Appendix A — Answer Key, Chapter 9: Linear Inequalities in Two Variables and Their Systems

SOL A.EI.2 (d, e, f, g, h) · Covers textbook Chapter 9 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 118 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Conventions used in every answer below. A solution of a linear inequality in two variables is an ordered pair, and the solution set is a shaded half plane. A strict symbol ($<$, $>$) gets a dashed boundary; an inclusive symbol ($\le$, $\ge$) gets a solid one. The shaded side is decided by a test point, the origin whenever the boundary misses it, and every region reported here has been confirmed by at least one substitution. A system is satisfied only by pairs that satisfy both inequalities, and its corner is the solution of the system of the two boundary equations. Contextual models carry $x \ge 0$ and $y \ge 0$ whenever the quantities cannot be negative. Nothing in this chapter is calculator-free; technology is used as a third check, as A.EI.2h requires.

The situations and figures used repeatedly, for reference:

• Figure 1 contrasts $x > 2$ on a number line with $y > x + 1$ on a plane
• Figures 2 and 3 are $y > x + 1$ and $y \lessgtr \tfrac12 x + 2$, used for testing points and for the boundary styles
• Figure 4 is $y \ge -x + 3$ beside $y \le -x + 3$; Figure 5 is $3x + 2y \le 12$; Figure 6 is $y < 3$ beside $x \ge -2$
• Figure 7 is the fair budget, $5x + 8y \le 80$, with $x$ ride bands at $\$5$ and $y$ game passes at $\$8$
• Figures 8 and 9 are the system $y \le -x + 5$, $y > 2x - 4$, with corner $(3,2)$
• Figures 10 and 11 are the work-schedule system $x + y \le 12$, $8x + 12y \ge 96$, with $x$ hours babysitting at $\$8$ and $y$ hours tutoring at $\$12$, whose feasible region has corners $(0,8)$, $(0,12)$, and $(12,0)$

Lesson 9.1 — From an Interval to a Region

Guided practice

1. A solution of $x > 2$ is a single number greater than $2$, such as $2.5$ or $100$, and the solution set is an interval drawn as a shaded ray on a number line with an open circle at $2$. A solution of $y > x + 1$ is an ordered pair $(x,y)$, such as $(0,4)$, and the solution set is a region of the plane — the half plane above the dashed line $y = x + 1$. The difference is dimensional: one variable needs one axis and produces an interval; two variables need two axes and produce a region.
2. $(-2,4)$: substitute to get $4 > -2 + 1$, that is $4 > -1$, which is true. Yes, it is a solution, and the point lies in the shaded region.
3. Solutions: $(-4,4)$ and $(1,5)$. Non-solutions: $(-1,-3)$ and $(3,-1)$. The picture says so by position: the two solutions lie in the shaded half plane above the boundary and are drawn as solid dots; the two non-solutions lie in the unshaded region below it and are drawn hollow.
4. Substituting $(1,2)$ gives $2 > 1 + 1$, that is $2 > 2$, which is false. The boundary is where the sentence changes from true to false, and a strict symbol excludes that changeover line entirely — which is exactly why the line is drawn dashed rather than solid.
5. $5 > 2 + 1$, so $5 > 3$, which is true. Yes.
6. $2(4) + 3(1) = 8 + 3 = 11$, and $11 \le 10$ is false. No.

Independent practice

7. a) $2 \le 3(2) - 4 = 2$: true, since $\le$ admits equality — a boundary solution b) $0 \le 3(0) - 4 = -4$: false — not a solution c) $-8 \le 3(-1) - 4 = -7$: true — a solution d) $12 \le 3(5) - 4 = 11$: false — not a solution
8. a) $0 + 4(3) = 12$, and $12 > 8$: true — a solution b) $8 + 4(0) = 8$, and $8 > 8$: false — a boundary point, excluded by the strict symbol c) $4 + 4(1) = 8$, and $8 > 8$: false — also on the boundary, also excluded d) $-2 + 4(4) = 14$, and $14 > 8$: true — a solution
9. The right panel, $y \le \tfrac12 x + 2$, has $(2,3)$ as a solution: $3 \le 3$ is true. In the left panel, $y < \tfrac12 x + 2$, the same point gives $3 < 3$, which is false. A dashed boundary means its points are not solutions; a solid boundary means they are. They grew from Chapter 3's open circle (boundary excluded, strict symbol) and closed circle (boundary included, inclusive symbol) on a number line — the same question asked about a whole line instead of one number.
10. The solution set is infinite in two directions at once: there are infinitely many $x$-values, and for each of them infinitely many $y$-values that work. Even a partial list would misrepresent it, because the set is a solid patch of the plane rather than a sequence of points. A linear equation in one variable has, in the ordinary case, one solution, which can simply be written down. Shading is the only practical way to report "all of these pairs and no others."
11. Any three pairs with $y < 2x$, for instance $(1,0)$, $(3,1)$, and $(0,-2)$. Confirmation for $(3,1)$: $1 < 2(3) = 6$ is true. (Note that $(0,0)$ does not work, since $0 < 0$ is false.)
12. Testing $(0,1)$ gives $1 > 0 + 1$, that is $1 > 1$, which is false, so it is not a solution. The error is treating a dashed boundary as part of the graph. The line is drawn only to show where the region stops. To make the student right, the symbol would have to be changed from $>$ to $\ge$: in $y \ge x + 1$, the sentence at $(0,1)$ reads $1 \ge 1$, which is true, and the boundary would then be drawn solid.
13. $2(4) + 3(3) = 8 + 9 = 17$, and $17 \ge 20$ is false, so $(4,3)$ is not a solution. It means four two-point baskets and three three-point baskets produce only $17$ points, which is $3$ points short of the $20$ the team needs.
14. $5 > 1 + 1$, that is $5 > 2$, which is true.

Exit ticket 9.1

15. $0 \ge -(-3) - 4 = 3 - 4 = -1$, and $0 \ge -1$ is true. Yes.
16. $2 < \tfrac13(6) = 2$, and $2 < 2$ is false. No — the pair lies exactly on the boundary, which a strict symbol excludes.
17. A boundary is dashed when the symbol is strict ($<$ or $>$), because points on the line make the sentence false; it is solid when the symbol is inclusive ($\le$ or $\ge$), because points on the line make it true. These correspond to Chapter 3's open circle and closed circle on a number line.
18. The inequality is a sentence about two unknowns at once, so it takes two numbers to say what has been substituted — one for $x$ and one for $y$. A single number leaves the sentence half-substituted and unable to be true or false. Geometrically, the answer lives in a plane, and a location in a plane needs two coordinates.

Lesson 9.2 — Graphing a Linear Inequality in Two Variables

Guided practice

19. The region above the line is shaded. At $(0,0)$, $0 \ge -0 + 3$ reads $0 \ge 3$, which is false, so the origin's side is the wrong side and the opposite side is shaded.
20. The region below the line is shaded. At $(0,0)$, $0 \le 3$ is true, so the origin's own side is shaded.
21. Boundary $y = 2x - 3$ through $(0,-3)$ and $(1,-1)$, drawn dashed because $>$ is strict. Test $(0,0)$: $0 > -3$ is true, so shade the origin's side, which is above the line.
22. Boundary $y = -\tfrac12 x + 4$ through $(0,4)$ and $(4,2)$, drawn solid because $\le$ is inclusive. Test $(0,0)$: $0 \le 4$ is true, so shade below the line.
23. Let $y = 0$: $3x = 12$, so the $x$-intercept is $(4,0)$. Let $x = 0$: $2y = 12$, so the $y$-intercept is $(0,6)$. Test $(0,0)$: $3(0) + 2(0) = 0$ and $0 \le 12$ is true, so the region containing the origin — below and left of the boundary — is shaded, and the boundary is solid.
24. $y < 3$: a horizontal dashed boundary at height $3$, with everything below it shaded; the $x$-coordinate is irrelevant because no $x$ appears. $x \ge -2$: a vertical solid boundary at $x = -2$, with everything to its right shaded, including the line itself.

Independent practice

25. a) Boundary $y = x + 2$, dashed; test $(0,0)$: $0 < 2$ true; shade below. b) Boundary $y = -3x$, solid; the origin is on the boundary, so test $(1,0)$: $0 \ge -3$ true; shade the side containing $(1,0)$, which is above the line. c) Boundary $y = \tfrac14 x - 1$, solid; test $(0,0)$: $0 \le -1$ is false; shade the side away from the origin, which is below the line. d) Boundary $y = -x - 5$, dashed; test $(0,0)$: $0 > -5$ true; shade above.
26. The boundary $y = -3x$ passes through the origin, so $(0,0)$ makes the sentence read $0 \ge 0$, which tells you the point is on the line and says nothing about either side. A test point must be off the boundary. $(1,0)$ works, and so does $(0,1)$ or $(-1,0)$.
27. $-y > -2x + 4$ by the subtraction property of inequality; then dividing both sides by $-1$ gives $y < 2x - 4$, with the symbol reversed by the division property of inequality with a negative divisor. Boundary $y = 2x - 4$ through $(0,-4)$ and $(2,0)$, dashed. Test $(0,0)$ in the original: $2(0) - 0 = 0$ and $0 > 4$ is false, so shade the side away from the origin, which is below the line. Confirm with the shaded point $(0,-6)$: $2(0) - (-6) = 6$ and $6 > 4$ is true.
28. Boundary $4x + 5y = 20$, solid. Let $y = 0$: $4x = 20$, so $(5,0)$. Let $x = 0$: $5y = 20$, so $(0,4)$. Test $(0,0)$: $0 \ge 20$ is false, so shade the side away from the origin — above and to the right of the line.
29. $x < 4$: a vertical dashed boundary at $x = 4$, with everything to its left shaded. $y \ge -2$: a horizontal solid boundary at $y = -2$, with everything above it shaded, the line included.
30. Slope $\dfrac{4-2}{4-0} = \dfrac{2}{4} = \tfrac12$, so the boundary is $y = \tfrac12 x + 2$. Dashed means strict and shaded below means "less than," so the inequality is $y < \tfrac12 x + 2$. Check with the shaded point $(0,0)$: $0 < 2$ is true.
31. Slope $\dfrac{3 - (-1)}{2 - 0} = \dfrac{4}{2} = 2$, so the boundary is $y = 2x - 1$. Solid means inclusive and shaded above means "greater than or equal," so the inequality is $y \ge 2x - 1$. Check with the shaded point $(1,5)$: $5 \ge 2(1) - 1 = 1$ is true.
32. Substituting $0$ for both variables is the easiest arithmetic there is — every variable term vanishes, so the test reduces to comparing the constant term with $C$, with no chance of a sign slip. The one situation in which it cannot be used is a boundary that passes through the origin, as in item 25b: then the origin is on the line and belongs to neither side, so it cannot decide between them. Any other point off the line will do.
33. Error 1: the symbol $>$ is strict, so the boundary must be dashed, not solid. Error 2: the shading is on the wrong side. Test $(0,0)$: $0 > -2(0) + 1$ reads $0 > 1$, which is false, so the origin — which lies below the boundary — is not a solution, and the region above the line should have been shaded instead.
34. Boundary $25x + 20y = 200$. Let $y = 0$: $25x = 200$, so $(8,0)$. Let $x = 0$: $20y = 200$, so $(0,10)$. Draw that line solid and test $(0,0)$: $0 \le 200$ is true, so shade toward the origin. Only the first-quadrant part is drawn because $x$ and $y$ count calculators and boxes: a negative number of either is not a purchase, so the situation adds $x \ge 0$ and $y \ge 0$ and the picture is a triangle with vertices $(0,0)$, $(8,0)$, and $(0,10)$.

Exit ticket 9.2

35. Boundary $y = -x + 1$, dashed because the symbol is strict; test $(0,0)$: $0 < 1$ is true, so shade below the line.
36. $3x - y \le 6$ gives $-y \le -3x + 6$, and dividing both sides by $-1$ reverses the symbol: $y \ge 3x - 6$. Boundary through $(0,-6)$ and $(2,0)$, solid. Test $(0,0)$ in the original: $3(0) - 0 = 0$ and $0 \le 6$ is true, so shade the origin's side, which is above the line — consistent with the $\ge$ obtained after the reversal.
37. Boundary the vertical solid line $x = 1$, with everything to its right shaded, the line included because the symbol is $\ge$.
38. (1) Solve for $y$ if necessary, reversing the symbol at any multiplication or division by a negative number. (2) Graph the boundary, dashed for $<$ or $>$ and solid for $\le$ or $\ge$. (3) Test a point that is not on the boundary, using the origin whenever the boundary misses it. (4) Shade the test point's side if the test was true, the other side if it was false. The reversal rule belongs to step 1, and a missed reversal is exactly what step 3 catches.

Lesson 9.3 — Writing an Inequality from a Situation

Guided practice

39. It says the total spent — $\$5$ for each ride band plus $\$8$ for each game pass — is at most the $\$80$ brought to the fair. $x$ is the number of ride bands bought; $y$ is the number of game passes bought. Both are counts, so both are whole numbers.
40. The inequality by itself is satisfied by pairs such as $(-20,5)$ and $(3,-40)$, but nobody buys a negative number of bands or passes. The situation therefore adds $x \ge 0$ and $y \ge 0$, and those two restrictions cut the half plane down to its first-quadrant part, a triangle with vertices $(0,0)$, $(16,0)$, and $(0,10)$.
41. $5(10) + 8(6) = 50 + 48 = 98$, and $98 \le 80$ is false. Not affordable: ten ride bands and six game passes would cost $\$98$, which is $\$18$ more than the student has.
42. "no more than" $\to \le$, solid. "at least" $\to \ge$, solid. "fewer than" $\to <$, dashed. "at most" $\to \le$, solid. The first, second, and fourth admit the exact value, which is why their boundaries are drawn solid.
43. $12x + 9y \ge 180$.
44. $x$ is the number of hours spent tutoring, at $\$12$ per hour. $y$ is the number of hours worked at the shop, at $\$9$ per hour. The situation adds $x \ge 0$ and $y \ge 0$, because a number of hours worked cannot be negative.

Independent practice

45. a) $x$ is the number of pens bought and $y$ the number of notebooks bought: $2x + 3y \le 60$, with $x \ge 0$ and $y \ge 0$. b) $x$ is the number of crates loaded and $y$ the number of boxes loaded: $40x + 25y \le 900$, with $x \ge 0$ and $y \ge 0$. c) $x$ is the number of three-point baskets made and $y$ the number of two-point baskets made: $3x + 2y > 45$, with $x \ge 0$ and $y \ge 0$. d) $x$ is the number of hours worked at the first job and $y$ the number worked at the second: $x + y \ge 30$, with $x \ge 0$ and $y \ge 0$.
46. $5x + 8y \le 80$, where $x$ is the number of ride bands bought at $\$5$ each and $y$ the number of game passes bought at $\$8$ each, with $x \ge 0$ and $y \ge 0$. Two purchases that work: $(4,5)$, since $20 + 40 = 60 \le 80$; and $(16,0)$, since $80 \le 80$ exactly — a boundary purchase, allowed because the symbol is inclusive. One that does not: $(10,6)$, since $50 + 48 = 98$, and $98 \le 80$ is false.
47. With $y = 8$, the inequality becomes $5x + 8(8) \le 80$, that is $5x + 64 \le 80$, so $5x \le 16$ and $x \le 3.2$. Ride bands come in whole numbers, so the greatest possible number is $3$. Check: $5(3) + 64 = 79 \le 80$ is true, while $5(4) + 64 = 84 \le 80$ is false.
48. In these models $x$ and $y$ count things or measure amounts — passes bought, kilograms loaded, hours worked — and none of those can be negative. The inequality alone does not know that; it would happily accept $(-20,5)$. Adding $x \ge 0$ and $y \ge 0$ restricts the half plane to its first-quadrant part, which usually turns an infinite region into a bounded triangle whose corners are exactly the extreme purchases worth discussing.
49. "At most $\$80$" is $5x + 8y \le 80$, with a solid boundary, because spending exactly $\$80$ is permitted. "Less than $\$80$" is $5x + 8y < 80$, with a dashed boundary, because spending exactly $\$80$ is not. They disagree about every purchase costing exactly $\$80$ — for instance $(16,0)$, or $(8,5)$, since $40 + 40 = 80$: allowed by the first model, forbidden by the second.
50. a) solid, from "at most" · b) solid, from "at most" · c) dashed, from "more than" · d) solid, from "at least."
51. The student added the two prices instead of building expressions for the money actually spent. The price $\$5$ has to be multiplied by the number of bands and $\$8$ by the number of passes, so that the model reports a total for a specific purchase rather than the cost of one of each. The correct inequality is $5x + 8y \le 80$, with $x$ the number of ride bands and $y$ the number of game passes. (The student's sentence $13 \le 80$ is true, but it only says that buying one of each is affordable.)
52. $x$-intercept: let $y = 0$, so $5x = 80$ and the point is $(16,0)$ — spending the entire $\$80$ on ride bands buys $16$ of them, with no game passes. $y$-intercept: let $x = 0$, so $8y = 80$ and the point is $(0,10)$ — spending the entire budget on game passes buys $10$ of them, with no ride bands. Each intercept answers the question "what if I buy only one kind?"

Exit ticket 9.3

53. $x + y \le 10$, where $x$ is the number of hours spent practicing piano in a week and $y$ is the number of hours spent practicing guitar, with $x \ge 0$ and $y \ge 0$.
54. $4x + 7y > 140$.
55. $x$ is the number of cookies sold, at $\$4$ each; $y$ is the number of pies sold, at $\$7$ each. The boundary is dashed, because "more than $\$140$" excludes raising exactly $\$140$ — a sale totaling precisely $\$140$ does not satisfy the requirement.
56. $6 + 5 = 11$, and $11 \le 10$ is false, so $(6,5)$ is not possible. Six hours of piano and five of guitar is $11$ hours of practice, one hour more than the $10$ allowed.

Lesson 9.4 — Systems of Two Linear Inequalities

Guided practice

57. $y \le -x + 5$ and $y > 2x - 4$. The first shades everything on and below the solid line through $(0,5)$ and $(5,0)$; the second shades everything strictly above the dashed line through $(0,-4)$ and $(2,0)$. They overlap in a wedge that opens leftward from the point where the two boundaries cross, and every point of that wedge — and only those points — satisfies both sentences.
58. First: $0 \le -0 + 5$, that is $0 \le 5$, true. Second: $0 > 2(0) - 4$, that is $0 > -4$, true. Both true, so $(0,0)$ is a solution of the system.
59. Set the boundary expressions equal: $-x + 5 = 2x - 4$, so $9 = 3x$ and $x = 3$; then $y = -3 + 5 = 2$. The corner is $(3,2)$. It is not a solution: it satisfies $2 \le 2$ but fails $2 > 2$. The dashed boundary of the second inequality is what shows the exclusion.
60. First: $-1 \le -4 + 5 = 1$, true. Second: $-1 > 2(4) - 4 = 4$, false. It is not a solution; it fails the second inequality, $y > 2x - 4$.
61. Graph $y = -x + 5$ solid and shade below it, then $y = 2x - 4$ dashed and shade above it. The solution set is the overlap: an unbounded wedge opening to the left, bounded above by the solid line and below by the dashed line, with its point at the corner $(3,2)$. The upper edge is included, the lower edge is not, and the corner is excluded.
62. $x + y \le 12$ and $8x + 12y \ge 96$. $x$ is the number of hours spent babysitting, paid at $\$8$ per hour; $y$ is the number of hours spent tutoring, paid at $\$12$ per hour. The situation also requires $x \ge 0$ and $y \ge 0$.

Independent practice

63. a) $y = x - 2$ solid with the region above shaded, and $y = -x + 4$ dashed with the region below shaded; the origin satisfies both, since $0 \ge -2$ and $0 < 4$. The solution set is the wedge containing the origin, and the corner is $(3,1)$, excluded. b) $y = 2x$ dashed with the region below shaded (test $(1,0)$: $0 < 2$ true), and $y = -1$ solid with the region above shaded. The solution set is the wedge opening to the right, and the corner is $\left(-\tfrac12, -1\right)$, since $2x = -1$ gives $x = -\tfrac12$. It is excluded, because it lies on the dashed boundary. c) $x = 0$ solid with everything to the right shaded, and $y = 3$ solid with everything below shaded. The solution set is the quarter plane to the right of the $y$-axis and below the height $3$, and the corner is $(0,3)$, which is included, since both boundaries are solid. d) The boundaries $y = \tfrac12 x + 3$ and $y = \tfrac12 x - 1$ are parallel, both of slope $\tfrac12$, so they never meet and there is no corner. The solution set is the infinite strip between them, with the upper edge solid and included and the lower edge dashed and excluded.
64. Setting the boundary expressions equal: $x - 2 = -x + 4$, so $2x = 6$, $x = 3$, and $y = 1$. The corner is $(3,1)$, and it is not in the solution set. Testing it: $1 \ge 3 - 2 = 1$ is true, so the first, inclusive, inequality accepts it; but $1 < -3 + 4 = 1$ is false, so the second, strict, one rejects it. A corner belongs to the solution set only when every boundary through it is solid.
65. The system asks for both sentences to be true of the same pair, which is a logical "and." The overlap is precisely the set of points where both shadings agree. Everything either shading covers would be a logical "or," and it would include points such as $(4,-1)$ that satisfy only one inequality — an affordable purchase that exceeds the weight limit, or a schedule that fits the hours but misses the pay. In a real situation, meeting one constraint while breaking the other is a failure, not a partial success.
66. For example $y > x + 1$ and $y < x - 2$. The two boundaries are parallel — both have slope $1$ — and the first region lies entirely above the upper line while the second lies entirely below the lower one. The two shadings never touch, so no point lies in both and the solution set is empty. (Any pair of parallel boundaries shaded away from each other does the same.)
67. a) $(0,0)$: $0 \le 5$ true and $0 > -4$ true — a solution b) $(3,2)$: $2 \le 2$ true, but $2 > 2$ false — not a solution; this is the corner, excluded by the strict inequality c) $(-4,1)$: $1 \le 9$ true and $1 > -12$ true — a solution d) $(5,3)$: $3 \le 0$ false — not a solution, and there is no need to check the second inequality once one has failed
68. Hours: $5 + 5 = 10$, and $10 \le 12$ is true. Earnings: $8(5) + 12(5) = 40 + 60 = 100$, and $100 \ge 96$ is true. Yes, it is a workable week: ten hours of work, two under the limit, earning $\$100$, which is $\$4$ more than the goal.
69. For example $(10,1)$ — ten hours babysitting and one hour tutoring. Hours: $11 \le 12$, true. Earnings: $8(10) + 12(1) = 80 + 12 = 92$, and $92 \ge 96$ is false. The week fits inside the hour limit but earns $\$4$ too little. (Other answers are possible, such as $(11,0)$, which earns $\$88$.)
70. At $(4,-1)$ the first inequality gives $-1 \le -4 + 5 = 1$, true, so the point does lie in one of the two shaded regions. But the second gives $-1 > 2(4) - 4 = 4$, which is false. A solution of a system must satisfy both inequalities, so being in one shading is not enough; only the overlap counts, and $(4,-1)$ is outside it.
71. $x + y \le 12$ and $8x + 12y \ge 96$, where $x$ is the number of hours spent babysitting at $\$8$ per hour and $y$ is the number of hours spent tutoring at $\$12$ per hour. The situation adds $x \ge 0$ and $y \ge 0$, since hours worked cannot be negative.
72. The corners are $(0,8)$, $(0,12)$, and $(12,0)$. Interpretations, any one of which is acceptable: $(0,8)$ means eight hours of tutoring and no babysitting, which earns exactly $\$96$ — the least work that reaches the goal. $(0,12)$ means twelve hours of tutoring and no babysitting, using the whole hour budget and earning $\$144$. $(12,0)$ means twelve hours of babysitting and no tutoring, earning exactly $\$96$ in exactly $12$ hours, so both limits bind at once.

Exit ticket 9.4

73. Graph $y = -2x + 6$ dashed with the region above shaded, and $y = x$ solid with the region below shaded. (Test $(4,0)$: $0 > -2$ true and $0 \le 4$ true, so the shadings overlap to the right.) The boundaries meet where $-2x + 6 = x$, so $6 = 3x$, $x = 2$, and $y = 2$: the corner is $(2,2)$, excluded, because $2 > 2$ is false.
74. First: $0 > -2(4) + 6 = -2$, true. Second: $0 \le 4$, true. Yes, $(4,0)$ is a solution of the system.
75. A system of two linear equations asks where two lines meet, and its solution is normally a single point — one pair of numbers. A system of two linear inequalities asks where two half planes overlap, and its solution is normally a region containing infinitely many pairs. The boundaries of that region are the same two lines the system of equations would use, and the corner of the region is exactly the point that system of equations would produce. In short, Chapter 8 finds the point; this chapter finds the region whose corner that point is.
76. $9x + 5y \le 50$ and $x \ge 2$, where $x$ is the number of movie tickets bought at $\$9$ each and $y$ is the number of snacks bought at $\$5$ each, with $y \ge 0$. Both boundaries are solid, since "at most" and "at least" both admit equality.

Lesson 9.5 — Verifying and Interpreting Solutions

Guided practice

77. Hours: $4 + 6 = 10$, and $10 \le 12$ is true. Earnings: $8(4) + 12(6) = 32 + 72 = 104$, and $104 \ge 96$ is true. Both pass, so A is a solution. On the graph it sits inside the shaded triangle, off both boundaries, which is the picture of a plan with room to spare.
78. Hours: $12 + 0 = 12$, and $12 \le 12$ is true. Earnings: $8(12) + 12(0) = 96$, and $96 \ge 96$ is true. B is a solution. Lying on both boundaries means both limits are met exactly and neither has any slack: the student works precisely the maximum $12$ hours and earns precisely the minimum $\$96$. It counts as a solution only because both symbols are inclusive; if either were strict, this corner would be excluded.
79. Hours: $10 + 1 = 11 \le 12$, true. Earnings: $8(10) + 12(1) = 80 + 12 = 92$, and $92 \ge 96$ is false. C fails the earnings inequality $8x + 12y \ge 96$, and it fails it by $\$4$. Graphically it sits just below the earnings boundary, outside the triangle.
80. $y \le 3x$ at $(2,5)$: $5 \le 6$, true. $y > x - 4$ at $(2,5)$: $5 > -2$, true. Both pass, so $(2,5)$ is a solution. On a graph, the point lies below the solid line $y = 3x$ and above the dashed line $y = x - 4$, that is, in the region covered by both shadings.
81. Enter y <= -x + 5 on one line and y > 2x - 4 on the next; the tool shades each half plane, and the doubly shaded region is the solution set. Compare its boundaries, its line styles, and the position of its overlap with your own graph, then plot a candidate point and see which region it lands in. If the shadings disagree, first re-read what was typed — a mistyped symbol or a lost negative sign explains most disagreements. Then check whether a reversal was missed when solving for $y$, by testing the origin in the original inequality, which requires no solving at all. Only after those two should you question the line styles or conclude anything about which picture is right.
82. Four hours of babysitting and six hours of tutoring is a workable week: it is $10$ hours of work, two hours under the $12$-hour limit, and it earns $\$104$, which is $\$8$ more than the $\$96$ goal.

Independent practice

83. a) $3(4) + 4(3) = 12 + 12 = 24$, and $24 \le 24$ is true — a solution, and it lies on the boundary b) $3(0) + 4(7) = 28$, and $28 \le 24$ is false — not a solution c) $3(8) + 4(0) = 24$, and $24 \le 24$ is true — a solution, also on the boundary d) $3(-2) + 4(6) = -6 + 24 = 18$, and $18 \le 24$ is true — a solution, strictly inside the region
84. a) $(1,1)$: $1 < 1 + 3 = 4$ true, and $1 \ge 2(1) - 6 = -4$ true — a solution b) $(6,2)$: $2 < 9$ true, but $2 \ge 2(6) - 6 = 6$ false — not a solution; it satisfies the first inequality only, so it lies in one shading but outside the overlap c) $(0,5)$: $5 < 0 + 3 = 3$ is false — not a solution
85. Type y <= -x + 5 on the first line and y > 2x - 4 on the second, using <= for $\le$ and >= for $\ge$. The tool draws each boundary in the correct style — dashed for the strict one — and shades each half plane in its own color. The solution set of the system is the region covered by both shadings, which appears darker where they overlap. To test a candidate point, enter it as an ordered pair, for example (0,0), and see whether the plotted point lands inside the doubly shaded region; then confirm by substituting it into both inequalities by hand, since a picture cannot decide a point lying exactly on a boundary.
86. Hours: $3 + 8 = 11$, and $11 \le 12$ is true. Earnings: $8(3) + 12(8) = 24 + 96 = 120$, and $120 \ge 96$ is true. Three hours of babysitting and eight hours of tutoring is a workable week: $11$ hours of work, one hour under the limit, earning $\$120$, which is $\$24$ more than the goal.
87. Substituting a point and getting true proves that that one point satisfies the inequality. It says nothing about the other side of the boundary, so a region shaded backwards can still contain points that test successfully — you simply happened to test a point that lies in both the correct region and your incorrect one, or a point you chose from your own shading without checking. The graphical check is what audits the region: it asks whether the whole shaded set is the right one, and the fastest version of it is testing the origin in the original inequality and confirming the origin is shaded exactly when the test comes out true.
88. The likely error is a missed reversal. Solving $5x - 2y < 10$ for $y$ gives $-2y < -5x + 10$, and dividing both sides by $-2$ reverses the symbol: $y > \tfrac52 x - 5$, which is the region above the boundary, as the tool shows. The one-line test that would have caught it: substitute $(0,0)$ into the original inequality — $5(0) - 2(0) = 0$ and $0 < 10$ is true, so the origin is a solution; and the origin lies above the boundary, since the boundary is at $-5$ when $x = 0$. So the shading must include the region above.
89. In $y = x - 2$: $3 - 2 = 1$, and the point's $y$-coordinate is $1$, so it lies on that boundary. In $y = -x + 4$: $-3 + 4 = 1$, so it lies on that boundary too — which is what makes it the corner. As a solution of the system, however, it fails: $1 \ge 1$ is true for $y \ge x - 2$, but $1 < 1$ is false for $y < -x + 4$. The corner is therefore not a solution, because one of the two boundaries through it is dashed.
90. Twelve hours of babysitting and no tutoring earns exactly $\$96$ in exactly $12$ hours, so the student meets the earnings goal precisely while using every hour available. Corners are worth asking about because they are where two limits bind at once — they are the most extreme plans that still work, and they are the plans people actually want named when they ask "what is the most I can do?"
91. An acceptable explanation: "I solved for $y$. Subtracting $2x$ from both sides gave $-y > -2x + 4$, and then dividing both sides by $-1$ reversed the symbol, by the division property of inequality with a negative divisor, so the inequality is $y < 2x - 4$. I drew the boundary $y = 2x - 4$ through $(0,-4)$ and $(2,0)$, and I drew it dashed, because the original symbol is strict and points on the line are not solutions. To decide the side, I tested $(0,0)$ in the original inequality: $2(0) - 0 = 0$, and $0 > 4$ is false, so the origin is not a solution and I shaded the side away from it, below the line. I confirmed with the shaded point $(0,-6)$: $2(0) - (-6) = 6$, and $6 > 4$ is true."
92. $5(8) + 8(5) = 40 + 40 = 80$, and $80 \le 80$ is true, so $(8,5)$ is a solution — and it lies exactly on the boundary. Eight ride bands and five game passes cost precisely $\$80$, spending the whole budget with nothing left over. It is allowed only because the symbol is $\le$; under a "less than $\$80$" rule this purchase would be excluded.

Exit ticket 9.5

93. $y \ge x - 1$ at $(2,3)$: $3 \ge 1$, true. $y < 4$ at $(2,3)$: $3 < 4$, true. Yes, a solution of the system.
94. $y \ge x - 1$ at $(-1,5)$: $5 \ge -2$, true. $y < 4$ at $(-1,5)$: $5 < 4$, false. Not a solution; it fails $y < 4$.
95. Algebraically, by substituting the pair into every inequality and reading true or false — this catches arithmetic slips and settles the strict-versus-inclusive question at a boundary point, which no picture can. Graphically, by locating the point on the graph and seeing whether it is in the shaded region — this audits the shading itself, catching a boundary drawn correctly but shaded on the wrong side, usually because a reversal was missed. With technology, by entering the inequalities into a graphing tool and comparing its shading with yours — this catches an error you made consistently in both of the first two checks, and a disagreement tells you to re-examine the typing, then the reversal, then the line style.
96. Any pair satisfying $12x + 9y \ge 180$, for example $(10,8)$: $12(10) + 9(8) = 120 + 72 = 192$, and $192 \ge 180$ is true. Interpreted: ten hours of tutoring at $\$12$ an hour together with eight hours at the shop at $\$9$ an hour earns $\$192$, which meets the $\$180$ goal with $\$12$ to spare. ($(15,0)$ works too, earning exactly $\$180$ from tutoring alone — a boundary solution.)

Chapter 9 Review

Part A — Graphing one linear inequality

97. Boundary $y = -2x + 4$ through $(0,4)$ and $(2,0)$, solid because $\le$ is inclusive. Test $(0,0)$: $0 \le 4$ is true, so shade the origin's side — below the line.
98. $5x - 2y < 10$ gives $-2y < -5x + 10$; dividing both sides by $-2$ reverses the symbol, by the division property of inequality with a negative divisor, so $y > \tfrac52 x - 5$. Boundary through $(0,-5)$ and $(2,0)$, dashed. Test $(0,0)$ in the original: $5(0) - 2(0) = 0$ and $0 < 10$ is true, so the origin is a solution; the origin lies above the boundary, so the region above is shaded — matching the $>$ produced by the reversal.
99. Boundary the vertical solid line $x = 3$, with everything to its left shaded, the line itself included. This is a case slope-intercept form cannot write, since a vertical line has no slope.
100. Slope $\dfrac{0-4}{2-0} = -2$, so the boundary is $y = -2x + 4$. Solid means inclusive, and the origin is shaded, so the inequality is $y \le -2x + 4$. Check: $0 \le -2(0) + 4 = 4$ is true.
101. The two panels shade the same side of the same line and differ only in whether the line belongs to the answer. At $(2,3)$ the left panel reads $3 < 3$, which is false, so that boundary is dashed and its points are excluded; the right panel reads $3 \le 3$, which is true, so that boundary is solid and its points are included. The line is drawn either way, because it shows where the region stops.
102. The student divided by $-1$ without reversing the symbol. Solving correctly, $-y > -2x + 4$ becomes $y < 2x - 4$, so the region below the boundary is the solution set. Testing $(0,0)$ in the original settles it: $2(0) - 0 = 0$, and $0 > 4$ is false, so the origin is not a solution — but the origin does lie above the line $y = 2x - 4$, since $2(0) - 4 = -4$ and $0 > -4$. The student's shading therefore contains the origin, which is not a solution, so it is wrong.

Part B — Creating an inequality from a situation

103. $x$ is the number of crates loaded, each weighing $50$ pounds; $y$ is the number of bags loaded, each weighing $30$ pounds. The inequality is $50x + 30y \le 1{,}200$, with $x \ge 0$ and $y \ge 0$, and the boundary is solid because "at most" permits exactly $1{,}200$ pounds.
104. $50(20) + 30(8) = 1{,}000 + 240 = 1{,}240$, and $1{,}240 \le 1{,}200$ is false. That load cannot be carried: it is $40$ pounds over the limit. Removing one crate, to $(19,8)$, gives $950 + 240 = 1{,}190 \le 1{,}200$, which works.
105. $(16,0)$: spending the whole $\$80$ on ride bands buys $16$ bands and no game passes. $(0,10)$: spending the whole $\$80$ on game passes buys $10$ passes and no ride bands. An interior point such as $(4,5)$: four ride bands and five game passes cost $5(4) + 8(5) = \$60$, which is affordable and leaves $\$20$ unspent.
106. The variables count or measure real quantities — passes bought, pounds loaded, hours worked — and none can be negative, so the situation adds $x \ge 0$ and $y \ge 0$. Those two restrictions discard three quadrants of the half plane, usually turning an infinite region into a bounded triangle whose corners are the extreme plans: spend everything on one item, spend everything on the other, or spend nothing. The restriction is a fact about the situation, not a rule about graphing.
107. $25x + 40y > 500$, where $x$ is the number of tickets sold at $\$25$ each and $y$ is the number of sponsorships sold at $\$40$ each, with $x \ge 0$ and $y \ge 0$. The boundary is dashed, because "more than $\$500$" excludes raising exactly $\$500$.
108. "At least $\$500$" is $\ge$, with a solid boundary, because raising exactly $\$500$ satisfies it. "More than $\$500$" is $>$, with a dashed boundary, because raising exactly $\$500$ does not. They disagree about every sale totaling precisely $\$500$ — for example $20$ tickets and no sponsorships, since $25(20) = 500$: a success under the first model and a failure under the second.

Part C — Systems of two linear inequalities

109. Graph $y = -x + 2$ solid with the region above shaded, and $y = 3x - 2$ dashed with the region below and to the right shaded. (Test $(4,0)$: $0 \ge -2$ true and $0 < 10$ true.) The boundaries meet where $-x + 2 = 3x - 2$, so $4 = 4x$, $x = 1$, and $y = 1$: the corner is $(1,1)$. It is excluded, since $1 < 1$ is false.
110. $y \ge -x + 2$ at $(4,2)$: $2 \ge -2$, true. $y < 3x - 2$ at $(4,2)$: $2 < 10$, true. Yes, a solution of the system.
111. The system is $y \le -x + 5$ and $y > 2x - 4$. The corner is $(3,2)$, found by solving $-x + 5 = 2x - 4$, so $x = 3$ and $y = 2$. It is not a solution: $2 \le 2$ is true but $2 > 2$ is false, and the dashed boundary is what shows the exclusion.
112. The system is $x + y \le 12$ and $8x + 12y \ge 96$, where $x$ is the number of hours spent babysitting at $\$8$ per hour and $y$ is the number of hours spent tutoring at $\$12$ per hour, with $x \ge 0$ and $y \ge 0$. The corners are $(0,8)$, $(0,12)$, and $(12,0)$. The corner $(0,8)$ means eight hours of tutoring and no babysitting: $8$ hours of work, four under the limit, earning $8(0) + 12(8) = \$96$ — exactly the goal, and the least total work that reaches it.
113. For example $y > x + 1$ and $y < x - 2$. The boundaries are parallel, both of slope $1$, and the regions are shaded away from each other — everything above the higher line, and everything below the lower one. The two shadings never overlap, so the solution set is empty: no pair can be both above the upper line and below the lower one at the same time.
114. Each linear equation describes a line, and two lines ordinarily cross in exactly one point, so the system's solution is that single pair. Each linear inequality describes a half plane — infinitely many points at once — and the overlap of two half planes is ordinarily still a two-dimensional region. The region's edges are the two boundary lines, and where those edges meet is the corner, which is precisely the solution of the system of equations formed by the same two boundaries. So the point Chapter 8 finds is the corner of the region this chapter finds.

Part D — Verifying and interpreting

115. Hours: $6 + 4 = 10$, and $10 \le 12$ is true. Earnings: $8(6) + 12(4) = 48 + 48 = 96$, and $96 \ge 96$ is true. Yes, a solution. Six hours of babysitting and four hours of tutoring is $10$ hours of work, two hours under the limit, earning exactly $\$96$ — the goal met precisely, with no dollars to spare. The pair lies on the earnings boundary, which counts only because "at least" is inclusive.
116. $y < 5x$ at $(2,9)$: $9 < 10$, true. $y \ge x + 2$ at $(2,9)$: $9 \ge 4$, true. Yes, a solution of the system. A graph would confirm it by showing the point below the dashed line $y = 5x$ and above the solid line $y = x + 2$, that is, inside the region where the two shadings overlap.
117. Enter y > 2.5x - 5, or the original 5x - 2y < 10 if the tool accepts it, and compare the tool's boundary, line style, and shaded side with your own graph. A disagreement about the side points almost certainly at a missed reversal: dividing by $-2$ while solving for $y$ must flip $<$ to $>$. Confirm independently by testing $(0,0)$ in the original inequality — $0 < 10$ is true, so the origin is a solution and the shaded region must contain it.
118. With $y = 5$, the inequality becomes $5x + 8(5) \le 80$, that is $5x + 40 \le 80$, so $5x \le 40$ and $x \le 8$. Since ride bands are counted in whole numbers and cannot be negative, the possible values are $x = 0, 1, 2, 3, 4, 5, 6, 7, 8$ — nine possibilities in all. Interpreted: a student who buys exactly five game passes, costing $\$40$, has $\$40$ left, which buys anywhere from zero to eight ride bands. The largest, $x = 8$, spends the budget exactly: $5(8) + 8(5) = 80$.