Algebra 1Algebra 1Lesson 5-6

Suppose y varies inversely with x, and a point on

the graph of the equation is (8, 9). 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

Inverse VariationInverse Variation

Additional Examples

Algebra 1Algebra 1Lesson 5-6

The 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).

Inverse VariationInverse Variation

Additional Examples

Algebra 1Algebra 1Lesson 5-6

A 120-lb weight is placed 5 ft from a fulcrum. How

far from the fulcrum should an 80-lb weight be placed to

balance the lever?

Relate:  A weight of 120 lb is 5 ft from the fulcrum.

A weight of 80 lb is x ft from the fulcrum.

Weight and distance vary inversely.

Define:  Let weight1 = 120 lb

Let weight2 = 80 lb

Let distance1 = 5 ft

Let distance2 = x ft

Inverse VariationInverse Variation

Additional Examples

Algebra 1Algebra 1Lesson 5-6

(continued)

600 = 80x Simplify.

The weight should be placed 7.5 ft from the fulcrum to balance the lever.

Write: weight1 • distance1 = weight2 • distance2

120 • 5 = 80 • x Substitute.

7.5 = x Simplify.

= x Solve for x.60080

Inverse VariationInverse Variation

Additional Examples

Algebra 1Algebra 1Lesson 5-6

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.

Inverse VariationInverse Variation

Additional Examples

Algebra 1Algebra 1Lesson 5-6

(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.

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

Check each ratio .yx

Inverse VariationInverse Variation

Additional Examples

64 = 1.5 = 1.5

128

yx = 1.5

32

Algebra 1Algebra 1

The cost per person times the number of people equals the total cost of the gift. Since the total cost is a constant product of $25, this is an inverse variation.

Lesson 5-6

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.

a. You buy several souvenirs for $10 each.

The cost per souvenir times the number of souvenirs equals the total cost of the souvenirs.Since the ratio is constant at $10 each,

cost souvenirs

this is a direct variation.

Inverse VariationInverse Variation

Additional Examples