 9$: $23 \le 15$, false.
9. $x \ge -2$. One multistep inequality with the same solution set is $3x + 5 \ge x + 1$: subtracting $x$ gives $2x + 5 \ge 1$, then $2x \ge -4$, then $x \ge -2$. Check $x = -2$: $-1 \ge -1$, true; $x = -3$: $-4 \ge -2$, false.
10. Solving gives $2x > -6$, so $x > -3$. Any three values above $-3$ work, for example $-2.5$, $0$, and $4$. Check $-2.5$: $2(-2.5) + 9 = 4$, and $4 > 3$ is true. Check $0$: $9 > 3$, true. Check $4$: $17 > 3$, true. The boundary fails: at $x = -3$, $3 > 3$ is false.
11. Let $s$ = the number of sensors. $45 + 5s \le 120$, so $5s \le 75$, so $s \le 15$. Check $s = 15$: $45 + 75 = 120 \le 120$, true; $s = 16$: $125 \le 120$, false. The club can buy at most 15 sensors, and 15 exactly spends the budget.
12. The equation $2x + 1 = 7$ asks which single number makes the two sides equal, and only $x = 3$ does. The inequality $2x + 1 \ge 7$ asks which numbers make the left side at least as large as the right, and once $x$ passes 3 the left side keeps growing, so every number from 3 upward qualifies — infinitely many, which is a ray rather than a point. What the two share is the boundary 3, and it comes from the equation: solving the matching equation locates exactly where the sentence changes from true to false.

Exit ticket 3.1

13. $5x < 10$, so $x < 2$; open circle at 2, shaded left. Check $x = 1$: $-3 < 2$, true; $x = 2$: $2 < 2$, false.
14. Yes. Left: $4(3) - 5 = 7$. Right: $2(3) + 1 = 7$. The sentence $7 \ge 7$ is true.
15. Subtracting $5x$: $3x + 3 \ge 18$, so $3x \ge 15$, so $x \ge 5$; closed circle at 5, shaded right. Check $x = 5$: $43 \ge 43$, true; $x = 4$: $35 \ge 38$, false.
16. 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: $3 > 3$ is false, so 3 is not in the set and the circle is hollow, while $3 \ge 3$ is true, so 3 is in the set and the circle is filled. Substituting the boundary into the original inequality settles it every time.

Lesson 3.2 — The Reversal Rule and Why It Happens

Guided practice

17. $2(-3) = -6$ and $5(-3) = -15$. On a number line $-6$ is to the right of $-15$, so the true statement is $-6 > -15$. The direction reversed, because multiplying by a negative number reflects both points across zero.
18. $2 + (-8) = -6$ and $5 + (-8) = -3$. The true statement is $-6 < -3$. The direction did not change, because both points slid 8 units left together and a slide never changes which point is on the left.
19. $x \le -5$; closed circle at $-5$, shaded left. Subtracting 7 gives $-3x \ge 15$; the reversal happens at the next step, dividing both sides by $-3$. Inside, $x = -6$: $25 \ge 22$, true. Boundary, $x = -5$: $22 \ge 22$, true. Outside, $x = -4$: $19 \ge 22$, false.
20. $x > -15$; open circle at $-15$, shaded right. Multiplying both sides by $-\tfrac{5}{2}$ reverses the symbol. Boundary, $x = -15$: $-\tfrac{2}{5}(-15) = 6$, and $6 < 6$ is false, so $-15$ is excluded. Inside, $x = 0$: $0 < 6$, true. Outside, $x = -20$: $8 < 6$, false.

Independent practice

21. a) Dividing by $-5$ reverses: $x < -4$; open circle at $-4$, shaded left. Check $x = -5$: $25 > 20$, true; $x = -4$: $20 > 20$, false. b) $-x \le 7$, and dividing by $-1$ reverses: $x \ge -7$; closed circle at $-7$, shaded right. Check $x = -7$: $10 \le 10$, true; $x = -8$: $11 \le 10$, false. c) $-\tfrac{1}{2}x \ge 5$, and multiplying by $-2$ reverses: $x \le -10$; closed circle at $-10$, shaded left. Check $x = -10$: $5 - 4 = 1 \ge 1$, true; $x = -8$: $4 - 4 = 0 \ge 1$, false. d) $-3x < 15$, and dividing by $-3$ reverses: $x > -5$; open circle at $-5$, shaded right. Check $x = -4$: $20 < 23$, true; $x = -5$: $23 < 23$, false.
22. a) $x \ge 5$, by adding 8 to both sides — a slide, so no reversal. Check $x = 5$: $-3 \ge -3$, true; $x = 4$: $-4 \ge -3$, false. b) $x \le \tfrac{3}{8}$, by dividing both sides by $-8$ — a reflection, so the symbol reverses. Check $x = \tfrac{3}{8}$: $-3 \ge -3$, true; $x = 1$: $-8 \ge -3$, false; $x = 0$: $0 \ge -3$, true. Only the second multiplies or divides both sides by a negative number. In the first, the negative number is merely being added and subtracted, which slides both sides equally and leaves the order alone.
23. Distributing: $6 - 2x - 8 > 10$, so $-2 - 2x > 10$, so $-2x > 12$, and dividing by $-2$ reverses to $x < -6$; open circle at $-6$, shaded left. Check $x = -7$: $6 - 2(-3) = 12 > 10$, true; $x = -6$: $6 - 2(-2) = 10 > 10$, false.
24. Subtracting $3x$ and 4: $-10x \le -30$, and dividing by $-10$ reverses to $x \ge 3$. Boundary, $x = 3$: $4 - 21 = -17$ and $9 - 26 = -17$, and $-17 \le -17$ is true. Outside, $x = 2$: $-10 \le -20$, false. Inside, $x = 4$: $-24 \le -14$, true.
25. a) No — subtracting a term from both sides is a slide. b) Yes — multiplying both sides by a negative number reflects both sides across zero. c) No — distributing changes one side only; it never scales both sides. d) No — $0.5$ is positive, so no point crosses zero.
26. Let $t$ = the number of seconds. $320 - 25t < 145$, so $-25t < -175$, and dividing by $-25$ reverses to $t > 7$. Check $t = 8$: $320 - 200 = 120 < 145$, true; $t = 7$: $145 < 145$, false. The drone is below 145 feet after 7 seconds of descent — not at 7 seconds, when it is exactly at 145 feet.
27. $-8 \cdot \left(-\tfrac{1}{2}\right) = 4$ and $-2 \cdot \left(-\tfrac{1}{2}\right) = 1$, so the true statement is $4 > 1$. Multiplying by a negative number reflects both points across zero: $-8$ sat far to the left of zero and lands far to the right, while $-2$ sat nearer zero and lands nearer zero on the right. The point that was farther left becomes the point that is farther right, so the order swapped and the symbol had to swap with it.
28. Test $x = 0$: $-5(0) + 2 = 2$, and $2 > 17$ is false — yet Kai's answer $x > -3$ includes 0. That single test proves his set contains non-solutions. He divided both sides by $-5$ without reversing. Correct: $-5x > 15$ gives $x < -3$. Check $x = -4$: $20 + 2 = 22 > 17$, true; $x = -3$: $17 > 17$, false.

Exit ticket 3.2

29. $-4x \le -16$, and dividing by $-4$ reverses: $x \ge 4$; closed circle at 4, shaded right. Check $x = 4$: $-16 + 5 = -11 \le -11$, true; $x = 3$: $-7 \le -11$, false.
30. $-\tfrac{x}{3} > 2$, and multiplying by $-3$ reverses: $x < -6$; open circle at $-6$, shaded left. Check $x = -7$: $10 + \tfrac{7}{3} \approx 12.33 > 12$, true; $x = -6$: $12 > 12$, false.
31. $-9 \div (-3) = 3$ and $6 \div (-3) = -2$, so the true statement is $3 > -2$. The direction reversed.
32. Multiplying both sides by a negative number reflects both quantities across zero, and reflection swaps which one is larger, so the symbol must be reversed to keep the sentence true. Distributing a negative number rewrites one side as an equivalent expression — the two sides are not both being scaled, so nothing about their comparison changes and the symbol stays put.

Lesson 3.3 — Multistep Inequalities with Every Step Named

Guided practice

33. $x \le 6$. Distributive property: $6x - 15 + 4 \le 25$. Combining like terms: $6x - 11 \le 25$. Addition property of inequality: $6x \le 36$. Division property of inequality, positive divisor: $x \le 6$. Boundary, $x = 6$: $3(7) + 4 = 25 \le 25$, true. Outside, $x = 7$: $31 \le 25$, false. Inside, $x = 0$: $-11 \le 25$, true.
34. $x > 5$. Distributing: $5x - 2x - 12 > 3$, so $3x - 12 > 3$, so $3x > 15$. Boundary, $x = 5$: $25 - 22 = 3$, and $3 > 3$ is false, so 5 is correctly excluded. Inside, $x = 6$: $30 - 24 = 6 > 3$, true.
35. Both routes give $x \le 5$. Subtracting $7x$: $4 \ge 3x - 11$, then $15 \ge 3x$, then $5 \ge x$, that is $x \le 5$, with no reversal. Subtracting $10x$: $-3x + 4 \ge -11$, then $-3x \ge -15$, then, dividing by $-3$ and reversing, $x \le 5$ — the second route reversed. Boundary, $x = 5$: $39 \ge 39$, true; outside, $x = 6$: $46 \ge 49$, false.
36. $x \le 0$; closed circle at 0, shaded left. Distributing: $6 - 2x \ge 4x + 6$, so $-6x \ge 0$, and dividing by $-6$ reverses. Boundary, $x = 0$: $6 \ge 6$, true. Inside, $x = -1$: $8 \ge 2$, true. Outside, $x = 1$: $4 \ge 10$, false.

Independent practice

37. a) $4x - 12 < 8$, so $4x < 20$, so $x < 5$; open circle at 5, shaded left. Check $x = 4$: $4 < 8$, true; $x = 5$: $8 < 8$, false. b) $-2x - 10 \ge 4$, so $-2x \ge 14$, and dividing by $-2$ reverses: $x \le -7$; closed circle at $-7$, shaded left. Check $x = -7$: $-2(-2) = 4 \ge 4$, true; $x = -6$: $2 \ge 4$, false. c) $6x - 3x + 6 \le 18$, so $3x + 6 \le 18$, so $x \le 4$; closed circle at 4, shaded left. Check $x = 4$: $24 - 6 = 18 \le 18$, true; $x = 5$: $30 - 9 = 21 \le 18$, false. d) $6x - 4 > 20$, so $6x > 24$, so $x > 4$; open circle at 4, shaded right. Check $x = 5$: $\tfrac{2}{3}(39) = 26 > 20$, true; $x = 4$: $\tfrac{2}{3}(30) = 20 > 20$, false.
38. $x \le 3$. Distributive property: $8 - 10x + 5 \ge -17$. Combining like terms: $13 - 10x \ge -17$. Subtraction property of inequality: $-10x \ge -30$. Division property of inequality, negative divisor, so the symbol reverses: $x \le 3$. Boundary, $x = 3$: $8 - 5(5) = -17 \ge -17$, true. Outside, $x = 4$: $-27 \ge -17$, false. Inside, $x = 0$: $13 \ge -17$, true.
39. Subtracting $5x$: $4x - 4 < 12$, so $4x < 16$, so $x < 4$. Check $x = 3$: $23 < 27$, true; $x = 4$: $32 < 32$, false.
40. Subtracting $0.9x$ and $1.2$: $-0.5x \ge -2$, and dividing by $-0.5$ reverses: $x \le 4$. Boundary, $x = 4$: $2.8 \ge 2.8$, true. Outside, $x = 5$: $3.2 \ge 3.7$, false. Inside, $x = 0$: $1.2 \ge -0.8$, true.
41. $3x + 6 - 4x \le 8 - 2x$, so $-x + 6 \le 8 - 2x$, so adding $2x$ to both sides gives $x + 6 \le 8$, so $x \le 2$. No reversal was needed, because the variable was collected on the side that kept its coefficient positive. Boundary, $x = 2$: $12 - 8 = 4$ and $2(2) = 4$, and $4 \le 4$ is true. Outside, $x = 3$: $15 - 12 = 3$ and $2$, and $3 \le 2$ is false.
42. Dana distributed $-4$ to the first term only, and she also lost the sign change on the second term: $-4 \cdot (-3) = +12$, not $-12$. Correct: $-4x + 12 \le 20$, so $-4x \le 8$, and dividing by $-4$ reverses to $x \ge -2$. Dana's line gives $-4x \le 32$, so $x \ge -8$, a much larger set. Test $x = -3$ in the original: $-4(-6) = 24$, and $24 \le 20$ is false, so $-3$ is not a solution — yet Dana's answer includes it.
43. Let $g$ = the number of guests. $250 + 22g \le 400 + 16g$, so $6g \le 150$, so $g \le 25$. Check $g = 25$: Hall A costs $250 + 550 = 800$ dollars and Hall B costs $400 + 400 = 800$ dollars, and $800 \le 800$ is true. Check $g = 26$: $822$ versus $816$, false. Hall A costs no more than Hall B for parties of 25 guests or fewer; at exactly 25 guests the two halls cost the same, and the inclusive symbol keeps that tie in the answer.
44. $x \le 3$. Distributing: $7 - 3x + 6 \ge 4x - 8$, so $13 - 3x \ge 4x - 8$. Adding $3x$ and 8 to both sides: $21 \ge 7x$, so $3 \ge x$, that is $x \le 3$. Boundary, $x = 3$: $7 - 3 = 4$ and $12 - 8 = 4$, and $4 \ge 4$ is true. Outside, $x = 4$: $1 \ge 8$, false. The two sign hazards are distributing $-3$ across $-2$, which must produce $+6$ rather than $-6$, and the choice of side when collecting the variable. Guard against the first by writing the factor above each term before multiplying, and against the second by moving the variable to the side that keeps its coefficient positive, which removes the reversal step entirely.

Exit ticket 3.3

45. $6x - 6 + 2 \le 20$, so $6x - 4 \le 20$, so $6x \le 24$, so $x \le 4$. Check $x = 4$: $18 + 2 = 20 \le 20$, true; $x = 5$: $26 \le 20$, false.
46. $12 - 4x - 4 > 2x - 4$, so $8 - 4x > 2x - 4$, so $12 > 6x$, so $x < 2$; open circle at 2, shaded left. Check $x = 1$: $12 - 8 = 4$ and $-2$, and $4 > -2$ is true; $x = 2$: $0 > 0$, false.
47. Subtracting $9x$ and adding 7: $-4x \ge 12$, and dividing by $-4$ reverses: $x \le -3$. Check $x = -3$: $-22 \ge -22$, true; $x = -2$: $-17 \ge -13$, 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 and their order swapped. Distributing a $-3$ across parentheses reflects nothing — it just renames what one side already was, and a renamed side compares to the other side exactly as before.

Lesson 3.4 — Verifying and Interpreting Solutions

Guided practice

49. Solving: $2x - 7 \ge 1$, so $2x \ge 8$, so $x \ge 4$. Inside, $x = 6$: $17 \ge 13$, true. Boundary, $x = 4$: $9 \ge 9$, true, so the endpoint belongs to the set and the circle is closed. Outside, $x = 3$: $5 \ge 7$, false.
50. The lines $y = 2x - 1$ and $y = x + 1$ cross at $(2, 3)$. To the right of the crossing, $y = 2x - 1$ is the higher line, which is exactly where $2x - 1 > x + 1$ holds, so the solution set is $x > 2$. The crossing point is excluded because at $x = 2$ the two sides are equal, $3 = 3$, and a strict $>$ does not accept equality. Substitution agrees: $x = 3$ gives $5 > 4$, true; $x = 2$ gives $3 > 3$, false; $x = 1$ gives $1 > 2$, false.
51. Enter $Y_1 = -3x + 7$ and $Y_2 = 22$. In the graph, $Y_1$ is a line falling from left to right, $Y_2$ is horizontal, and the intersect feature reports the crossing at $(-5, 22)$. To the left of that crossing $Y_1$ lies above $Y_2$, which is where $-3x + 7 \ge 22$ is satisfied, so the solution set runs left from $-5$: $x \le -5$. In the table, the row $x = -6$ reads $Y_1 = 25$ and $Y_2 = 22$; the row $x = -5$ reads $22$ and $22$, an equality the $\ge$ symbol accepts, so the endpoint is included; the row $x = -4$ reads $19$ and $22$, which fails.
52. Let $t$ = the number of tables, a whole number with $t \ge 0$. $180 + 12t \le 500$, so $12t \le 320$, so $t \le \tfrac{80}{3} \approx 26.7$. Test $t = 26$: $180 + 312 = 492 \le 500$, true; $t = 27$: $180 + 324 = 504 \le 500$, false. The council can decorate at most 26 tables. The algebraic endpoint, about 26.7 tables, is not usable because tables come in whole numbers.

Independent practice

53. Subtracting $5x$ and 2: $-2x \le -10$, and dividing by $-2$ reverses: $x \ge 5$. Inside, $x = 6$: $20 \le 22$, true. Boundary, $x = 5$: $17 \le 17$, true. Outside, $x = 4$: $14 \le 12$, false.
54. $-2x > -8$, and dividing by $-2$ reverses: $x < 4$. The graphs of $y = -2x + 9$ and $y = 1$ cross at $(4, 1)$. Because $y = -2x + 9$ falls from left to right, it is the higher graph to the left of $x = 4$, which is exactly the solution set. Check $x = 3$: $3 > 1$, true; $x = 4$: $1 > 1$, false.
55. From the table, $Y_1 \ge Y_2$ in the columns $x = 0$, $1$, $2$, and $3$ — at $x = 3$ the two entries are both $-1$, which the $\ge$ symbol accepts — and it fails from $x = 4$ on, where $-3 \ge 0$ is false. Algebraically: $5 - 2x \ge x - 4$ gives $9 \ge 3x$, so $x \le 3$. The two agree, and the boundary happens to be one of the table's own $x$-values, so the table locates it exactly here.
56. Let $x$ = the number of extra texts. $35 + 0.05x \le 50$, so $0.05x \le 15$, so $x \le 300$. Check $x = 300$: $35 + 15 = 50 \le 50$, true; $x = 301$: $50.05 \le 50$, false. Amara can send at most 300 extra texts, and 300 exactly spends the $\$50$ limit. Texts are whole and cannot be negative, so the usable values are the whole numbers from 0 through 300.
57. Let $c$ = the number of crates. $180 + 65c \le 2{,}000$, so $65c \le 1{,}820$, so $c \le 28$. Boundary, $c = 28$: $180 + 1{,}820 = 2{,}000 \le 2{,}000$, true. Outside, $c = 29$: $2{,}065 \le 2{,}000$, false. At most 28 crates. The endpoint is usable here because 28 is already a whole number and the symbol is inclusive, so a load of exactly 28 crates reaches the limit without exceeding it — the elevator is full, not overloaded.
58. The subscription can run at most 13.75 months, but months are counted whole and cannot be negative, so the usable values are 0 through 13 and the greatest usable value is 13 months. The pair of tests: 13 satisfies the original condition, and 14 exceeds 13.75 and fails it, so the practical edge sits between them. Rounding down is forced by the ceiling, not chosen.
59. One inside test can be passed by a wrong answer, because a set that is too large still contains genuine solutions — testing $x = 10$ against a claimed $x > 5$ says nothing about whether the boundary should have been 5 or 7. The three convincing tests are: a value inside the set, which must make the original true; a value outside, which must make it false; and the boundary itself. The boundary test is the one that decides between an open and a closed endpoint, since it comes out true for $\le$ and $\ge$ and false for $<$ and $>$.
60. Test $x = 0$: $6 - 0 = 6$, and $6 \ge 14$ is false — yet Priya's answer $x \ge -4$ includes 0, so her set contains non-solutions. She divided by $-2$ without reversing. Correct: $-2x \ge 8$, and dividing by $-2$ reverses to $x \le -4$; closed circle at $-4$, shaded left. Check $x = -4$: $6 + 8 = 14 \ge 14$, true; $x = -5$: $16 \ge 14$, true; $x = -3$: $12 \ge 14$, false.

Exit ticket 3.4

61. $-3x < 15$, and dividing by $-3$ reverses: $x > -5$. Inside, $x = -4$: $4 + 12 = 16 < 19$, true. Boundary, $x = -5$: $19 < 19$, false, so the endpoint is excluded. Outside, $x = -6$: $22 < 19$, false.
62. Graph $Y_1 = 4 - 3x$ and $Y_2 = 19$. The falling line $Y_1$ crosses the horizontal line $Y_2$ at $(-5, 19)$. To the right of that crossing $Y_1$ lies below $Y_2$, which is where $4 - 3x < 19$ holds, so the solution set is $x > -5$, matching the algebra. The intersect feature or the table row $x = -5$, where both columns read 19, locates the boundary, and the strict symbol makes it an open circle.
63. Let $t$ = the number of tacos, a whole number with $t \ge 0$. $2.75t + 3.00 \le 25.00$, so $2.75t \le 22.00$, so $t \le 8$. Boundary, $t = 8$: $22.00 + 3.00 = 25.00 \le 25.00$, true; $t = 9$: $27.75 \le 25.00$, false. Devon can buy 8 tacos, which spends exactly $\$25.00$.
64. 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, checking whether the endpoint is usable, and giving the answer as a sentence rather than a symbol string. Example: a ticket budget that solves to $t \le 6.4$ has the practical answer "at most 6 tickets," because a tenth of a ticket cannot be bought; the algebraic endpoint 6.4 is a real number in the solution set but not a real number of tickets.

Chapter 3 Review

Part A — Solving inequalities algebraically and graphing the solution set (A.EI.1c)

65. a) $5x \ge 15$, so $x \ge 3$; closed circle at 3, shaded right. Check $x = 3$: $12 \ge 12$, true; $x = 2$: $7 \ge 12$, false. b) $2x < -8$, so $x < -4$; open circle at $-4$, shaded left. Check $x = -5$: $1 < 3$, true; $x = -4$: $3 < 3$, false. c) Dividing by $-6$ reverses: $x \ge -3$; closed circle at $-3$, shaded right. Check $x = -3$: $18 \le 18$, true; $x = -4$: $24 \le 18$, false. d) $\tfrac{x}{4} > 3$, so $x > 12$; open circle at 12, shaded right. Check $x = 16$: $3 > 2$, true; $x = 12$: $2 > 2$, false.
66. $x \le 4$; closed circle at 4, shaded left. Distributive property: $4x - 8 + 9 \le 17$. Combining like terms: $4x + 1 \le 17$. Subtraction property of inequality: $4x \le 16$. Division property of inequality, positive divisor: $x \le 4$. Check $x = 4$: $8 + 9 = 17 \le 17$, true; $x = 5$: $21 \le 17$, false.
67. $7 - 2x - 6 > 5$, so $1 - 2x > 5$, so $-2x > 4$, and dividing by $-2$ reverses: $x < -2$; open circle at $-2$, shaded left. Check $x = -3$: $7 - 2(0) = 7 > 5$, true; $x = -2$: $7 - 2(1) = 5 > 5$, false.
68. Subtracting $3x$ and 1: $5x \ge -15$, so $x \ge -3$; closed circle at $-3$, shaded right. Check $x = -3$: $-23 \ge -23$, true; $x = -4$: $-31 \ge -26$, false.
69. $-\tfrac{3}{4}x < -3$, and multiplying by $-\tfrac{4}{3}$ reverses: $x > 4$; open circle at 4, shaded right. Check $x = 5$: $-3.75 + 2 = -1.75 < -1$, true; $x = 4$: $-1 < -1$, false.
70. $10x - 15 \le 9x + 3$, so $x \le 18$; closed circle at 18, shaded left. Check $x = 18$: $165 \le 165$, true; $x = 19$: $175 \le 174$, false.
71. Adding $x$ and 9: $15 \ge 5x$, so $3 \ge x$, that is $x \le 3$; closed circle at 3, shaded left. Check $x = 3$: $3 \ge 3$, true; $x = 4$: $2 \ge 7$, false.
72. $20 > 5x$ gives $4 > x$, written variable-first as $x < 4$; open circle at 4, shaded left. Swapping the sides swaps the symbol. Check $x = 3$: $20 > 15$, true; $x = 4$: $20 > 20$, false.
73. $-8$ and $-3$. Solving: $-2x \ge 6$, and dividing by $-2$ reverses to $x \le -3$. Check $x = -3$: $6 + 5 = 11 \ge 11$, true; $x = 0$: $5 \ge 11$, false; $x = 2.5$: $0 \ge 11$, false; $x = 7$: $-9 \ge 11$, false.
74. $x > \tfrac{3}{2}$. One multistep inequality with the same solution set is $4x - 1 > 2x + 2$: subtracting $2x$ gives $2x - 1 > 2$, then $2x > 3$, then $x > \tfrac{3}{2}$. Check $x = 2$: $7 > 6$, true; $x = \tfrac{3}{2}$: $5 > 5$, false.
75. Let $h$ = the number of hours. $2{,}400 - 150h < 1{,}500$, so $-150h < -900$, and dividing by $-150$ reverses to $h > 6$; open circle at 6, shaded right. Check $h = 7$: $2{,}400 - 1{,}050 = 1{,}350 < 1{,}500$, true; $h = 6$: $1{,}500 < 1{,}500$, false. The hiker is below 1{,}500 feet after 6 hours of descending. The part of the graph to the right of 6 is real, but the ray as drawn continues forever, and in context the descent also cannot run backward in time — no part of the graph left of $h = 0$ describes a real moment, and the practical answer is $h > 6$ with $h$ measured forward from the start.
76. Reverse the symbol exactly when you multiply or divide both sides by a negative number, because that reflects both sides across zero and swaps which is larger. Every other move — adding or subtracting any number, multiplying or dividing by a positive number, distributing, combining like terms — leaves the direction alone. Example requiring a reversal: $-6x \le 18$ gives $x \ge -3$, since dividing by $-6$ reflects both sides; check $x = -3$: $18 \le 18$, true, and $x = -4$: $24 \le 18$, false. Example with many minus signs and no reversal: $-3x - 8 \le -2x - 2$ becomes $-8 \le x - 2$ after adding $3x$ to both sides, then $-6 \le x$, that is $x \ge -6$; every move was an addition, so the symbol never turned. Check $x = -6$: $18 - 8 = 10$ and $12 - 2 = 10$, and $10 \le 10$ is true; $x = -7$: $13$ and $12$, and $13 \le 12$ is false.

Part B — Verifying solutions algebraically, graphically, and with technology, and interpreting in context (A.EI.1f)

77. $-4x \le 16$, and dividing by $-4$ reverses: $x \ge -4$; closed circle at $-4$, shaded right. Inside, $x = 0$: $9 \le 25$, true. Boundary, $x = -4$: $9 + 16 = 25 \le 25$, true. Outside, $x = -5$: $29 \le 25$, false.
78. Graph $y = 3x - 2$ and $y = x + 4$. They cross where $3x - 2 = x + 4$, that is at $(3, 7)$. To the left of the crossing the line $y = 3x - 2$ is the lower of the two, which is exactly where $3x - 2 < x + 4$ holds, so the solution set is $x < 3$, and the crossing is excluded because there the two sides are equal. Check $x = 2$: $4 < 6$, true; $x = 3$: $7 < 7$, false.
79. From the table, $Y_1 \ge Y_2$ from $x = 3$ onward, and it fails at $x = 2$, where $5 \ge 6$ is false, so the boundary lies somewhere between 2 and 3. Algebraically: $2x + 1 \ge 8 - x$ gives $3x \ge 7$, so $x \ge \tfrac{7}{3} \approx 2.33$. The table alone could not give the exact boundary because the table lists only whole-number inputs and the boundary is not a whole number; it can bracket the boundary between two rows but not name it. Check $x = \tfrac{7}{3}$: both sides equal $\tfrac{17}{3}$, and $\ge$ accepts the tie; check $x = 2$: $5 \ge 6$, false.
80. Enter the left side as $Y_1$ and the right side as $Y_2$. Graph both and use the intersect feature to find the boundary value, which is the $x$-coordinate of the crossing; then look at which graph is higher on each side to decide the direction, since the inequality asks where one side exceeds the other. The table gives the same information numerically: read down the two columns and find the row where the comparison flips, and read the row at the boundary to decide whether an equality is accepted, which settles open versus closed. If the calculator and your algebra disagree, one of them is wrong, and the tiebreaker is substitution into the original inequality — check the boundary and one value on each side of it, then find the step where the disagreement began.
81. Let $c$ = the number of classes. $40 + 6c \le 70 + 2c$, so $4c \le 30$, so $c \le 7.5$. Check $c = 7$: Gym A costs $\$82$ and Gym B costs $\$84$, and $82 \le 84$ is true; $c = 8$: $\$88$ versus $\$86$, false. Classes are whole, so Gym A costs no more for 7 classes or fewer per month; from 8 classes on, Gym B is cheaper. The algebraic endpoint 7.5 is not a usable number of classes.
82. Let $d$ = the number of days. $90 - 6d > 30$, so $-6d > -60$, and dividing by $-6$ reverses to $d < 10$; open circle at 10, shaded left. Check $d = 9$: $90 - 54 = 36 > 30$, true; $d = 10$: $30 > 30$, false. The barrel holds more than 30 gallons for the first 10 days. The part of the graph to the left of 0 describes no real day, since days since the watering began cannot be negative, so in context the answer is $0 \le d < 10$.
83. Let $b$ = the number of boxes. $7.50b + 12.00 \le 60.00$, so $7.50b \le 48.00$, so $b \le 6.4$. Check $b = 6$: $45.00 + 12.00 = 57.00 \le 60.00$, true; $b = 7$: $52.50 + 12.00 = 64.50 \le 60.00$, false. The shop may ship 6 whole boxes.
84. In item 57 the endpoint 28 is itself a whole number of crates and the symbol is inclusive, so 28 is part of the answer: a load of exactly 28 crates weighs exactly 2{,}000 pounds, which the limit allows. In item 83 the endpoint 6.4 is not a whole number of boxes, so it cannot be shipped at all; the answer is the largest whole number below it, 6. The test that settles each is the same: substitute the two whole numbers nearest the boundary into the original inequality. For crates, 28 passes and 29 fails, so 28 is the edge; for boxes, 6 passes and 7 fails, so 6 is the edge.
85. The hiker is below 1{,}500 feet after 6 hours have passed, that is, from just after $h = 6$ onward. At exactly $h = 6$ the elevation is $2{,}400 - 150(6) = 1{,}500$ feet, which is not below 1{,}500, so the strict symbol correctly excludes that instant and the endpoint is drawn as an open circle.
86. A single inside test only shows that his set contains at least one genuine solution; it cannot show that the set contains nothing but solutions, or that the boundary is in the right place. He is missing an outside test, which must come out false, and a boundary test, which decides open versus closed. A wrong answer his test would not catch: if the true solution set were $x > 8$, then $x = 10$ still makes the original true, so the single test passes while the claimed set $x > 5$ wrongly includes every value from 5 to 8.
87. Test $x = 0$: $5 - 0 = 5$, and $5 > 21$ is false — yet Marcus's answer $x > -4$ includes 0, so his set contains non-solutions. He divided both sides by $-4$ without reversing. Correct: $-4x > 16$ gives $x < -4$; open circle at $-4$, shaded left. Check $x = -5$: $5 + 20 = 25 > 21$, true; $x = -4$: $21 > 21$, false.
88. Sample situation: A summer camp charges a $\$25$ registration fee plus $\$4$ for each craft session, and a family will spend at most $\$81$. How many sessions can a camper attend? Solving: $25 + 4x \le 81$, so $4x \le 56$, so $x \le 14$. Boundary, $x = 14$: $25 + 56 = 81 \le 81$, true. Outside, $x = 15$: $85 \le 81$, false. Sessions are whole and cannot be negative, so the camper can attend at most 14 sessions, and 14 sessions spends the budget exactly.