Appendix A — Answer Key, Chapter 11: One- and Two-Step Inequalities

SOL 7.PFA.4 · Covers textbook Chapter 11 and the companion workbook. Item numbers match the textbook and run continuously from 1 to 130; workbook items are the same problems with the same numbers, so this key serves both books. Reasoning answers show an acceptable response, not the only wording; for "create a situation" items the wording will vary, and a sample is given.

Reading the graph answers. A key cannot draw, so every graphing answer states three things: the endpoint value, whether the circle is open (for $<$ or $>$) or closed (for $\le$ or $\ge$), and the shading direction (right for $>$ or $\ge$, left for $<$ or $\le$), always with an arrowhead showing the shading continues.

Lesson 11.1 — What an Inequality Says

Guided practice

1. "$x$ is less than or equal to 7." Yes, $x = 7$ is a solution, because $7 \le 7$ is true — the symbol $\le$ includes equality.
2. $x = 5$: $9 > 10$ is false. $x = 6$: $10 > 10$ is false. $x = 7$: $11 > 10$ is true.
3. $x = -2$: $-5 \le 5$ true. $x = 0$: $-1 \le 5$ true. $x = 3$: $5 \le 5$ true. $x = 4$: $7 \le 5$ false. Solutions: $-2$, $0$, and $3$.
4. Let $w$ = the weight of the backpack in pounds. $w \le 15$. Yes, a 15-pound backpack is allowed, because "at most" includes the boundary and $15 \le 15$ is true.

Independent practice

5. a) $n > -4$ b) $n \ge 9$ c) $n \le 2.5$ d) $n < 30$
6. a) $4(3) + 1 = 13$ and $13 > 13$ is false, so no. b) $13 \ge 13$ is true, so yes. c) $2(-5) + 3 = -7$ and $-7 \ge -7$ is true, so yes. d) $6\left(\tfrac{1}{2}\right) - 2 = 1$ and $1 < 1$ is false, so no.
7. $x = -4$: $-10 < 8$ true. $x = -1$: $-1 < 8$ true. $x = 0$: $2 < 8$ true. $x = 2$: $8 < 8$ false. $x = 5$: $17 < 8$ false. Solutions: $-4$, $-1$, and $0$.
8. Sample: $-2$, $0$, and $3.5$. Any three values greater than or equal to $-2$ are acceptable, provided one is not an integer; $-2$ itself is allowed because the symbol is inclusive.
9. The equation $x = 6$ names one exact number, so only 6 satisfies it. The inequality $x < 6$ describes a whole region of the number line, and between any two numbers below 6 there is always another one, so the solutions never run out.
10. Let $h$ = a rider's height in inches. $h \ge 48$. Yes, a rider exactly 48 inches tall may ride, because "at least 48" includes 48 and $48 \ge 48$ is true.
11. The number is 5. It is a solution of $x \ge 5$ because $5 \ge 5$ is true, and it is not a solution of $x > 5$ because $5 > 5$ is false. Every other value behaves the same way in both.
12. Miguel computed the left side correctly but then read the symbol wrongly: the sentence he produced is $17 < 17$, which is false, not true. So $x = 4$ is not a solution of $5x - 3 < 17$. It would be a solution of $5x - 3 \le 17$, since $17 \le 17$ is true.

Exit ticket 11.1

13. $s \le 28$
14. $x = -3$: $1 \ge 6$ false. $x = 1$: $5 \ge 6$ false. $x = 2$: $6 \ge 6$ true. $x = 6$: $10 \ge 6$ true. Solutions: $2$ and $6$.
15. $-3(-2) = 6$, and $6 < 6$ is false. So $x = -2$ is not a solution.
16. The solution set is every value of the variable that makes the inequality true. It usually contains infinitely many numbers, so it is described with symbols or drawn on a number line rather than listed.

Lesson 11.2 — Solving One-Step Inequalities

Every solution below has been checked with one value from inside the solution set and one from outside; those two tests are shown wherever the item asks for them.

Guided practice

17. Subtraction property of inequality. $x > 6$. Inside test $x = 10$: $19 > 15$ is true. Outside test $x = 6$: $15 > 15$ is false. (Any inside value greater than 6 and any outside value 6 or less is acceptable.)
18. $y \le 7$. Boundary test $y = 7$: $7 - 3 = 4$ and $4 \le 4$ is true, so 7 belongs to the solution set.
19. $x < 7$. Inside test $x = 0$: $0 < 42$ is true. Outside test $x = 7$: $42 < 42$ is false.
20. $x \ge 15$. Boundary test $x = 15$: $\tfrac{15}{5} = 3$ and $3 \ge 3$ is true.

Independent practice

21. a) $x \le 8$ b) $m > 5$ c) $x < 5.5$ d) $w \ge \tfrac{3}{4}$
22. a) $x \ge 7$ b) $n < 24$ c) $x > 14$ d) $x \le 12$
23. $8 > x$, which written with the variable first is $x < 8$.
24. $x \le 0$. Sample solutions: $0$, $-1$, and $-7.5$.
25. $x \ge -5$. Inside test $x = -5$: $2(-5) = -10$ and $-10 \ge -10$ is true. Outside test $x = -6$: $-12 \ge -10$ is false.
26. Let $c$ = the number of crates. $60c \le 1500$, so $c \le 25$. Crates come in whole numbers, so the van can carry at most 25 crates. Check: $60(25) = 1500 \le 1500$ is true, and $60(26) = 1560 \le 1500$ is false.
27. Both begin by subtracting 7 from both sides, because subtraction undoes addition either way. The equation gives the single answer $x = 5$; the inequality gives $x < 5$, which is every number below 5 and does not include 5 itself.
28. The variable was divided by 2, so it must be multiplied by 2, not divided again. The correct solution is $x > 12$. Nadia's answer includes $x = 8$, but $\tfrac{8}{2} = 4$ and $4 > 6$ is false, so 8 is not a solution — her solution set is far too large.

Exit ticket 11.2

29. $x \ge 8$. Boundary test $x = 8$: $8 - 6 = 2$ and $2 \ge 2$ is true.
30. $x < 7$
31. $x \le -12$
32. The subtraction property of inequality (subtract 4 from both sides). The solution is $x > 5$.

Lesson 11.3 — The Reversal Rule

Guided practice

33. $-3 > -7$
34. $2 > -5$
35. $-2x < 14$ gives $x > -7$ after dividing by $-2$ and reversing. Test $x = 0$: $-2(0) = 0$ and $0 < 14$ is true, and 0 is in the solution set. Test $x = -7$: $-2(-7) = 14$ and $14 < 14$ is false, and $-7$ is correctly excluded.
36. $-\dfrac{x}{4} \ge 3$ gives $x \le -12$ after multiplying by $-4$ and reversing. Boundary test $x = -12$: $-\tfrac{-12}{4} = 3$ and $3 \ge 3$ is true. Outside test $x = -8$: $-\tfrac{-8}{4} = 2$ and $2 \ge 3$ is false.

Independent practice

37. a) $-2 > -5$ b) $-80 < -50$ c) Only part b reversed. Adding $-10$ slid both numbers 10 units left, which cannot change which one is farther right. Multiplying by $-10$ reflected both across zero, which swaps their order, so the symbol had to reverse to keep the statement true.
38. a) $x \ge -7$ b) $n < 7$ c) $x \le -10$ d) $x > -8$
39. a) $x > 5$ (no reversal — you add 9 to both sides) b) $x < \tfrac{4}{9}$ (reversal — you divide by $-9$) c) $x \le 5$ (no reversal — you add 6 to both sides) d) $x \ge -12$ (reversal — you multiply by $-\tfrac{3}{2}$)
40. $x > -3$. Inside test $x = 0$: $-0 = 0$ and $0 < 3$ is true. Outside test $x = -3$: $-(-3) = 3$ and $3 < 3$ is false.
41. a) No. Subtraction slides both sides equally and never reflects across zero. b) Yes. Dividing by a negative reflects both sides across zero, which swaps their order. c) No. Adding a negative is still addition; both sides slide the same distance in the same direction. d) No. $\tfrac{1}{2}$ is positive, so both sides shrink toward zero without either crossing it.
42. Let $t$ = the number of seconds, with $t \ge 0$. $-3t < -45$. Divide both sides by $-3$ and reverse: $t > 15$. Check $t = 16$: $-48 < -45$ is true. Check $t = 15$: $-45 < -45$ is false. The submarine's elevation is below $-45$ meters after more than 15 seconds of descending.
43. $-4 \cdot \left(-\tfrac{1}{2}\right) = 2$ and $6 \cdot \left(-\tfrac{1}{2}\right) = -3$, so the true statement is $2 > -3$. On the number line $-4$ started to the left of 6, but after multiplying by a negative the results are 2 and $-3$, and 2 is now to the right of $-3$. Multiplying by a negative reflects both numbers across zero, which reverses their left-to-right order, so the symbol must reverse as well.
44. Test $x = -10$: $-5(-10) = 50$ and $50 \ge 30$ is true, so $-10$ is a solution — but Cleo's answer $x \ge -6$ leaves it out. Test $x = 0$: $-5(0) = 0$ and $0 \ge 30$ is false, so 0 is not a solution — but Cleo's answer includes it. Her solution set is wrong in both directions because she did not reverse the symbol. The correct solution is $x \le -6$.

Exit ticket 11.3

45. $x > -6$
46. $x \le 6$
47. $-30 < -6$
48. Subtracting a negative number slides both sides the same distance along the number line in the same direction, and sliding two points together never changes which one is farther right, so the comparison stays the same. Dividing both sides by a negative number reflects both sides across zero, and a reflection swaps which one is farther right, so the symbol must reverse or the statement would become false.

Lesson 11.4 — Solving Two-Step Inequalities

Guided practice

49. Subtract 3 from both sides (subtraction property of inequality) to get $2x < 8$, then divide both sides by 2, which is positive, so no reversal (division property of inequality): $x < 4$. Test $x = 0$: $2(0) + 3 = 3$ and $3 < 11$ is true. Test $x = 4$: $2(4) + 3 = 11$ and $11 < 11$ is false.
50. $5x \ge 20$, so $x \ge 4$. Boundary test $x = 4$: $5(4) - 4 = 16$ and $16 \ge 16$ is true. Outside test $x = 3$: $15 - 4 = 11$ and $11 \ge 16$ is false.
51. $-3x \le 12$, then dividing both sides by $-3$ reverses the symbol: $x \ge -4$. The reversal happens at the second step, the division by $-3$. Boundary test $x = -4$: $-3(-4) + 2 = 12 + 2 = 14$ and $14 \le 14$ is true.
52. $\dfrac{x}{2} > -2$, so $x > -4$. Test $x = 0$: $0 + 6 = 6$ and $6 > 4$ is true. Test $x = -4$: $-2 + 6 = 4$ and $4 > 4$ is false.

Independent practice

53. a) $x \le 6$ b) $x > 3$ c) $x \ge -6$ d) $x < 9$
54. a) $x < -4$ b) $x \ge -6$ c) $x \le 3$ d) $x > -4$. In each part the reversal happens at the second step, when both sides are divided or multiplied by the negative coefficient.
55. $4x > 7$, so $x > \tfrac{7}{4}$. Check $x = 2$: $8 + 3 = 11$ and $11 > 10$ is true. Check $x = \tfrac{7}{4}$: $7 + 3 = 10$ and $10 > 10$ is false, so the boundary is correctly excluded.
56. $x \le 7$
57. $24 \ge 4x$, so $6 \ge x$, which written with the variable first is $x \le 6$.
58. Let $m$ = the number of months. $12m + 20 \le 100$, so $12m \le 80$ and $m \le \tfrac{20}{3} = 6\tfrac{2}{3}$. Months are counted in whole numbers, so Marisol can afford at most 6 months. Check: $12(6) + 20 = 92$ and $92 \le 100$ is true; $12(7) + 20 = 104$ and $104 \le 100$ is false.
59. $-5x \ge 25$, then $x \le -5$. The direction reverses at the second step, where both sides are divided by $-5$. It does not reverse at the first step because subtracting 1 from both sides only slides both sides the same distance and does not reflect anything across zero.
60. Test $x = 0$ in the original: $-3(0) + 4 = 4$ and $4 < 19$ is true, so 0 is a solution — yet Theo's answer $x < -5$ excludes it. He divided by $-3$ without reversing the symbol. The correct solution is $x > -5$.

Exit ticket 11.4

61. $x \le 5$
62. $x < -4$
63. $x \ge -8$
64. Same: you undo the constant term first and the coefficient second, using the same inverse operations and the same order, and the boundary number you get is the solution of the matching equation. Different: an equation gives one number while an inequality gives a whole range, and dividing or multiplying by a negative number reverses the inequality symbol, something that never happens with an equation.

Lesson 11.5 — Graphing Solution Sets on a Number Line

Guided practice

65. Endpoint 4, open circle, shade left.
66. Endpoint $-2$, closed circle, shade right.
67. $x \le 4$. Endpoint 4, closed circle, shade left.
68. $x < 6$. Endpoint 6, open circle, shade left.

Independent practice

69. a) Endpoint $-5$, open circle, shade right. b) Endpoint 1, closed circle, shade left. c) Endpoint 0, closed circle, shade right. d) Endpoint $-1$, open circle, shade left.
70. a) $x < 4$: endpoint 4, open circle, shade left. b) $x \ge -2$: endpoint $-2$, closed circle, shade right.
71. $x \ge -5$. Endpoint $-5$, closed circle, shade right. (Dividing by $-2$ reverses the symbol.) Check $x = -5$: $-2(-5) = 10$ and $10 \le 10$ is true.
72. $x > 4$. Endpoint 4, open circle, shade right. Check $x = 4$: $8 - 3 = 5$ and $5 > 5$ is false, so the open circle is correct.
73. $x > -3$. Sample values in the set: $-2$ and $0$. Sample value not in the set: $-3$ itself, or $-5$.
74. Let $h$ = a car's height in feet. $h \le 6.5$. Endpoint 6.5, closed circle, shade left. The part of the graph at or below zero does not describe any real car, since a car cannot have a height of zero or a negative height; the meaningful part is $0 < h \le 6.5$.
75. The equation $x = 3$ names one exact number, so its graph is the single point at 3. The inequality $x \ge 3$ describes 3 together with every number above it, so its graph is a ray starting at 3 and continuing forever to the right. In the equation graph the mark at 3 is the entire solution; in the inequality graph the filled circle at 3 marks where the solution set begins and shows that 3 is included in it.
76. Error 1: the circle should be closed, not open, because $\le$ includes the boundary — test $x = -2$, which gives $-2 \le -2$, true, so $-2$ is a solution and cannot be shown hollow. Error 2: the shading should run left, not right, because the solutions are the numbers $-2$ and below — test $x = 0$, which gives $0 \le -2$, false, so nothing to the right of $-2$ belongs. The correct graph is a closed circle at $-2$ with shading to the left.

Exit ticket 11.5

77. Endpoint $-4$, open circle, shade right.
78. $x \le 3$. Endpoint 3, closed circle, shade left.
79. $x \le 2$
80. Use an open circle for $<$ or $>$ and a closed circle for $\le$ or $\ge$. The reason is that the circle records whether the boundary number itself passes the test. For $x > 3$, substituting 3 gives $3 > 3$, which is false, so 3 is excluded and the circle is hollow. For $x \ge 3$, substituting 3 gives $3 \ge 3$, which is true, so 3 is included and the circle is filled.

Lesson 11.6 — Writing and Solving Inequalities in Context

Guided practice

81. Let $t$ = the number of tickets. $9t + 6 \le 42$, so $9t \le 36$ and $t \le 4$. Tickets are whole, so Nia can buy at most 4 tickets. Check: $9(4) + 6 = 42 \le 42$ is true; $9(5) + 6 = 51 \le 42$ is false.
82. $t > -5$
83. Let $b$ = the number of books. $30 - 4b \ge 10$, so $-4b \ge -20$ and, reversing when dividing by $-4$, $b \le 5$. Books are whole, so Rosa can buy at most 5 books. Check: $30 - 4(5) = 10 \ge 10$ is true; $30 - 4(6) = 6 \ge 10$ is false.
84. Sample: A movie ticket costs $2 and one bag of popcorn costs $3. Dev has at most $15 to spend on tickets and one popcorn. How many tickets can he buy? Wording will vary. The solution is $2x \le 12$, so $x \le 6$; with whole tickets, at most 6. Check: $2(6) + 3 = 15 \le 15$ is true.

Independent practice

85. Let $h$ = the number of hours parked. $2h + 4 \le 18$, so $2h \le 14$ and $h \le 7$. Jamal can park for at most 7 hours. Check: $2(7) + 4 = 18 \le 18$ is true; $2(8) + 4 = 20 \le 18$ is false.
86. $3n + 6 \ge 27$, so $3n \ge 21$ and $n \ge 7$. Check: $3(7) + 6 = 27 \ge 27$ is true; $3(6) + 6 = 24 \ge 27$ is false.
87. $2n - 5 < 9$, so $2n < 14$ and $n < 7$. Check: $2(6) - 5 = 7 < 9$ is true; $2(7) - 5 = 9 < 9$ is false.
88. Let $b$ = the number of boxes. $55b + 180 \le 1500$, so $55b \le 1320$ and $b \le 24$. Boxes are whole, so at most 24 boxes may be loaded. Check: $55(24) + 180 = 1320 + 180 = 1500 \le 1500$ is true; $55(25) + 180 = 1555 \le 1500$ is false.
89. Let $t$ = the number of tacos. $2.5t + 1 \le 14$, so $2.5t \le 13$ and $t \le 5.2$. Tacos come in whole numbers and cannot be negative, so test the two whole numbers nearest the boundary: $2.5(5) + 1 = 13.50$ and $\$13.50 \le \$14.00$ is true, while $2.5(6) + 1 = 16$ and $\$16.00 \le \$14.00$ is false. Priya can buy at most 5 tacos.
90. Sample: A 50-gallon tank drains 4 gallons every minute. For how many minutes does it still hold at least 10 gallons? Wording will vary. Solving: $-4x \ge -40$, and dividing by $-4$ reverses the symbol, giving $x \le 10$. With $x \ge 0$, the tank holds at least 10 gallons for the first 10 minutes. Check: $-4(10) + 50 = 10 \ge 10$ is true; $-4(11) + 50 = 6 \ge 10$ is false.
91. Sample: A fundraiser has $15 in donations and earns $6 for each wristband sold. How many wristbands must be sold for the total to be more than $51? Wording will vary. Solving: $6x > 36$, so $x > 6$. The symbol is strict, so 6 itself does not work — check: $6(6) + 15 = 51$ and $51 > 51$ is false, while $6(7) + 15 = 57$ and $57 > 51$ is true. Since wristbands are whole objects, the smallest number that works is 7, so the practical answer is at least 7, not at least 6.
92. Bo's version multiplies the $5 shipping fee by the number of shirts, but shipping is charged once for the whole order, not once per shirt. The correct inequality is $12x + 5 \le 65$, so $12x \le 60$ and $x \le 5$. Check the largest whole number: $12(5) + 5 = 65$ and $\$65 \le \$65$ is true; $12(6) + 5 = 77$ and $\$77 \le \$65$ is false. Ana can buy at most 5 shirts.

Exit ticket 11.6

93. Let $m$ = the number of miles. $2m + 3 \le 21$, so $2m \le 18$ and $m \le 9$. Lena can ride at most 9 miles. Check: $2(9) + 3 = 21 \le 21$ is true.
94. $5n + 4 \le 39$, so $5n \le 35$ and $n \le 7$. Check: $5(7) + 4 = 39 \le 39$ is true; $5(8) + 4 = 44 \le 39$ is false.
95. Sample: A caterer charges $7 per guest plus a $10 setup fee, and a family can spend at most $80. How many guests can they invite? Wording will vary. The solution is $7x \le 70$, so $x \le 10$; with whole guests, at most 10. Check: $7(10) + 10 = 80 \le 80$ is true.
96. You round down when the variable counts objects that cannot be split and the inequality puts a ceiling on it, so that the next whole number up would break the limit. Sample: with $26 at a carnival charging $7 admission plus $3 per ride, $3r + 7 \le 26$ gives $r \le 6\tfrac{1}{3}$; since $3(7) + 7 = 28$ exceeds $26 but $3(6) + 7 = 25$ does not, the answer is at most 6 rides.

Chapter 11 Review

Part A — Applying properties of inequality (7.PFA.4a)

97. a) $x \le 7$ b) $m > 4$ c) $x \ge 7$ d) $x < -15$
98. a) $x \le 6$ b) $x > 3$ c) $x \ge -6$ d) $x < 12$
99. $0.4x \ge 2.4$, so $x \ge 6$. Check: $0.4(6) + 1.2 = 2.4 + 1.2 = 3.6 \ge 3.6$ is true; $0.4(5) + 1.2 = 3.2 \ge 3.6$ is false.
100. $30 \ge 6x$, so $5 \ge x$, which written with the variable first is $x \le 5$.
101. $5x < 7$, so $x < \tfrac{7}{5}$. Check: $x = 1$ gives $7 < 9$, true; $x = \tfrac{7}{5}$ gives $7 + 2 = 9$ and $9 < 9$, false.

Part B — Multiplying or dividing by a negative (7.PFA.4b)

102. a) $-9 < -6$ b) $-18 > -27$ c) Only part b reversed. Adding $-15$ slid both numbers 15 units to the left, leaving 6 still to the left of 9. Multiplying by $-3$ reflected both across zero, which swapped their order, so the symbol had to reverse.
103. a) $x \ge -4$ b) $x < -7$ c) $x \le -10$ d) $x > -15$
104. $-6x \ge -24$, so $x \le 4$. Boundary test $x = 4$: $-6(4) + 5 = -24 + 5 = -19$ and $-19 \ge -19$ is true. Outside test $x = 5$: $-30 + 5 = -25$ and $-25 \ge -19$ is false.
105. Start with $2 < 6$. Multiplying both sides by $-1$ gives $-2$ and $-6$. On the number line, multiplying by a negative reflects every point across zero, and a reflection turns left into right: 2 was to the left of 6, so $-2$ lands to the right of $-6$. Since $-2$ is now the larger number, the true statement is $-2 > -6$, with the symbol reversed. Adding a negative does not reverse anything because it slides both numbers the same distance in the same direction instead of reflecting them: adding $-10$ to $2 < 6$ gives $-8 < -4$, and $-8$ is still to the left of $-4$.

Part C — Representing solutions algebraically and graphically (7.PFA.4c)

106. Endpoint $-1$, closed circle, shade left.
107. Endpoint 3.5, open circle, shade right.
108. $4x \le 12$, so $x \le 3$. Endpoint 3, closed circle, shade left. Check $x = 3$: $12 - 6 = 6 \le 6$ is true.
109. $-3x < 9$, and dividing by $-3$ reverses the symbol, so $x > -3$. Endpoint $-3$, open circle, shade right. Check $x = -3$: $9 + 1 = 10$ and $10 < 10$ is false, so the open circle is correct; check $x = 0$: $1 < 10$ is true.
110. $x \ge -4$. Sample solutions: $-4$ and $0$. (Any two values greater than or equal to $-4$ are acceptable.)

Part D — Writing an inequality from a situation (7.PFA.4d)

111. $w \le 50$
112. $p + 40 \le 250$, so $p \le 210$. At most 210 additional people may be seated. Check: $210 + 40 = 250 \le 250$ is true; $211 + 40 = 251 \le 250$ is false.
113. Let $h$ = the number of hours. $4h + 6 \le 30$, so $4h \le 24$ and $h \le 6$. Check: $4(6) + 6 = 30 \le 30$ is true; $4(7) + 6 = 34 \le 30$ is false.

Part E — Creating a situation from an inequality (7.PFA.4e)

114. Sample: A class buys notebooks at $3 each plus a one-time $12 shipping charge, and it can spend at most $45. How many notebooks can it buy? Wording will vary. Solution: $3x \le 33$, so $x \le 11$; with whole notebooks, at most 11. Check: $3(11) + 12 = 45 \le 45$ is true.
115. Sample: A student earns $10 for each hour of yard work and needs to earn at least $250 for a trip. How many hours must the student work? Wording will vary. Solution: $x \ge 25$; with whole hours, at least 25. Check: $10(25) = 250 \ge 250$ is true; $10(24) = 240 \ge 250$ is false.
116. Sample: A 60-gallon rain barrel loses 5 gallons each day to watering. For how many days does it still hold more than 20 gallons? Wording will vary. Solution: $-5x > -40$, and dividing by $-5$ reverses the symbol, giving $x < 8$; with $x \ge 0$, the barrel holds more than 20 gallons for the first 8 days. Check: $-5(7) + 60 = 25 > 20$ is true; $-5(8) + 60 = 20 > 20$ is false.

Part F — Solving problems in context (7.PFA.4f)

117. Let $p$ = the number of pizzas. $13p + 8 \le 150$, so $13p \le 142$ and $p \le \tfrac{142}{13} = 10\tfrac{12}{13}$. Pizzas are whole, so the club can order at most 10 pizzas. Check: $13(10) + 8 = 138$ and $\$138 \le \$150$ is true; $13(11) + 8 = 151$ and $\$151 \le \$150$ is false.
118. Let $L$ = the number of lawns. $14L + 62 \ge 160$, so $14L \ge 98$ and $L \ge 7$. Lawns are whole, so Mia must mow at least 7 lawns. Check: $14(7) + 62 = 98 + 62 = 160 \ge 160$ is true; $14(6) + 62 = 146 \ge 160$ is false.
119. Let $m$ = the number of minutes, with $m \ge 0$. $40 - 6m > 10$, so $-6m > -30$ and, reversing when dividing by $-6$, $m < 5$. The tank holds more than 10 gallons for the first 5 minutes, that is, for $0 \le m < 5$. Check: at $m = 4$, $40 - 24 = 16 > 10$ is true; at $m = 5$, $40 - 30 = 10 > 10$ is false.

Part G — Identifying values in a solution set (7.PFA.4g)

120. $x = -6$: $2(-6) + 5 = -7$ and $-7 \le 11$ true. $x = -2$: $1 \le 11$ true. $x = 0$: $5 \le 11$ true. $x = 3$: $11 \le 11$ true. $x = 8$: $21 \le 11$ false. Solutions: $-6$, $-2$, $0$, and $3$.
121. $x = -5$: $-4(-5) = 20$ and $20 > 12$ true. $x = -3$: $12 > 12$ false. $x = 0$: $0 > 12$ false. $x = 4$: $-16 > 12$ false. Solution: $-5$ only.
122. The solution set is $x \ge 3$. Sample values: $3$, $4.5$, and $10$. Check each: $3(3) - 2 = 7 \ge 7$ true; $3(4.5) - 2 = 11.5 \ge 7$ true; $3(10) - 2 = 28 \ge 7$ true.

Part H — Comparing inequalities and equations (7.PFA.4h)

123. $4x + 6 = 26$ gives $x = 5$. $4x + 6 \le 26$ gives $x \le 5$. Related: both have the same boundary number, 5, and both are solved by the same two moves — subtract 6, then divide by 4. Different: the equation has exactly one solution, graphed as a single point at 5, while the inequality has infinitely many, graphed as a closed circle at 5 with shading to the left.
124. In common: both are solved by undoing the constant term first and the coefficient second, and both use the same inverse operations applied to both sides; also, the same properties of real numbers, such as the additive inverse and multiplicative identity properties, justify the steps in either case. Different: an equation of this type has exactly one solution while an inequality has infinitely many, and multiplying or dividing both sides by a negative number reverses the inequality symbol, which has no counterpart when solving an equation. A further difference is in how the answer is displayed: one point versus a shaded ray.
125. Solving $2x = 8$ asks which single number doubles to exactly 8, and only 4 does, so its graph is one point at 4. Solving $2x \le 8$ asks which numbers double to 8 or less, and every number 4 or below qualifies — 4, 3, 0, $-11$, $2.5$, and so on without end — so its graph is a closed circle at 4 with shading running left forever.

Part I — Mixed application and reasoning

126. $-2x \le -6$, and dividing by $-2$ reverses the symbol, so $x \ge 3$. Endpoint 3, closed circle, shade right. Boundary test $x = 3$: $-2(3) + 9 = -6 + 9 = 3$ and $3 \le 3$ is true. Outside test $x = 2$: $-4 + 9 = 5$ and $5 \le 3$ is false.
127. Test $x = 0$: $-3(0) = 0$ and $0 < 9$ is true, so 0 is a solution — but the student's answer $x < -3$ leaves 0 out, which proves the answer is wrong. The student divided by $-3$ without reversing the symbol. The correct solution is $x > -3$.
128. The one-step inequality is $x \ge -2$. A different two-step inequality with the same solution set is $3x + 4 \ge -2$: subtracting 4 gives $3x \ge -6$, and dividing by 3 gives $x \ge -2$. Check the boundary: $3(-2) + 4 = -6 + 4 = -2$ and $-2 \ge -2$ is true; check outside: $3(-3) + 4 = -5$ and $-5 \ge -2$ is false. (Other correct answers exist, such as $-2x + 1 \le 7$.)
129. Let $b$ = the number of benches of students. $4b + 8 \le 44$, so $4b \le 36$ and $b \le 9$. Benches are whole, so at most 9 benches may be filled. Check: $4(9) + 8 = 36 + 8 = 44 \le 44$ is true; $4(10) + 8 = 48 \le 44$ is false.
130. The number that separates them is 4. Their graphs are identical except at the endpoint: $x > 4$ has an open circle at 4, and $x \ge 4$ has a closed circle at 4; both shade to the right. Sample situations: a sign reading "children over 4 years old may enter" is $x > 4$, since a child who is exactly 4 is turned away, while "you must be at least 4 years old to enter" is $x \ge 4$, since a child who is exactly 4 is admitted.

Workbook-only items

Page 2, symbol table. $<$ is less than, strict. $>$ is greater than, strict. $\le$ is less than or equal to, inclusive. $\ge$ is greater than or equal to, inclusive.

Page 2, fill in the blanks. $<$ and $>$ are strict; the boundary number is not a solution. $\le$ and $\ge$ are inclusive; the boundary number is a solution. The wide end opens toward the larger amount.

Page 2, words into symbols. at least → $\ge$, included. at most → $\le$, included. more than → $>$, not included. fewer than → $<$, not included. no more than → $\le$, included. exceeds → $>$, not included.

Page 3, fill in the blanks. A solution; the solution set; the equation has one solution, the inequality infinitely many; substitute into the original inequality and decide whether the sentence is true.

Page 3, test table. $-7 \ge -4$ false. $-4 \ge -4$ true. $0 \ge -4$ true. $5 \ge -4$ true.

Page 6, properties. Addition: the same direction. Subtraction: the direction stays the same. Multiplication: by the same positive number. Division: by the same positive number.

Page 6, why the moves are safe. Adding the same number slides both points the same distance, so which one is on the left does not change. Multiplying by a positive number never moves a point across zero.

Page 6, the two-sided check. Substitute one value from inside the solution set (it should be true) and one from outside (it should be false).

Page 9, run the experiment. The left becomes $-2$ and the right becomes $-6$. On the number line $-2$ is to the right of $-6$, so $-2$ is the larger number. The true statement is $-2 > -6$; the direction reversed. Multiplying by $-1$ reflects every point across zero, and a reflection turns left into right.

Page 9, table. $4 < 10$ times $-2$ gives $-8$ and $-20$, so $-8 > -20$. $9 > 3$ times $-1$ gives $-9$ and $-3$, so $-9 < -3$. $-1 < 5$ times $-3$ gives $3$ and $-15$, so $3 > -15$. $12 > -4$ times $-\tfrac{1}{2}$ gives $-6$ and $2$, so $-6 < 2$.

Page 10, fill in the blanks. $2 < 6$ plus $-10$ gives $-8 < -4$. Both points slid 10 units left, so the direction did not change. Adding a negative slides the points; multiplying by a negative reflects them across zero.

Page 10, the reversal rule. Reverse only when you multiply or divide both sides by a negative number. In every other case the direction stays the same.

Page 10, reverses or not. add $-8$ → no. divide by $-4$ → yes. subtract 8 → no. multiply by $\tfrac{1}{2}$ → no. multiply by $-\tfrac{1}{2}$ → yes.

Page 13, the procedure. Name the coefficient and the constant term. Undo the addition or subtraction first; the direction never changes at this step. Then undo the multiplication or division; ask whether that number is negative, and if so reverse the symbol. Check with a value from inside the solution set and one from outside.

Page 13, where the boundary comes from. The boundary value is the solution of the matching equation.

Page 16, fill in the blanks. The three pieces are the endpoint, the type of circle, and the shading direction. Use an open circle for $<$ or $>$; the boundary is not a solution. Use a closed circle for $\le$ or $\ge$; the boundary is a solution. Shade to the right for $>$ or $\ge$ and to the left for $<$ or $\le$.

Page 16, summary table. $x < 1$: open circle, shade left. $x \le 1$: closed circle, shade left. $x > -2$: open circle, shade right. $x \ge -2$: closed circle, shade right.

Page 16, variable first. $6 > x$ means the same as $x < 6$. Swap the sides and swap the symbol.

Page 20, the path. 2. Name the unknown, with units. 3. Find the per-unit amount and the fixed amount. 4. Choose the symbol. 5. Check against the story.

Page 20, whole-number constraints. The value must be a whole number and cannot be negative. $t \le 5.2$ gives at most 5 tacos. $w \ge 8.4$ gives at least 9 weeks. The rounding went in different directions, so test the two whole numbers nearest the boundary.