Direct and Inverse Variation Problems: Definition & Examples

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  • 0:02 Variation
  • 2:05 Example 1
  • 3:56 Example 2
  • 5:36 Lesson Summary
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Lesson Transcript
Elizabeth Foster

Elizabeth has been involved with tutoring since high school and has a B.A. in Classics.

Expert Contributor
Kathryn Boddie

Kathryn earned her Ph.D. in Mathematics from UW-Milwaukee in 2019. She has over 10 years of teaching experience at high school and university level.

In this lesson, you'll learn how to approach questions about direct and inverse variation with a simple explanation of what the terms mean and how to apply them to problems.


Equations with direct and inverse variation sound a little intimidating, but really, they're just two different ways of talking about how one number changes relative to another number.

In direct variation, as one number increases, so does the other. This is also called direct proportion: they're the same thing. An example of this is relationship between age and height. As the age in years of a child increases, the height will also increase.

In the abstract, we can express direct variation by using the equation y = kx.

x and y are the two quantities - in our example, they'd be the age and the height of the child. k is called the constant of proportionality: it tells you specifically how much bigger y will get for every increase in x. For example, maybe y = 2x: this means that for every increase in x, y will increase by double that amount.

You can see that the bigger the number you plug in for x, the bigger the resulting value of y will be.

In inverse variation, it's exactly the opposite: as one number increases, the other decreases. This is also called inverse proportion. An example would be the relationship between time spent goofing off in class and your grade on the midterm. The more you goof off, the lower your score on the test.

If we wanted to give this one an equation, we would say:

y = k/x, where x and y are the two quantities, and k is still the constant of proportionality, telling how much one varies when the other changes.

You can see that in this equation, you divide a constant number by x to get the value of y. So the bigger the value of x, the smaller the value of y will be. That's inverse variation: as one goes up, the other goes down.

Example 1

This might seem really complicated and confusing, but just remember the two formulas: y = kx for direct variation, and y = k/x for inverse variation. As you practice with example problems, you'll learn how to apply them to specific problems.

A very simple example is a problem like this one:

The population of a certain species of bacteria varies directly with the temperature. When the temperature is 35 degrees Celsius, there are 7 million bacteria. How many millions of bacteria are there when the temperature is 38 degrees Celsius?

First of all, we can see that we'll be using the direct variation equation, y = kx.

Now let's plug in what we have from the problem:

The problem gives us two values: temperature and number of bacteria. We'll plug in the temperature for x and the number of bacteria for y. This gives us 7 = k(35).

Now all we have to do is divide to find the value of k for this particular problem: it turns out to be 0.2.

The next step is to use that value to find out how many millions of bacteria there are at 38 degrees Celsius. So, we use the equation again.

This time, we'll plug in x and k, since we're looking for y. We find that y = (0.2)(38). Do the multiplication, and we learn that y, or the value of the population in millions is 7.6. So the answer to this question would be 7.6 million bacteria.

That wasn't so painful, right? It's just about using the equations properly. Now let's try one that's a little harder.

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Additional Activities

Further Study into Direct and Inverse Variation

In the video lesson, we learned that when one quantity varies directly with another, we can represent the situation by the equation y = kx and k is called the constant of proportionality. If one quantity varies inversely with another, we can represent the situation by the equation y = k/x. But what if the situation is a little more complicated than that?


y varies directly with the square of x. If y = 5 when x = 3, what is the value of y when x = 9?


In this example, y doesn't vary directly with x, but with the square of x. The idea for solving the problem is still the same. As x2 gets bigger, so will y, and we can represent this with the equation y = kx2 where k is the constant of proportionality. Now, use the given values for x and y to solve for the constant of proportionality.

y = kx2

5 = k(3)2

5 = 9k

5/9 = k

Now that the constant of proportionality is found, we can write the direct variation relationship between our variables as y = 5/9 x2

To find what y is when x = 9, substitute into the equation.

y = 5/9 (9)2

y = 5/9 (81)

y = 45

Challenge Exercise

The circulation time is the amount of time it takes for blood to fully circulate in a body. In mammals, the circulation time is directly proportional to the fourth root of the body mass of the animal. The circulation time for a 450 kg horse is 80 seconds. What is the body mass of a human whose circulation time is 50 seconds?


This is a direct variation problem, but instead of one variable being directly proportional to the other, it is directly proportional to the fourth root of the other variable. We can set this up as an equation T = k(m)1/4 where T is the circulation time of the mammal in seconds and m is the body mass of the animal in kilograms. Now we can substitute the information about the horse into the equation to find the constant of proportionality.

T = k m1/4

80 = k 4501/4

80/(450)1/4 = k

k is approximately 17.369

So our direct variation equation is T=17.369 m1/4 . Now to find the body mass of a human whose circulation time is 50 seconds, we substitute T = 50 into the equation.

T = 17.369 m1/4

50 = 17.369 m1/4

50/17.369 = m1/4

(50/17.369)4 = m

m is approximately 68.672

The human has a mass of approximately 68.672 kilograms.

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