Appendix A — Answer Key, Chapter 7: Proportional Relationships: Tables, Graphs, and $y = mx$

SOL 7.PFA.1 · Covers textbook Chapter 7 and the companion workbook. Item numbers run continuously through the chapter and match the workbook, so this key serves both. Reasoning answers show an acceptable response, not the only wording. Graphing answers name two points the line must pass through and its slope, since a key cannot show a drawing.

Lesson 7.1 — Rate of Change as Slope

Guided practice

1. $m = \dfrac{8}{4} = 2$
2. $m = \dfrac{15}{5} = 3$
3. $m = \dfrac{6}{4} = \dfrac{3}{2}$
4. $m = \dfrac{-3}{1} = -3$
5. $12 \div 4 = 3$, so 3 gallons per minute.

Independent practice

6. a) $5$ b) $\dfrac{3}{12} = \dfrac{1}{4}$ c) $\dfrac{-8}{4} = -2$ d) $\dfrac{7}{7} = 1$
7. $m = \dfrac{9}{6} = \dfrac{3}{2}$
8. $m = \dfrac{-10}{2} = -5$
9. $m = \dfrac{6 - 0}{8 - 0} = \dfrac{6}{8} = \dfrac{3}{4}$
10. The unit rate is the amount of $y$ that goes with exactly 1 unit of $x$. Slope is the change in $y$ for a run of 1 in $x$. Those are the same measurement, so they are the same number.
11. $1{,}500 \div 6 = 250$ pages per minute. In 10 minutes: $10 \times 250 = 2{,}500$ pages.
12. Jae divided the run by the rise. Slope is rise over run, so it is $\tfrac{10}{4} = \tfrac{5}{2}$, not $\tfrac{4}{10}$. The correct slope is $\dfrac{5}{2}$.

Exit ticket 7.1

13. $m = \dfrac{12}{3} = 4$
14. $m = \dfrac{-6}{2} = -3$
15. $45 \div 5 = 9$, so 9 liters per minute.
16. Rate of change tells how much one quantity changes for each one-unit change in the other. In a proportional relationship $y$ is always the same multiple of $x$, so that change is the same everywhere on the line, which is why the graph is straight and the rate stays constant.

Lesson 7.2 — Finding Slope from a Table

Guided practice

17. $9 \div 1 = 9$, $18 \div 2 = 9$, $27 \div 3 = 9$, $36 \div 4 = 9$. $m = 9$
18. $1 \div 2 = 0.5$, $2 \div 4 = 0.5$, $3 \div 6 = 0.5$. $m = \dfrac{1}{2}$
19. $-5 \div 1 = -5$, $-10 \div 2 = -5$, $-15 \div 3 = -5$. $m = -5$
20. $12 \div 3 = 4$, $24 \div 6 = 4$, $36 \div 9 = 4$. $m = 4$
21. Not proportional. $4 \div 1 = 4$ but $7 \div 2 = 3.5$, and the quotients must all match.

Independent practice

22. a) $7 \div 2 = 3.5$, so $m = 3.5$ (or $\tfrac72$) b) $-15 \div 5 = -3$, so $m = -3$ c) $3 \div 4 = 0.75$, so $m = \dfrac{3}{4}$
23. $2.5 \div 1 = 2.5$, $5 \div 2 = 2.5$, $7.5 \div 3 = 2.5$. $m = 2.5$. (The row $x = 0$, $y = 0$ is the origin and cannot be used for the quotient, since dividing by zero is undefined.)
24. $m = 6 \div 1 = 6$. The missing value is $3 \times 6 = 18$.
25. Table 1 is proportional: $3 \div 1 = 3$, $6 \div 2 = 3$, $9 \div 3 = 3$, so $m = 3$. Table 2 is not: $3 \div 1 = 3$ but $5 \div 2 = 2.5$. Table 2's $y$-values rise by a steady 2, which makes the points fall on a line, but that line does not pass through the origin.
26. By dividing: $8 \div 6 = \tfrac{4}{3}$ and $12 \div 9 = \tfrac{4}{3}$. By change over change: $\dfrac{12 - 8}{9 - 6} = \dfrac{4}{3}$. Both give $m = \dfrac{4}{3}$.
27. $46.50 \div 3 = 15.50$, $77.50 \div 5 = 15.50$, $124.00 \div 8 = 15.50$. The quotients match, so it is proportional with $m = 15.5$, or $15.50 per hour. For 12 hours: $12 \times 15.50 = \$186.00$.
28. Divide $y$ by $x$ in every row. If every quotient is the same, the relationship is proportional and that quotient is the slope. One row is not enough because a single quotient always exists — it takes at least two matching quotients to show the multiplier never changes, and a table like $x$: 1, 2 with $y$: 5, 9 passes a one-row check but fails the second.

Exit ticket 7.2

29. $11 \div 1 = 11$, $22 \div 2 = 11$, $33 \div 3 = 11$, $44 \div 4 = 11$. $m = 11$
30. $-2 \div 4 = -0.5$, $-4 \div 8 = -0.5$, $-6 \div 12 = -0.5$. $m = -\dfrac{1}{2}$
31. Not proportional. $5 \div 2 = 2.5$ and $10 \div 4 = 2.5$, but $16 \div 6 = \tfrac{8}{3} \approx 2.67$. The last row breaks the pattern; it would have to be 15 for the table to be proportional.
32. A relationship is proportional exactly when $y$ is the same fixed multiple of $x$ in every pair. Dividing $y$ by $x$ recovers that multiple, so equal quotients in every row is precisely the definition being checked.

Lesson 7.3 — Writing $y = mx$ from a Table, Graph, or Situation

Guided practice

33. $m = 4 \div 1 = 4$, so $y = 4x$.
34. $m = \dfrac{21}{3} = 7$, so $y = 7x$.
35. $y = 14x$, where $x$ is hours worked and $y$ is earnings in dollars.
36. $m = \dfrac{-12}{4} = -3$, so $y = -3x$.
37. $y = 6(9) = 54$

Independent practice

38. a) $9 \div 2 = 4.5$, so $y = 4.5x$ b) $2 \div 5 = 0.4$, so $y = \dfrac{2}{5}x$ (or $y = 0.4x$) c) $-7 \div 1 = -7$, so $y = -7x$
39. $m = \dfrac{4}{6} = \dfrac{2}{3}$, so $y = \dfrac{2}{3}x$.
40. $m = \dfrac{-9}{1} = -9$, so $y = -9x$.
41. $y = 2.5(14) = 35$. For $y = 60$: $60 = 2.5x$, so $x = 60 \div 2.5 = 24$.
42. $y = 55x$, with $x$ in hours and $y$ in miles. In 4.5 hours: $55(4.5) = 247.5$ miles. For 302.5 miles: $302.5 \div 55 = 5.5$ hours.
43. $y = 6.40x$, with $x$ in pounds and $y$ in dollars. For 3.5 pounds: $6.40(3.5) = \$22.40$. For $40: $40 \div 6.40 = 6.25$ pounds.
44. The total is $y = 5x + 20$. The $20 is charged even when $x = 0$, so the cost at zero gigabytes is $20, not $0. The form $y = mx$ requires that $x = 0$ give $y = 0$ — the graph must pass through the origin — and this plan does not.

Exit ticket 7.3

45. $m = 15$, so $y = 15x$.
46. $m = \dfrac{2}{8} = \dfrac{1}{4}$, so $y = \dfrac{1}{4}x$.
47. $y = 40x$, with $x$ in minutes and $y$ in copies. In 7 minutes: $40(7) = 280$ copies.
48. Find one point on the line other than the origin, preferably where the line crosses a grid intersection. Divide its $y$-coordinate by its $x$-coordinate; that quotient is $m$. Equivalently, draw a slope triangle from the origin to that point and compute rise over run.

Lesson 7.4 — Positive, Negative, and Zero Slope

Guided practice

49. Positive.
50. Negative.
51. Zero.
52. $m = \dfrac{-6}{2} = -3$; negative.
53. $m = \dfrac{0}{7} = 0$.

Independent practice

54. a) positive b) negative c) zero d) positive, since $m = \tfrac{20}{5} = 4$ e) negative, since $m = \tfrac{8}{-4} = -2$
55. Least steep to steepest: $y = \tfrac{1}{2}x$, $y = 2x$, $y = 6x$.
56. $m$ is negative. A line that falls from left to right has a negative rise for a positive run.
57. The graph of $y = 0x$ is a horizontal line lying exactly along the $x$-axis, because $y = 0$ for every value of $x$. Three points: $(0, 0)$, $(3, 0)$, $(-5, 0)$. (Any points with $y$-coordinate 0 are acceptable.)
58. $m = \dfrac{-10}{-2} = 5$; positive. Both the rise and the run are negative, and a negative divided by a negative is positive — the line still climbs from left to right.
59. Positive. As hours increase, inches of snow increase, so the graph rises from left to right. Contextual relationships in this chapter pair quantities that grow together, so their slopes are positive and only the first-quadrant portion of the graph has meaning.
60. Zero slope means zero rise for any run, which describes a horizontal line. A vertical line has zero run instead, and dividing by a run of zero is undefined, so a vertical line has no slope at all. A zero-slope line through the origin lies flat along the $x$-axis.

Exit ticket 7.4

61. Negative.
62. $m = \dfrac{12}{3} = 4$; positive.
63. A horizontal line lying along the $x$-axis, passing through the origin, with every point having a $y$-coordinate of 0.
64. Read the graph from left to right. If the line goes up, the slope is positive; if it goes down, the slope is negative; if it stays flat, the slope is zero.

Lesson 7.5 — Graphing a Proportional Relationship

Guided practice

65. Slope 3. The line passes through $(0, 0)$, $(1, 3)$, and $(2, 6)$. Two points other than the origin: $(1, 3)$ and $(2, 6)$.
66. Slope $-2$. The line passes through $(0, 0)$, $(1, -2)$, and $(2, -4)$. Two points: $(1, -2)$ and $(2, -4)$.
67. Slope $\tfrac12$. The line passes through $(0, 0)$, $(2, 1)$, and $(4, 2)$. Two whole-number points: $(2, 1)$ and $(4, 2)$.
68. Slope 2 through $(3, 6)$. Stepping back left 1 and down 2 repeatedly reaches $(0, 0)$, so yes, it passes through the origin. Its equation is $y = 2x$, and it also passes through $(1, 2)$.
69. Slope $\tfrac25$. The line passes through $(0, 0)$, $(5, 2)$, and $(10, 4)$. Two whole-number points: $(5, 2)$ and $(10, 4)$.

Independent practice

70. a) Slope 4, through $(0, 0)$, $(1, 4)$, $(2, 8)$; two points: $(1, 4)$ and $(2, 8)$. b) Slope $-1$, through $(0, 0)$, $(1, -1)$, $(2, -2)$; two points: $(1, -1)$ and $(2, -2)$. c) Slope $\tfrac13$, through $(0, 0)$, $(3, 1)$, $(6, 2)$; two points: $(3, 1)$ and $(6, 2)$.
71. $m = \dfrac{10}{4} = 2.5$, so $y = 2.5x$. Another point: $(2, 5)$.
72. Check: $-3 \times (-2) = 6$, which matches the $y$-coordinate, so $(-2, 6)$ fits slope $-3$ on a line through the origin. The equation is $y = -3x$, and another point is $(1, -3)$.
73. Slope $\tfrac52$; the line passes through $(0, 0)$, $(2, 5)$, and $(4, 10)$. When $x = 6$: $y = \tfrac52 (6) = 15$, giving the point $(6, 15)$.
74. It is a horizontal line lying along the $x$-axis, since $y = 0x$ means $y = 0$ everywhere. A point other than the origin: $(4, 0)$.
75. $y = 1.5x$, with $x$ the number of cookies and $y$ the cost in dollars. For 2 cookies: $(2, 3)$, meaning $3.00. For 6 cookies: $(6, 9)$, meaning $9.00. Only the first quadrant is used because you cannot buy a negative number of cookies and the cost is never negative.
76. Two points determine a line, and a proportional relationship hands you the origin as one of them for free. So one more point is all that is missing. From those two points you can also compute the slope, $m = \tfrac{y}{x}$, and extend the line in both directions.

Exit ticket 7.5

77. Slope 5, through $(0, 0)$, $(1, 5)$, $(2, 10)$. Two points: $(1, 5)$ and $(2, 10)$.
78. Slope $-4$, through $(0, 0)$, $(1, -4)$, $(2, -8)$. Two points: $(1, -4)$ and $(2, -8)$.
79. $m = \dfrac{4}{6} = \dfrac{2}{3}$, so $y = \dfrac{2}{3}x$.
80. Substituting $x = 0$ into $y = mx$ gives $y = 0$, so $(0, 0)$ is always on the line. You never have to compute it, which means one more plotted point is enough to draw the whole line.

Lesson 7.6 — Connecting All Four Representations

Guided practice

81. $y = 5x$, with $x$ in minutes and $y$ in pages.

82. Multiply each $x$ by 7.

83. $m = \dfrac{10}{2} = 5$, so $y = 5x$.
84. $21 \div 3 = 7$, $42 \div 6 = 7$, $63 \div 9 = 7$, so $m = 7$ and $y = 7x$. A point other than the origin: $(3, 21)$.
85. $y = 1.5(4) = 6$, so the ordered pair is $(4, 6)$.

Independent practice

86. $y = 28x$, with $x$ in gallons and $y$ in miles.

The graph is a straight line through the origin with slope 28, rising to the right, drawn only in the first quadrant; it passes through $(2, 56)$ and $(4, 112)$.

87. $m = \dfrac{-15}{5} = -3$, so $y = -3x$.

88. $3 \div 2 = 1.5$, $7.5 \div 5 = 1.5$, $15 \div 10 = 1.5$, so $m = 1.5$ and $y = 1.5x$. When $x = 20$: $y = 1.5(20) = 30$.
89. a) $6$ b) $2$, since $2 \div 1 = 2$, $4 \div 2 = 2$, $6 \div 3 = 2$ c) $-1$
90. $y = 15x$, with $x$ in hours and $y$ in kilowatt-hours.

The point $(6, 90)$ means that in 6 hours the turbine produces 90 kilowatt-hours of electricity.

91. $y = 2.80x$, with $x$ in kilograms and $y$ in dollars.

For 4.25 kilograms: $2.80(4.25) = \$11.90$.

92. Find the slope of each. From the table, divide $y$ by $x$ in a row; from the graph, read a point other than the origin and divide its $y$ by its $x$. If the slopes match and both pass through the origin, they describe the same relationship. As a further check, pick a pair from the table and confirm that point lies on the graph.

Exit ticket 7.6

93. Multiply each $x$ by 9.

94. $m = \dfrac{10}{4} = 2.5$, so $y = 2.5x$.
95. $4.5 \div 3 = 1.5$ and $9 \div 6 = 1.5$, so $m = 1.5$ and $y = 1.5x$.
96. The equation. Substituting $x = 250$ takes one multiplication, while a table would need hundreds of rows and a graph would need a grid far too large to read accurately.

Chapter 7 Review

Part A — Slope as rate of change and writing $y = mx$ (7.PFA.1a)

97. $m = \dfrac{6}{4} = \dfrac{3}{2}$
98. $11 \div 2 = 5.5$, $22 \div 4 = 5.5$, $33 \div 6 = 5.5$, so $m = 5.5$ and $y = 5.5x$.
99. $y = 24x$, with $x$ in minutes and $y$ in bottles. In 15 minutes: $24(15) = 360$ bottles.
100. $-8 \div 1 = -8$, $-16 \div 2 = -8$, $-24 \div 3 = -8$, so $m = -8$ and $y = -8x$.
101. $m = \dfrac{4}{10} = \dfrac{2}{5}$, so $y = \dfrac{2}{5}x$ (or $y = 0.4x$).
102. $y = 13x$, with $x$ in hours and $y$ in miles. In 3.5 hours: $13(3.5) = 45.5$ miles.

Part B — Positive, negative, and zero slope (7.PFA.1b)

103. a) positive b) negative c) zero d) negative, since $m = \dfrac{15}{-5} = -3$
104. It is a horizontal line lying along the $x$-axis; every point has a $y$-coordinate of 0. Two points: $(2, 0)$ and $(-6, 0)$. (Any two points with $y = 0$ are acceptable.)
105. Positive.
106. $y = 5x$ is steeper. It rises 5 units for each unit right, while $y = \tfrac15 x$ rises only $\tfrac15$ of a unit, and 5 is farther from zero than $\tfrac15$.

Part C — Graphing from a point and the slope (7.PFA.1c)

107. $m = \dfrac{8}{2} = 4$; $y = 4x$. The line passes through $(0, 0)$ and $(2, 8)$, and also $(1, 4)$.
108. $m = \dfrac{-6}{3} = -2$; $y = -2x$. The line passes through $(0, 0)$ and $(3, -6)$, and also $(1, -2)$. It falls from left to right.
109. $m = \dfrac{6}{8} = \dfrac{3}{4}$; $y = \dfrac{3}{4}x$. The line passes through $(0, 0)$ and $(8, 6)$, and also $(4, 3)$.

Part D — Graphing from an equation (7.PFA.1d)

110. Slope 3, through $(0, 0)$, $(1, 3)$, $(2, 6)$. Two points other than the origin: $(1, 3)$ and $(2, 6)$.
111. Slope $-\tfrac32$, through $(0, 0)$, $(2, -3)$, $(4, -6)$. Two whole-number points: $(2, -3)$ and $(4, -6)$.
112. Slope $\tfrac14$, through $(0, 0)$, $(4, 1)$, $(8, 2)$. Two whole-number points: $(4, 1)$ and $(8, 2)$.

Part E — Mixed application and reasoning (7.PFA.1e)

113. $y = 35x$, with $x$ in minutes and $y$ in square feet.

The graph is a straight line through the origin with slope 35, rising to the right, drawn only in the first quadrant. The point $(10, 350)$ means that in 10 minutes the landscaper spreads 350 square feet of mulch.

114. $9 \div 2 = 4.5$, $13.5 \div 3 = 4.5$, $22.5 \div 5 = 4.5$, so $m = 4.5$ and $y = 4.5x$. When $x = 11$: $y = 4.5(11) = 49.5$.
115. The quotients are not equal: $6 \div 1 = 6$, $11 \div 2 = 5.5$, $16 \div 3 \approx 5.33$. The $y$-values rise by a steady 5, so the points do lie on a line, but that line crosses the $y$-axis at 1 rather than at the origin. Since $y = mx$ forces $y = 0$ when $x = 0$, no value of $m$ can produce this table.
116. Pump B. Pump A moves 18 gallons per minute; Pump B moves $95 \div 5 = 19$ gallons per minute, and $19 > 18$. On one grid, Pump B's line would be the steeper of the two, rising above Pump A's line for every positive number of minutes.
117. Substitute $x = 0$ into $y = mx$: $y = m \cdot 0 = 0$. So the ordered pair $(0, 0)$ satisfies the equation no matter what $m$ is, which means the origin is on every such graph. In context this says that zero of one quantity always pairs with zero of the other.
118. $y = 2x$ has slope 2 and rises from left to right, through $(1, 2)$ and $(2, 4)$. $y = -2x$ has slope $-2$ and falls from left to right, through $(1, -2)$ and $(2, -4)$. $y = 0x$ has slope 0 and is horizontal, lying along the $x$-axis through $(1, 0)$ and $(2, 0)$. All three pass through the origin, because all three are proportional relationships.

Workbook-only items

Page 2, fill in the blanks. Slope is the ratio of rise to run. In a proportional relationship the rate of change is constant. The graph is a straight line through the origin, the point $(0, 0)$.

Page 2, slope triangle read-off. Rise $= 3$, run $= 2$, $m = \tfrac{3}{2}$.

Page 3, origin test. Left graph: proportional, because the line passes through $(0, 0)$. Right graph: not proportional, because it crosses the $y$-axis at 3 instead of at the origin.

Page 3, two triangles. Small: rise 2 over run 3 $= \tfrac{2}{3}$. Large: rise 4 over run 6 $= \tfrac{2}{3}$. The slope of a line is the same no matter where on the line you measure it, which is what "constant rate of change" means.

Page 6, quotient check table. $2 \div 1 = 2$; $4 \div 2 = 2$; $6 \div 3 = 2$; $8 \div 4 = 2$. Proportional, $m = 2$.

Page 6, watch out. Not proportional. $5 \div 1 = 5$ but $9 \div 2 = 4.5$. The steady jump of 4 makes the points fall on a line, but that line crosses the $y$-axis at 1, not at the origin.

Page 10, fill in the blanks. $m$ stands for the slope. Substituting $x = 0$ gives $y = \mathbf{0}$, so the graph always passes through the origin. The relationship $y = 3x + 5$ is not proportional, because at $x = 0$ it gives $y = \mathbf{5}$.

Page 10, three ways in. From a table: divide $y$ by $x$ in a row and confirm every row gives the same quotient. From a graph: read one point other than the origin and divide its $y$ by its $x$. From a situation: find the unit rate, and state what $x$ and $y$ represent with units.

Page 14, three directions table. Positive: the line rises from left to right; example $y = 2x$. Negative: the line falls from left to right; example $y = -2x$. Zero: the line is horizontal; example $y = 0x$. (Any correct example equations are acceptable.)

Page 14, zero slope. $y = 0x$ simplifies to $y = \mathbf{0}$. Every point has a $y$-coordinate of 0. The graph is a horizontal line lying along the $x$-axis.

Page 17, five steps. 1. Plot the origin. 2. A whole number $m$ becomes $\tfrac{m}{1}$. 3. Move right by the run and up or down by the rise. 4. Plot a third point as a check. 5. Draw the line. Going backward from a given point should land you on the origin.

Pages 18, 19, 20, and 26, blank grids. The grids are for student graphing. The correct line for each item is described in the numbered answers above: items 65–80 for pages 18–20, and items 107–112 for page 26.

Page 21, where the slope hides. Situation: the unit rate stated in words. Table: the quotient $y \div x$, the same in every row. Equation: the coefficient of $x$. Graph: the rise over the run, read from a slope triangle.

Page 21, say the point out loud. The point $(5, 160)$ means that 5 gallons of gasoline carries the car 160 miles.