ALGEBRA 1

Lesson 4-6 Warm-Up

ALGEBRA 1

“Inverse Variation” (4-6)What is an “inverse variation”?

inverse variation (sometimes called an indirect proportion): an inverse (“opposite”) relationship in the form of xy = k, or y = k / x (when both sides are divided by x) where k ≠ 0 and coefficient k is called the “constant of variation for an inverse variation”. This means that the y varies, or “changes”, indirectly, or in opposite proportion, with changes in x (in other words, as x gets bigger y gets smaller and vice-versa).

Note: Since y = “undefined” (meaning impossible) when x = 0, no inverse variations pass through the origin (0, 0)

Examples: y = 4 / x y = -½ / x

Note: All indirect variations have the same curved shape when graphed as you can see from the following graphs.

ALGEBRA 1

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 VariationLESSON 4-6

Additional Examples

ALGEBRA 1

“Inverse Variation” (4-6)

How can you tell if an equation is an inverse variation?

How can you find a missing coordinate for an inverse variation?

You can tell an equation is an inverse variation if two ordered pairs (x and y) have the same constant, k, when multiplied together by the equation k = yx

Example: If (x1, y1), (x2,, y2), and (x3, y3), are two ordered pairs of an inverse variation (in other words, they lie on the graph of the inverse variation), then k = x1y1 = x2y2 = x3y3

Since k = yx and the k is constant (doesn’t change), the product of all x and y in the varation must equal each other. Therefore, we can use x1y1 = x2y2 to find a missing value if we know one coordinate an part of another.

Example: The points (3,8) and (2, y) are two points that lie on the graph of a direct variation. Find the missing value y.

The missing value is 12.

ALGEBRA 1

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 VariationLESSON 4-6

Additional Examples

ALGEBRA 1

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?

Words:  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 VariationLESSON 4-6

Additional Examples

ALGEBRA 1

(continued)

600 = 80x Simplify.

The 80-lb weight should be placed 7.5 ft from the fulcrum to balance the lever.

Equation: weight1 • distance1 = weight2 • distance2

120 • 5 = 80 • x Substitute.

7.5 = x Simplify.

= x Solve for x.60080

Inverse VariationLESSON 4-6

Additional Examples

ALGEBRA 1

“Inverse Variation” (4-6)

How can you tell if an equation is a direct or indirect variation in written or graph form?

Summary: You can tell an equation is a direct or indirect variation if the graph forms a line (linear function) or a curve (inverse function) or by determining how x and y are related (the ratio of y/x is constant = direct variation; the product xy is constant = inverse variation).

Direct Variation Indirect Variation

1. y is directly proportional to x (as y gets bigger, x gets bigger and vice-versa)

2. The ratio y / x is constant (the same for every coordinate on the graph of the equation)

1. y is inversely proportional to x (as y gets bigger, x gets smaller and vice-versa)

2. The product xy is constant (the same for every coordinate on the graph of the equation)

ALGEBRA 1

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 aninverse variation, and k = 30. The equation is xy = 30.

Inverse VariationLESSON 4-6

Additional Examples

ALGEBRA 1

(continued)

x y

2 3

4 6

8 12

b.

The ratio is the same for all pairs of data. So, this is a direct variation, and k = 1.5. The equation is y = 1.5x.

yx

The values of y seem to vary directly with the values of x.Check each ratio .y

x

64 = 1.5 = 1.5

128

yx = 1.5

32

Inverse VariationLESSON 4-6

Additional Examples

ALGEBRA 1

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.

Since the total cost is a constant product of $25,

The cost per person times the number of people equals the total cost of the gift. this is an inverse variation.

Inverse VariationLESSON 4-6

Additional Examples

ALGEBRA 1

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

xy =13 3

1 18

Inverse VariationLESSON 4-6

Lesson Quiz

x y

1

2

6

131619