Inverse VariationALGEBRA 1 LESSON 8-10ALGEBRA 1 LESSON 8-10

(For help, go to Lesson 5-5.)

Suppose y varies directly with x. Find each constant of variation.

1. y = 5x 2. y = –7x 3. 3y = x 4. 0.25y = x

Write an equation of the direct variation that includes the given point.

5. (2, 4) 6. (3, 1.5) 7. (–4, 1) 8. (–5, –2)

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Inverse VariationALGEBRA 1 LESSON 8-10ALGEBRA 1 LESSON 8-10

1. y = 5x; constant of variation = 5

2. y = –7x; constant of variation = –7

3. 3y = x

• 3y = • x

y = x; constant of variation =

4. 0.25y = x

y = x

4 • y = 4 • x

y = 4x; constant of variation = 4

13

1313

13

1414

Solutions

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Inverse VariationALGEBRA 1 LESSON 8-10ALGEBRA 1 LESSON 8-10

Solutions (continued)

5. Point (2, 4) in y = kx: 4 = k(2), so k = 2 and y = 2x.

6. Point (3, 1.5) in y = kx: 1.5 = k(3), so k = 0.5 and y = 0.5x.

7. Point (–4, 1) in y = kx: 1 = k(–4), so k = – and y = – x.

8. Point (–5, –2) in y = kx: –2 = k(–5), so k = and y = x.

14

14

25

25

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ALGEBRA 1 LESSON 8-10ALGEBRA 1 LESSON 8-10

Inverse VariationSuppose y varies inversely with x, and y = 9 when x = 8.

Write an equation for the inverse variation.

xy = k Use the general form for an inverse variation.

(8)(9) = k Substitute 8 for x and 9 for y.

72 = k Multiply to solve for k.

xy = 72 Write an equation. Substitute 72 for k in xy = k.

The equation of the inverse variation is xy = 72 or y = .72x

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ALGEBRA 1 LESSON 8-10ALGEBRA 1 LESSON 8-10

Inverse VariationThe points (5, 6) and (3, y) are two points on the graph of an

inverse variation. Find the missing value.

x1 • y1 = x2 • y2 Use the equation x1 • y1 = x2 • y2 since you know coordinates, but not the constant of variation.

5(6) = 3y2 Substitute 5 for x1, 6 for y1, and 3 for x2.

30 = 3y2 Simplify.

10 = y2 Solve for y2.

The missing value is 10. The point (3, 10) is on the graph of the inverse variation that includes the point (5, 6).

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ALGEBRA 1 LESSON 8-10ALGEBRA 1 LESSON 8-10

Inverse VariationJeff weighs 130 pounds and is 5 ft from the lever’s fulcrum. If

Tracy weighs 93 pounds, how far from the fulcrum should she sit in

order to balance the lever?

Relate:  A weight of 130 lb is 5 ft from the fulcrum.A weight of 93 lb is x ft from the fulcrum.Weight and distance vary inversely.

Define:  Let weight1 = 130 lbLet weight2 = 93 lbLet distance1 = 5 ftLet distance2 = x ft

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ALGEBRA 1 LESSON 8-10ALGEBRA 1 LESSON 8-10

Inverse Variation(continued)

650 = 93x Simplify.

Tracy should sit 6.99, or 7 ft, from the fulcrum to balance the lever.

Write: weight1 • distance1 = weight2 • distance2

130 • 5 = 93 • x Substitute.

6.99 = x Simplify.

= x Solve for x.65093

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Inverse VariationALGEBRA 1 LESSON 8-10ALGEBRA 1 LESSON 8-10

Decide if each data set represents a direct variation or an

inverse variation. Then write an equation to model the data.

x y

3 10

5 6

10 3

a. The values of y seem to vary inversely with the values of x.

Check each product xy.

xy: 3(10) = 30    5(6) = 30    10(3) = 30

The product of xy is the same for all pairs of data.

So, this is an inverse variation, and k = 30.

The equation is xy = 30.

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Inverse VariationALGEBRA 1 LESSON 8-10ALGEBRA 1 LESSON 8-10

(continued)

x y

2 3

4 6

8 12

b. The values of y seem to vary directly with the values of x.

So, this is a direct variation, and k = 1.5.

The equation is y = 1.5x.

64 = 1.5 = 1.5

128

yx = 1.5

32

The ratio is the same for all pairs of data.yx

Check each ratio .yx

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Inverse VariationALGEBRA 1 LESSON 8-10ALGEBRA 1 LESSON 8-10

Explain whether each situation represents a direct variation

or an inverse variation.

b. The cost of a $25 birthday present is split among several friends.

Since the total cost is a constant product of $25, this is an inverse variation.

The cost per souvenir times the number of souvenirs equals the total cost of the souvenirs.

Since the ratio is constant at $10 each, this is a direct variation. cost souvenirs

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The cost per person times the number of people equals the total cost of the gift.

a. You buy several souvenirs for $10 each.

Inverse VariationALGEBRA 1 LESSON 8-10ALGEBRA 1 LESSON 8-10

1. The points (5, 1) and (10, y) are on the graph of an inverse variation. Find y.

2. Find the constant of variation k for the inverse variation where a = 2.5 when b = 7.

4. Tell whether each situation represents a direct variation or an inverse variation.a. You buy several notebooks for $3 each.

b. The $45 cost of a dinner at a restaurant is split among several people.

3. Write an equation to model the data and complete the table.

0.5

17.5

direct variation

Inverse variation

x y

1

2

6

131619

xy =13 3

1 18

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