1. Direct Variation Solve each equation for the given variable.
ALGEBRA 1 LESSON 8-9
(For help, go to Lessons 2–7 and 4–1.)
Solve each equation for the given variable.
1. nq = m; q 2. d = rt; r 3. ax + by = 0; y
Solve each proportion.
= = =
= = = 5 8 x 12 4 9 n 45 25 15 y 3 7 n 35 50 8 d 20 36 14 18 63 n
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2. Direct Variation Solutions 1. nq = m 2. d = rt 3. ax + by = 0 = =
ALGEBRA 1 LESSON 8-9
Solutions
1. nq = m 2. d = rt
= =
q = = r
r = nq n m d t rt
ax + by = 0
ax – ax + by = 0 – ax
by = –ax =
y = – by b
–ax ax =
8x = 5(12)
8x = 60
x = 7.5 5 8 x 12
= =
9n = 4(45) 15y = 25(3)
9n = y = 75
n = y = 5 4 9 n 45 25 15 y 3
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3. Direct Variation Solutions (continued) 7. = 8. = 9. =
ALGEBRA 1 LESSON 8-9
Solutions (continued)
= =
35n = 7(50) 20d = 8(36)
35n = d = 288
n = d = 14.4 7 n 35 50 8 d 20 36 14 18 63 =
14n = 18(63)
14n = 1134
n = 81
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4. Direct Variation
ALGEBRA 1 LESSON 8-9
Is each equation a direct variation? If it is, find the constant of variation.
a. 2x – 3y = 1
–3y = 1 – 2x Subtract 2x from each side.
y = – x Divide each side by –3. 1 3 2
The equation does not have the form y = kx. It is not a direct variation.
b. 2x – 3y = 0
–3y = –2x Subtract 2x from each side.
y = x Divide each side by –3. 2 3
The equation has the form y = kx, so the equation is a direct variation. The constant of variation is . 2 3
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5. Direct Variation
ALGEBRA 1 LESSON 8-9
Write an equation for the direct variation that includes the point (–3, 2).
y = kx Use the general form of a direct variation.
2 = k(–3) Substitute –3 for x and 2 for y.
– = k Divide each side by –3 to solve for k. 2 3
y = – x Write an equation. Substitute – for k in y = kx. 2 3
The equation of the direct variation is y = – x . 2 3
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6. Direct Variation
ALGEBRA 1 LESSON 8-9
The weight an object exerts on a scale varies directly with the mass of the object. If a bowling ball has a mass of 6 kg, the scale reads 59. Write an equation for the relationship between weight and mass.
Relate: The weight varies directly with the mass.
When x = 6, y = 59.
Define: Let = the mass of an object.
Let = the weight of an object. x y
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7. Direct Variation (continued)
ALGEBRA 1 LESSON 8-9
(continued)
Write: = k Use the general form of a direct variation.
59 = k(6) Solve for k. Substitute 6 for x and 59 for y. x y
= k Divide each side by 6 to solve for k. 59 6
y = x Write an equation. Substitute for k in y = kx. 59 6
The equation y = x relates the weight of an object to its mass. 59 6
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8. Direct Variation
ALGEBRA 1 LESSON 8-9
For the data in each table, tell whether y varies directly with x. If it does, write an equation for the direct variation. 1 –2
= –0.5 –1 2 4 y x
x y
–2 1
2 –1 4 –2 2 –1
= –2 1 –4
= 2 y x
x y
–1 2 –4
No, the ratio is not the
same for each pair of data. x y
Yes, the constant of variation is –0.5. The equation is y = –0.5x.
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9. Direct Variation
ALGEBRA 1 LESSON 8-9
Suppose a windlass requires 0.75 lb of force to lift an object that weighs 48 lb. How much force would you need to lift 210 lb?
Relate: The force of 0.75 lb lifts 48 lb. The force of n lb lifts 210 lb.
Define: Let n = the force you need to lift 210 lb.
Write: = Use a proportion.
force1
weight1
force2
weight2
Substitute 0.75 for force1, 48 for weight1, and 210 for weight2.
0.75 48 n
210 =
0.75(210) = 48n
Use cross products.
Solve for n. n
You need about 3.3 lb of force to lift 210 lb.
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10. Direct Variation
ALGEBRA 1 LESSON 8-9
1. Is each equation a direct variation? If it is, find the constant of variation.
a. x + 5y = 10
b. 3y + 8x = 0 no
yes; – 8 3
2. Write an equation of the direct variation that includes the point (–5, –4).
y = x 4 5
3. For each table, tell whether y varies directly with x. If it does, write an equation for the direct variation.
x y
–1 3
0 0
2 –6
3 –9 a.
x y
–1 –2
0 0
1 2
3 –6 b.
yes; y = –3x no
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