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By the end of this section, you will be able to:
  • Solve direct variation problems
  • Solve inverse variation problems

Before you get started, take this readiness quiz.

If you miss a problem, go back to the section listed and review the material.

  1. Find the multiplicative inverse of −8 .
    If you missed this problem, review [link] .
  2. Solve for n : 45 = 20 n .
    If you missed this problem, review [link] .
  3. Evaluate 5 x 2 when x = 10 .
    If you missed this problem, review [link] .

When two quantities are related by a proportion, we say they are proportional to each other. Another way to express this relation is to talk about the variation of the two quantities. We will discuss direct variation and inverse variation in this section.

Solve direct variation problems

Lindsay gets paid $15 per hour at her job. If we let s be her salary and h be the number of hours she has worked, we could model this situation with the equation

s = 15 h

Lindsay’s salary is the product of a constant, 15, and the number of hours she works. We say that Lindsay’s salary varies directly with the number of hours she works. Two variables vary directly if one is the product of a constant and the other.

Direct variation

For any two variables x and y , y varies directly with x if

y = k x , where k 0

The constant k is called the constant of variation.

In applications using direct variation, generally we will know values of one pair of the variables and will be asked to find the equation that relates x and y . Then we can use that equation to find values of y for other values of x .

How to solve direct variation problems

If y varies directly with x and y = 20 when x = 8 , find the equation that relates x and y .

Solution

The above image has 3 columns. The table shows the steps to solve direct variation problems. Step one is to write the formula for the direct variation. The direct variation formula is y equals k x. Then we get y equals k times x. Step two is substitute the given values for the variables. We are given y equals 20 and x equals 8. Then we have 20 equals k times 8. Step three is to solve for the constant variation. Divide both sides of the equation by 8, then multiply. We now get 20 divided by 8 equals k. K equals 2.5. Step four is to write the equation that relates x and y. Rewrite the general equation with the value we found k to get y equals 2 and five-tenths times x.
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If y varies directly as x and y = 3 , when x = 10 . find the equation that relates x and y .

y = 3 10 x

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If y varies directly as x and y = 12 when x = 4 find the equation that relates x and y .

y = 3 x

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We’ll list the steps below.

Solve direct variation problems.

  1. Write the formula for direct variation.
  2. Substitute the given values for the variables.
  3. Solve for the constant of variation.
  4. Write the equation that relates x and y.

Now we’ll solve a few applications of direct variation.

When Raoul runs on the treadmill at the gym, the number of calories, c , he burns varies directly with the number of minutes, m , he uses the treadmill. He burned 315 calories when he used the treadmill for 18 minutes.

  1. Write the equation that relates c and m .
  2. How many calories would he burn if he ran on the treadmill for 25 minutes?

Solution


The number of calories, c , varies directly with
the number of minutes, m , on the treadmill,
and c = 315 when m = 18 .
Write the formula for direct variation. .
We will use c in place of y and m in place of x . .
Substitute the given values for the variables. .
Solve for the constant of variation. .
.
Write the equation that relates c and m . .
Substitute in the constant of variation. .


Find c when m = 25 .
Write the equation that relates c and m . .
Substitute the given value for m . .
Simplify. .
Raoul would burn 437.5 calories if he used the
treadmill for 25 minutes.

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The number of calories, c , burned varies directly with the amount of time, t, spent exercising. Arnold burned 312 calories in 65 minutes exercising.

  1. Write the equation that relates c and t .
  2. How many calories would he burn if he exercises for 90 minutes?

c = 4.8 t 432 calories

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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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