Appendix A — Answer Key, Chapter 8: Systems of Two Linear Equations

SOL A.EI.2 (a, b, c, h) · Covers textbook Chapter 8 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 132 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Conventions used in every answer below: a solution of a system is an ordered pair that makes both equations true, and it is always written $(x, y)$ — never as a single number. Verifying means substituting the pair into both original equations, not into the rewritten ones. The three possible counts are one solution, no solution, and infinitely many solutions, matching Chapter 2's language for a single equation. Technology is used freely throughout; no part of this course is calculator-free.

The figures used repeatedly in the chapter, for reference:

• Figure 1 is $x - y = -1$ with $2x + y = 7$, meeting at $(2,3)$
• Figure 2 is $y = 2x - 3$ with $y = -x + 3$, meeting at $(2,1)$ — the one-solution case
• Figure 3 is $y = \tfrac12 x + 3$ with $y = \tfrac12 x - 2$ — the no-solution case
• Figure 4 is $2x - y = 4$ with $4x - 2y = 8$, one line drawn twice — the infinitely-many case
• Figure 5 sets those same three cases side by side
• Figure 6 is $2x + y = 8$ with $x - y = 1$, meeting at $(3,2)$
• Figure 7 is $x + y = 5$ with $2x - y = 1$, meeting at $(2,3)$, plus the non-solution $(4,1)$
• Figure 8 is the ride plans $y = 2x + 3$ and $y = x + 7$, meeting at $(4,11)$
• Figure 9 is $y = 1.5x + 2$ with $y = 1.6x - 2$, meeting at $(40,62)$ far outside the standard window
• Figure 10 is the ticket sale $x + y = 12$ with $5x + 3y = 50$, meeting at $(7,5)$

Lesson 8.1 — What a System Is, and What a Solution Means

Guided practice

1. The solution is $(2,3)$. The feature that names it is the point where the two lines cross — the only point that lies on both graphs at once, and therefore the only ordered pair that satisfies both equations.
2. $x - y = -1$: $2 - 3 = -1$ ✓. $2x + y = 7$: $2(2) + 3 = 4 + 3 = 7$ ✓. Both original equations are true, so $(2,3)$ is verified.
3. No. $x - y = -1$: $0 - 1 = -1$ ✓, so $(0,1)$ is on the first line. But $2x + y = 7$: $2(0) + 1 = 1$, and $1 \ne 7$ ✗. Satisfying one equation is not enough; $(0,1)$ is on one line and not the other.
4. $(2,3)$ is the solution. Check: $2 + 3 = 5$ ✓ and $2(2) - 3 = 4 - 3 = 1$ ✓. $(4,1)$ is not: $4 + 1 = 5$ ✓ but $2(4) - 1 = 7$, and $7 \ne 1$ ✗. On the graph, $(4,1)$ sits on the blue line only.
5. Let $a$ be the number of adult tickets sold, and let $c$ be the number of child tickets sold. The count of tickets gives $a + c = 40$; the money taken in, in dollars, gives $12a + 8c = 408$.
6. Each equation on its own describes a whole line's worth of pairs. The system asks which pairs satisfy both conditions at the same time, so a pair that makes only one equation true answers only half the question — graphically, it is a point on one line that is not on the other.

Independent practice

7. The solution of the system is $(4,-1)$. a) $(4,-1)$: $3(4) + (-1) = 11$ ✓ and $4 - 2(-1) = 6$ ✓. Yes, a solution. b) $(2,5)$: $3(2) + 5 = 11$ ✓ but $2 - 2(5) = -8 \ne 6$ ✗. No. c) $(0,11)$: $3(0) + 11 = 11$ ✓ but $0 - 2(11) = -22 \ne 6$ ✗. No. d) $(-2,-4)$: $3(-2) + (-4) = -10 \ne 11$ ✗. No — it fails the first equation, so there is no need to test the second.
8. $2x + 5y = 4$: $2(-3) + 5(2) = -6 + 10 = 4$ ✓. $x - y = -5$: $-3 - 2 = -5$ ✓. Both hold, so $(-3,2)$ is a solution.
9. Let $n$ be the number of nickels in the jar and let $d$ be the number of dimes. Counting coins: $n + d = 12$. Counting cents: $5n + 10d = 95$.
10. Let $x$ be the number of miles driven and let $y$ be the cost of the ride in dollars. Ride Rite: $y = 2x + 3$. Cab Co: $y = x + 7$.
11. Let $L$ be the length of the rectangle in centimeters and let $W$ be its width in centimeters. Perimeter: $2L + 2W = 34$. Relationship: $L = W + 5$.
12. Let $b$ be the number of boys in the class and let $g$ be the number of girls. Total: $b + g = 27$. Relationship: $b = 2g + 3$.
13. Two distinct points determine exactly one line. If two lines shared two different points, then each line would be the line through those two points, so the two lines would be the same line — and then they would share not two points but infinitely many. So the only possibilities are no shared points, one shared point, or all of them.
14. The student checked only the first equation. $(1,4)$ does satisfy $x + y = 5$, but the second equation gives $2(1) - 4 = -2$, and $-2 \ne 4$ ✗, so $(1,4)$ is not a solution. Graphing $y = -x + 5$ and $y = 2x - 4$ shows a crossing at $(3,2)$, and $3 + 2 = 5$ ✓ with $2(3) - 2 = 4$ ✓. The solution is $(3,2)$.
15. Solve each equation for $y$: $y = -3x + 11$ and $y = \tfrac{x - 6}{2}$. Enter both, set a window that includes $x = 4$ and $y = -1$ (the standard $-10$ to $10$ window is enough). Trace or evaluate each function at $x = 4$: the first should return $-1$ and so should the second. Equivalently, use the intersect feature and confirm it reports $x = 4$, $y = -1$.
16. Let $n$ be the price of one notebook in dollars and let $p$ be the price of one pen in dollars. First purchase: $3n + 2p = 13$. Second purchase: $5n + 4p = 23$.

Exit ticket 8.1

17. Yes. $x + y = 3$: $5 + (-2) = 3$ ✓. $2x + 3y = 4$: $2(5) + 3(-2) = 10 - 6 = 4$ ✓.
18. No. $y = 4x + 2$: $4(1) + 2 = 6$ ✓. $3x + y = 10$: $3(1) + 6 = 9$, and $9 \ne 10$ ✗. It satisfies the first equation only.
19. Let $g$ be the number of guppies and let $t$ be the number of tetras. Total fish: $g + t = 9$. Relationship: $t = 2g$.
20. A solution of a system is a pair of values — one for each variable — that makes both equations true at the same time. It is written as an ordered pair because it takes two numbers to answer the question, and because the pair names the single point where the two lines cross.

Lesson 8.2 — Solving a System by Graphing

Guided practice

21. $2x + y = 8$ becomes $y = -2x + 8$, and $x - y = 1$ becomes $y = x - 1$. The graphs cross at $(3,2)$.
22. $2x + y = 8$: $2(3) + 2 = 8$ ✓. $x - y = 1$: $3 - 2 = 1$ ✓. Verified in the original equations.
23. The crossing is at $(2,1)$. Check: $y = 2x - 3$ gives $2(2) - 3 = 1$ ✓, and $y = -x + 3$ gives $-2 + 3 = 1$ ✓.
24. Graph $y = x + 2$ from $(0,2)$ with slope $1$, and $y = -2x + 5$ from $(0,5)$ with slope $-2$. They cross at $(1,3)$. Check: $1 + 2 = 3$ ✓ and $-2(1) + 5 = 3$ ✓.
25. Graph $y = \tfrac12 x - 1$ from $(0,-1)$ with slope $\tfrac12$, and $y = -x + 5$ from $(0,5)$ with slope $-1$. They cross at $(4,1)$. Check: $\tfrac12(4) - 1 = 1$ ✓ and $-4 + 5 = 1$ ✓.
26. The left panel appears to show two parallel lines with no crossing, which would mean no solution. That reading is wrong because the slopes are $1.5$ and $1.6$, and different slopes force exactly one crossing. The crossing is simply outside the window; widening to $0 \le x \le 80$ reveals it at $(40,62)$. The slopes told you a solution had to exist before any graph was drawn.

Independent practice

27. a) $(3,-1)$ — check $3 - 4 = -1$ ✓ and $-2(3) + 5 = -1$ ✓ b) $(2,-4)$ — check $-3(2) + 2 = -4$ ✓ and $2 - 6 = -4$ ✓ c) $(-1,2)$ — check $-1 + 2 = 1$ ✓ and $2(-1) + 4 = 2$ ✓ d) $(2,2)$ — check $\tfrac12(2) + 1 = 2$ ✓ and $-\tfrac12(2) + 3 = 2$ ✓
28. a) $y = -3x + 5$ and $y = x - 3$; crossing $(2,-1)$. In the originals: $3(2) + (-1) = 5$ ✓ and $2 - (-1) = 3$ ✓. b) $y = -\tfrac12 x + 4$ and $y = x - 2$; crossing $(4,2)$. In the originals: $4 + 2(2) = 8$ ✓ and $2 = 4 - 2$ ✓. c) $y = 2x - 4$ and $y = -x + 5$; crossing $(3,2)$. In the originals: $2(3) - 2 = 4$ ✓ and $3 + 2 = 5$ ✓.
29. No solution. Both lines have slope $2$, and their $y$-intercepts $(0,1)$ and $(0,-3)$ differ, so the graphs are parallel — the second sits $4$ units below the first at every input and the two never meet.
30. Infinitely many solutions. Dividing $2x + 2y = 6$ by $2$ gives $x + y = 3$, which is $y = -x + 3$ — the same line. The two graphs lie exactly on top of one another, so every point of that line solves both equations.
31. The solution is $(4,11)$: a $4$-mile trip costs $\$11$ with either company. For a $2$-mile trip, Ride Rite charges $2(2) + 3 = \$7$ and Cab Co charges $2 + 7 = \$9$, so Ride Rite is cheaper. For a $10$-mile trip, Ride Rite charges $\$23$ and Cab Co charges $\$17$, so Cab Co is cheaper. Short trips favor the lower starting fee; long trips favor the lower per-mile rate.
32. Let $x$ be the number of weeks and $y$ the amount saved in dollars: $y = 5x + 60$ and $y = 3x + 100$. Solving, $5x + 60 = 3x + 100$ gives $2x = 40$ and $x = 20$, then $y = 160$. The solution $(20,160)$ lies far outside a standard $-10$ to $10$ window, so a window of roughly $0 \le x \le 25$ and $0 \le y \le 200$ is needed. After $20$ weeks both have $\$160$.
33. A window such as $0 \le x \le 60$, $0 \le y \le 100$ shows the crossing. The intersect feature reports $(40, 62)$. Algebraically: $1.5(40) + 2 = 62$ ✓ and $1.6(40) - 2 = 62$ ✓, so the technology and the algebra agree.
34. A graph gives an exact answer only when both coordinates of the crossing are integers, because those are the only values you can read off grid corners with certainty. When the crossing falls between grid lines, the reading is an estimate — "about $(2.2, 1.8)$" — and you should switch to substitution or elimination, which produce exact values such as $\left(\tfrac94, \tfrac74\right)$, or use the intersect feature on technology.
35. Both lines have slope $3$ and different $y$-intercepts, so they are parallel and never cross. What the student saw near the corner was the two lines running close together and leaving the grid, not meeting. Correct answer: no solution.
36. The lines cross at $(20, 200)$. In context: making and selling $20$ shirts costs $\$200$ and brings in $\$200$, so $20$ shirts is the break-even point. Below $20$ shirts the cost line is above the revenue line and the shop loses money; above $20$ shirts revenue exceeds cost and the shop makes money.
37. The solution is $(-2, 4)$: the vertical line $x = -2$ and the horizontal line $y = 4$ meet at exactly one point. The line $x = -2$ cannot be written as $y = mx + b$ because it has no slope — the run between any two of its points is $0$, and a rise cannot be divided by $0$. In standard form it is $1x + 0y = -2$, as Chapter 5 showed.

Exit ticket 8.2

38. $(3,2)$. Check: $3 - 1 = 2$ ✓ and $-3 + 5 = 2$ ✓.
39. $(1,3)$. Check: $1 + 3 = 4$ ✓ and $2(1) + 1 = 3$ ✓.
40. No solution. "Falls four units for every one unit right" means both slopes are $-4$; equal slopes with different $y$-intercepts is the parallel case, and parallel lines never cross.
41. Different slopes guarantee exactly one crossing, so the system does have a solution — the window is simply too small to show it. Next step: zoom out or widen the window (try $-100$ to $100$ on both axes), or use the intersect feature, which searches beyond the displayed region. Then confirm the reported point algebraically in both equations.

Lesson 8.3 — Solving by Substitution

Guided practice

42. Substituting $y = 3x - 2$ into $2x + y = 8$ gives $2x + 3x - 2 = 8$, so $5x = 10$ and $x = 2$; then $y = 3(2) - 2 = 4$. Solution $(2,4)$. Check: $4 = 3(2) - 2$ ✓ and $2(2) + 4 = 8$ ✓.
43. Substituting $x = y + 3$ into $2x + 5y = 13$ gives $2(y + 3) + 5y = 13$, so $7y + 6 = 13$, $7y = 7$, and $y = 1$; then $x = 4$. Solution $(4,1)$. Check: $4 = 1 + 3$ ✓ and $2(4) + 5(1) = 13$ ✓.
44. Both equations give $y$, so set the expressions equal: $-2x + 9 = 4x - 3$, so $12 = 6x$ and $x = 2$; then $y = 4(2) - 3 = 5$. Solution $(2,5)$. Check: $-2(2) + 9 = 5$ ✓.
45. The first equation has an $x$ with coefficient $1$, so solve it for $x$: $x = 7 - y$. Substituting into $3x - 2y = 6$ gives $3(7 - y) - 2y = 6$, so $21 - 5y = 6$, $-5y = -15$, and $y = 3$; then $x = 4$. Solution $(4,3)$.
46. $x + y = 7$: $4 + 3 = 7$ ✓. $3x - 2y = 6$: $3(4) - 2(3) = 12 - 6 = 6$ ✓.
47. A coefficient of $1$ or $-1$ means isolating that variable takes only addition or subtraction — no division — so no fractions enter the work. Isolating a variable with coefficient $7$ would produce an expression like $\tfrac{13 - 4y}{7}$, which then has to be distributed through the second equation, and every later step carries the denominator.

Independent practice

48. a) $3x + (2x + 1) = 11 \Rightarrow 5x = 10 \Rightarrow x = 2$, $y = 5$. $(2,5)$ b) $2(3y - 1) + y = 12 \Rightarrow 7y = 14 \Rightarrow y = 2$, $x = 5$. $(5,2)$ c) $4x - (-x + 6) = 9 \Rightarrow 5x = 15 \Rightarrow x = 3$, $y = 3$. $(3,3)$ d) $x + 5x = 18 \Rightarrow 6x = 18 \Rightarrow x = 3$, $y = 15$. $(3,15)$
49. a) Isolate $x$ in the first equation ($x$ has coefficient $1$): $x = 11 - 2y$. Then $3(11 - 2y) - y = 5 \Rightarrow 33 - 7y = 5 \Rightarrow y = 4$, $x = 3$. $(3,4)$ b) Isolate $y$ in the first equation ($y$ has coefficient $1$): $y = -1 - 2x$. Then $5x - 3(-1 - 2x) = 25 \Rightarrow 11x + 3 = 25 \Rightarrow x = 2$, $y = -5$. $(2,-5)$ c) Isolate $x$ in the first equation: $x = 4y - 2$. Then $3(4y - 2) + 2y = 8 \Rightarrow 14y = 14 \Rightarrow y = 1$, $x = 2$. $(2,1)$
50. $6x - 2(3x + 1) = 5$ becomes $6x - 6x - 2 = 5$, so $-2 = 5$. Both variables vanish and the statement is false, so the system has no solution. Graphically the lines are parallel: $6x - 2y = 5$ is $y = 3x - \tfrac52$, which has the same slope $3$ as $y = 3x + 1$ and a different intercept.
51. $8x + 2(-4x + 5) = 10$ becomes $8x - 8x + 10 = 10$, so $10 = 10$. Both variables vanish and the statement is true, so the system has infinitely many solutions. Graphically it is one line drawn twice: $8x + 2y = 10$ divided by $2$ is $4x + y = 5$, which is $y = -4x + 5$.
52. Let $x$ be the larger number and $y$ the smaller: $x + y = 46$ and $x - y = 12$. From the second, $x = y + 12$; substituting, $(y + 12) + y = 46$, so $2y = 34$ and $y = 17$, then $x = 29$. The numbers are $29$ and $17$. Check: $29 + 17 = 46$ ✓ and $29 - 17 = 12$ ✓.
53. $2L + 2W = 34$ with $L = W + 5$. Substituting: $2(W + 5) + 2W = 34$, so $4W + 10 = 34$, $4W = 24$, and $W = 6$; then $L = 11$. The width is $6$ cm and the length is $11$ cm. Check: $2(11) + 2(6) = 34$ ✓ and $11 = 6 + 5$ ✓.
54. $b + g = 27$ with $b = 2g + 3$. Substituting: $(2g + 3) + g = 27$, so $3g = 24$ and $g = 8$; then $b = 19$. There are $19$ boys and $8$ girls. Check: $19 + 8 = 27$ ✓ and $19 = 2(8) + 3$ ✓.
55. Let $x$ be the number of minutes used and $y$ the monthly cost in dollars: $y = 0.10x + 25$ and $y = 0.15x + 15$. Setting the expressions equal, $0.10x + 25 = 0.15x + 15$, so $10 = 0.05x$ and $x = 200$; then $y = 0.10(200) + 25 = 45$. At $200$ minutes both plans cost $\$45$. Below $200$ minutes Plan B is cheaper; above $200$ minutes Plan A is.
56. Substitution is clearly better when a variable is already isolated or has coefficient $1$ or $-1$ — for example $y = 3x - 2$ with $2x + y = 8$, where the replacement takes one step. It is painful when neither variable has a small coefficient, as in $7x + 4y = 2$ with $3x - 4y = 18$: isolating anything produces fractions, while elimination cancels the $y$-terms by simple addition. The feature to look at is the coefficients.
57. The method broke at step 2. Substituting an expression back into the same equation it came from always produces a true statement like $2x - 1 = 2x - 1$, because it is that equation rewritten — it uses no new information. The substitution must go into the other equation. Doing it correctly: $4x - 2(2x - 1) = 2$ gives $4x - 4x + 2 = 2$, so $2 = 2$, true. So this particular system really does have infinitely many solutions — but the student's work never showed it, and the same faulty method would have reported "infinitely many" for a system with exactly one solution.
58. Substituting $y = x + 4$ into $3x + 2y = 23$ gives $3x + 2(x + 4) = 23$, so $5x + 8 = 23$, $5x = 15$, and $x = 3$; then $y = 7$. Solution $(3,7)$. Check: $7 = 3 + 4$ ✓ and $3(3) + 2(7) = 9 + 14 = 23$ ✓. Graphically, $y = x + 4$ and $y = -\tfrac32 x + \tfrac{23}{2}$ would be drawn on a grid reaching $x = 4$ and $y = 8$, and their crossing would sit exactly on the grid corner $(3,7)$.

Exit ticket 8.3

59. $2x + (4x - 7) = 11 \Rightarrow 6x = 18 \Rightarrow x = 3$, $y = 4(3) - 7 = 5$. $(3,5)$; check $2(3) + 5 = 11$ ✓.
60. $3(2y + 1) - y = 8 \Rightarrow 5y + 3 = 8 \Rightarrow y = 1$, $x = 3$. $(3,1)$; check $3 = 2(1) + 1$ ✓ and $3(3) - 1 = 8$ ✓.
61. Isolate $y$ in the first equation: $y = 9 - x$. Then $2x - (9 - x) = 3$, so $3x - 9 = 3$, $3x = 12$, and $x = 4$; then $y = 5$. $(4,5)$; check $4 + 5 = 9$ ✓ and $2(4) - 5 = 3$ ✓.
62. $g + t = 9$ with $t = 2g$. Substituting: $g + 2g = 9$, so $3g = 9$ and $g = 3$; then $t = 6$. There are $3$ guppies and $6$ tetras.

Lesson 8.4 — Solving by Elimination

Guided practice

63. The $y$-terms are opposites. Adding: $2x = 14$, so $x = 7$; then $7 + y = 10$ gives $y = 3$. $(7,3)$; check $7 + 3 = 10$ ✓ and $7 - 3 = 4$ ✓.
64. The $y$-terms $+2y$ and $-2y$ are opposites. Adding: $4x = 16$, so $x = 4$; then $4 - 2y = 0$ gives $y = 2$. $(4,2)$; check $3(4) + 2(2) = 16$ ✓ and $4 - 2(2) = 0$ ✓.
65. Multiply the second equation by $3$: $3x - 3y = 3$. Adding to $2x + 3y = 12$ gives $5x = 15$, so $x = 3$; then $3 - y = 1$ gives $y = 2$. $(3,2)$
66. Multiply the first equation by $3$ and the second by $4$: $9x + 12y = 30$ and $8x + 12y = 28$. Subtracting: $x = 2$; then $2(2) + 3y = 7$ gives $3y = 3$ and $y = 1$. $(2,1)$
67. $3x + 4y = 10$: $3(2) + 4(1) = 6 + 4 = 10$ ✓. $2x + 3y = 7$: $2(2) + 3(1) = 4 + 3 = 7$ ✓.
68. Each equation is a true statement that two quantities are equal. The addition property of equality says that adding equal amounts to both sides of a true equation keeps it true: if $a = b$ and $c = d$, then $a + c = b + d$. Adding the two equations adds the left side of the second to the left side of the first, and the right to the right — equal amounts on both sides — so the sum is a true equation about the same solution.

Independent practice

69. a) Add: $2x = 14$, $x = 7$, $y = 5$. $(7,5)$ b) Add: $5x = 20$, $x = 4$, then $2(4) + y = 9$ gives $y = 1$. $(4,1)$ c) Add: $8x = 24$, $x = 3$, then $15 + 2y = 11$ gives $y = -2$. $(3,-2)$ d) Add: $10x = 20$, $x = 2$, then $14 + 4y = 2$ gives $y = -3$. $(2,-3)$
70. a) Multiplier $3$ on the second equation: $6x - 3y = 21$. Adding to $x + 3y = 7$ gives $7x = 28$, $x = 4$, then $4 + 3y = 7$ gives $y = 1$. $(4,1)$ b) Multiplier $2$ on the first equation: $8x + 2y = 20$. Adding to $3x - 2y = 13$ gives $11x = 33$, $x = 3$, then $4(3) + y = 10$ gives $y = -2$. $(3,-2)$ c) Multiplier $2$ on the second equation: $6x + 2y = 18$. Adding to $5x - 2y = 4$ gives $11x = 22$, $x = 2$, then $3(2) + y = 9$ gives $y = 3$. $(2,3)$
71. a) Multipliers $3$ and $2$: $6x + 9y = 39$ and $6x + 4y = 24$. Subtracting: $5y = 15$, $y = 3$, then $2x + 9 = 13$ gives $x = 2$. $(2,3)$ b) Multipliers $2$ and $5$: $8x + 10y = 14$ and $15x - 10y = -60$. Adding: $23x = -46$, $x = -2$, then $4(-2) + 5y = 7$ gives $5y = 15$ and $y = 3$. $(-2,3)$
72. Multiplying $2x + y = 5$ by $2$ gives $4x + 2y = 10$; subtracting $4x + 2y = 3$ leaves $0 = 7$, a false statement. No solution. The graph is two parallel lines: both have slope $-2$, with $y$-intercepts $(0,5)$ and $\left(0,\tfrac32\right)$.
73. Multiplying $3x - y = 6$ by $2$ gives $6x - 2y = 12$, which is the second equation exactly; subtracting leaves $0 = 0$, a true statement. Infinitely many solutions. The graph is one line, $y = 3x - 6$, drawn twice.
74. $3n + 2p = 13$ and $5n + 4p = 23$. Multiply the first by $-2$: $-6n - 4p = -26$. Adding: $-n = -3$, so $n = 3$; then $3(3) + 2p = 13$ gives $2p = 4$ and $p = 2$. A notebook costs $\$3$ and a pen costs $\$2$. Check: $3(3) + 2(2) = 13$ ✓ and $5(3) + 4(2) = 23$ ✓.
75. $a + c = 40$ and $12a + 8c = 408$. Multiply the first by $-8$: $-8a - 8c = -320$. Adding: $4a = 88$, so $a = 22$; then $c = 18$. $22$ adult tickets and $18$ child tickets. Check: $22 + 18 = 40$ ✓ and $12(22) + 8(18) = 264 + 144 = 408$ ✓.
76. $n + d = 12$ and $5n + 10d = 95$. Multiply the first by $-5$: $-5n - 5d = -60$. Adding: $5d = 35$, so $d = 7$; then $n = 5$. $5$ nickels and $7$ dimes. Check: $5 + 7 = 12$ ✓ and $5(5) + 10(7) = 25 + 70 = 95$ ✓.
77. Let $x$ be the number of adult tickets and $y$ the number of student tickets: $x + y = 12$ and $5x + 3y = 50$. Multiply the first by $-3$: $-3x - 3y = -36$. Adding: $2x = 14$, so $x = 7$; then $y = 5$. $(7,5)$ — $7$ adult tickets and $5$ student tickets, which is exactly the point marked where the two lines cross in the figure.
78. Multiplying an equation by a nonzero number produces an equivalent equation: every pair that made the original true still makes the new one true, and no new pairs are admitted, because you can divide by the same number to undo it. The graph is the identical line. Multiplying by $0$ destroys the equation — $0 = 0$ is true for every pair, so all the information the equation carried is gone and the "system" no longer describes the original problem.
79. The student added the constants instead of subtracting them. Subtracting means subtracting every term of the second equation: $(5x + 2y) - (3x + 2y) = 11 - 5$ gives $2x = 6$, not $2x = 16$. So $x = 3$, and $5(3) + 2y = 11$ gives $2y = -4$ and $y = -2$. $(3,-2)$; check $3(3) + 2(-2) = 9 - 4 = 5$ ✓.

Exit ticket 8.4

80. Add: $2x = 10$, $x = 5$, then $5 + y = 8$ gives $y = 3$. $(5,3)$; check $5 - 3 = 2$ ✓.
81. Add: $6x = 12$, $x = 2$, then $2(2) + 3y = 7$ gives $3y = 3$ and $y = 1$. $(2,1)$; check $4(2) - 3(1) = 5$ ✓.
82. Multiplier $2$ on the second equation: $2x - 2y = 6$. Adding to $3x + 2y = 4$ gives $5x = 10$, $x = 2$; then $2 - y = 3$ gives $y = -1$. $(2,-1)$
83. Let $h$ be the price of a hot dog in dollars and $d$ the price of a drink in dollars: $5h + 2d = 17$ and $3h + 4d = 20$. Multiply the first by $-2$: $-10h - 4d = -34$. Adding: $-7h = -14$, so $h = 2$; then $5(2) + 2d = 17$ gives $2d = 7$ and $d = 3.50$. A hot dog costs $\$2$ and a drink costs $\$3.50$. Check: $5(2) + 2(3.5) = 17$ ✓ and $3(2) + 4(3.5) = 6 + 14 = 20$ ✓.

Lesson 8.5 — One Solution, No Solution, or Infinitely Many

Guided practice

84. Left panel: intersecting lines — one solution. Middle panel: parallel lines — no solution. Right panel: identical lines — infinitely many solutions.
85. One solution. The slopes are $2$ and $-1$, which differ, so the lines are not parallel and must cross exactly once. The figure marks the crossing at $(2,1)$.
86. No solution. Both slopes are $\tfrac12$, so the lines rise at the same rate, and the $y$-intercepts $(0,3)$ and $(0,-2)$ differ, so one sits $5$ units below the other at every input. A constant gap of $5$ never closes.
87. Infinitely many solutions. Multiplying $2x - y = 4$ by $2$ gives $4x - 2y = 8$, which is the second equation exactly. Solving either for $y$ gives $y = 2x - 4$, so the two graphs are the same line and every point on it satisfies both equations.
88. No solution. Both are already in slope-intercept form with slope $3$, and the intercepts $1$ and $-4$ differ — equal slopes with different intercepts is the parallel case.
89. Infinitely many solutions. Dividing $2x - 4y = 6$ by $2$ gives $x - 2y = 3$, the second equation. Both become $y = \tfrac12 x - \tfrac32$, so one line is written twice.

Independent practice

90. a) Slopes $-2$ and $5$ differ. One solution. b) Slopes both $\tfrac14$; intercepts $-3$ and $2$ differ. No solution. c) $3x + 3y = 24$ divided by $3$ is $x + y = 8$, i.e. $y = -x + 8$ — the same line. Infinitely many solutions. d) $4x + y = 9$ is $y = -4x + 9$; the other is $y = -4x + 1$. Slopes both $-4$, intercepts differ. No solution. e) $y = -x + 6$ and $y = x - 6$. Slopes $-1$ and $1$ differ. One solution, namely $(6,0)$.
91. a) Both become $y = -\tfrac23 x + 2$. Infinitely many solutions. b) $y = -\tfrac23 x + 2$ and $y = -\tfrac23 x + 3$. Same slope, different intercepts. No solution. c) $y = -\tfrac23 x + 2$ and $y = \tfrac23 x - 2$. Different slopes. One solution, namely $(3,0)$.
92. $k = 5$. Parallel requires equal slopes, and the slope of the second line is $5$; the intercepts $3$ and $-2$ already differ, so equal slopes give parallel lines and no solution. No value of $k$ produces infinitely many solutions, because that would require the intercepts to match as well, and $3 \ne -2$ no matter what $k$ is.
93. $4x - 2y = -6$ rearranges to $-2y = -4x - 6$, then $y = 2x + 3$. So $c = 3$ makes the two equations identical and gives infinitely many solutions. For any other $c$ the slopes still match at $2$ but the intercepts differ, which is exactly the parallel case: no solution.
94. Let $x$ be the number of shirts and $y$ the total cost in dollars: $y = 3x + 45$ and $y = 3x + 60$. Both slopes are $3$ and the intercepts differ, so the system has no solution. In context: the two shops charge the same $\$3$ per shirt, and Shop B's setup fee is $\$15$ higher, so B costs exactly $\$15$ more no matter how many shirts are ordered — there is no order size at which the prices are equal.
95. Two lines in a plane either cross, or do not, or coincide. If they crossed at three different points, then any two of those points would determine the same line for both, so the two lines would be identical — and identical lines share infinitely many points, not three. So exactly three shared points is impossible; the only counts are $0$, $1$, and infinitely many.
96. The student swapped the two endings. $0 = 0$ is a true statement, which means every ordered pair on the line makes both equations true: the correct conclusion is infinitely many solutions. The graph is one line drawn twice, the two equations describing the same set of points. "No solution" is what $0 = 7$ — a false statement — reports.
97. When both variables vanish, the two equations differ only in their constant terms, which is exactly what happens when both lines have the same slope: the $x$-terms cancel because the rates of change are identical. What is left, $-2 = 5$ or similar, compares the two constants and says they are unequal — that is the algebra's way of saying the two lines sit at different heights and stay a fixed distance apart. Equal slopes, unequal intercepts, no crossing, false statement: four descriptions of one situation.

Exit ticket 8.5

98. Infinitely many solutions. The two equations are literally identical, so the graph is one line drawn twice and every point of $y = -3x + 2$ solves both.
99. No solution. $x - y = 4$ is $y = x - 4$, and $2x - 2y = 9$ is $y = x - \tfrac92$. Same slope $1$, different intercepts, so the graph is two parallel lines.
100. One solution. The slopes are $-5$ and $1$, which differ. Adding the equations gives $6x = 6$, so $x = 1$, and then $1 - y = 3$ gives $y = -2$. The solution is $(1,-2)$; check $5(1) + (-2) = 3$ ✓.
101. $0 = 0$ is a true statement: the equation left standing is true no matter what $x$ and $y$ are, so every pair on the line works and there are infinitely many solutions. $0 = 7$ is a false statement: nothing can make it true, so no pair works at all and there is no solution. The difference is whether the surviving statement is true or false — not the fact that the variables disappeared, which happens in both.

Lesson 8.6 — Modeling with Systems, and Verifying Three Ways

Guided practice

102. Let $x$ be the number of adult tickets sold and let $y$ be the number of student tickets sold. Then $x + y = 12$ (tickets) and $5x + 3y = 50$ (dollars). Eliminating: multiply the first by $-3$ and add to get $2x = 14$, so $x = 7$ and $y = 5$. The club sold $7$ adult tickets and $5$ student tickets.
103. Let $x$ be the number of miles driven and let $y$ be the cost in dollars. Then $y = 2x + 3$ and $y = x + 7$. Setting them equal: $2x + 3 = x + 7$, so $x = 4$ and $y = 11$. A $4$-mile ride costs $\$11$ with either company. Left of the crossing (trips under $4$ miles) Ride Rite is cheaper because its starting fee is lower; right of it (trips over $4$ miles) Cab Co is cheaper because its per-mile rate is lower.
104. Let $r$ be the number of red marbles and let $b$ be the number of blue marbles. Then $r + b = 30$ and $b = r + 6$. Substituting: $r + (r + 6) = 30$, so $2r = 24$ and $r = 12$; then $b = 18$. $12$ red and $18$ blue.
105. Algebraically: $12 + 18 = 30$ ✓ and $18 = 12 + 6$ ✓ — both original equations are true. Graphically: with $r$ on the horizontal axis and $b$ on the vertical, $b = -r + 30$ and $b = r + 6$ cross at $(12,18)$, which is where the drawn lines meet. With technology: enter $y_1 = -x + 30$ and $y_2 = x + 6$, set a window of about $0 \le x \le 30$ and $0 \le y \le 35$, and use the intersect feature; it should report $x = 12$, $y = 18$.
106. Acceptable response: "I used substitution, because the second equation already had $b$ isolated, so I could replace $b$ in the first equation in one step. The key step was writing $r + (r + 6) = 30$, which reduced the problem to a one-variable equation; solving gave $r = 12$, and back-substituting gave $b = 18$. I checked $(12,18)$ in both original equations, confirmed the crossing on a graph, and confirmed it again with the intersect feature on a graphing calculator."
107. Let $v$ be the number of vans and let $b$ be the number of buses. Then $v + b = 5$ and $6v + 20b = 78$. Substituting $v = 5 - b$: $6(5 - b) + 20b = 78$, so $30 + 14b = 78$, $14b = 48$, and $b = \tfrac{24}{7}$; then $v = 5 - \tfrac{24}{7} = \tfrac{11}{7}$. The pair $\left(\tfrac{11}{7}, \tfrac{24}{7}\right)$ genuinely solves the system, but it cannot describe the trip, because vehicles come in whole numbers and there is no such thing as $\tfrac{24}{7}$ of a bus. The honest conclusion is that no combination of $5$ full vehicles of these two sizes carries exactly $78$ people, so one of the stated facts is wrong or the vehicles are not filled to capacity.