Lesson Objectives
  • Demonstrate an understanding of direct variation
  • Learn how to solve an inverse variation problem
  • Learn how to solve an inverse variation as a power problem
  • Learn how to solve an inverse variation word problem

How to Solve an Inverse Variation Problem


In the last lesson, we learned about direct variation. When we discuss direct variation, we generally say that y varies directly with x, if there is some constant k such that:
y = kx
In our equation, k is known as the constant of variation or the constant of proportionality. We can say that if:
k > 0
x ↑ , y ↑
x ↓ , y ↓
With direct variation, if k is positive, as x increases, y increases. When we look at inverse variation problems, the reverse will be true. For inverse variation, we can say that y varies inversely with x if there is some constant k such that: $$y = \frac{k}{x}$$ Again, in our equation k is known as the constant of variation or the constant of proportionality. We can say that if:
k > 0
x ↑ , y ↓
x ↓ , y ↑

Solving an Inverse Variation Problem

  • Write the variation equation: y = k/x or k = xy
  • Substitute in for the given values and find the value of k
  • Rewrite the variation equation: y = k/x with the known value of k
  • Substitute the remaining values and find the unknown
Let's look at a few examples.
Example 1: Solve each inverse variation problem
y varies inversely with x and y = 2 when x = 5. Find y when x = 30
Step 1) Write the variation equation: y = k/x or k = xy:
k = xy
Step 2) Substitute in for the given values of x and y:
k = (2)(5) = 10
Step 3) Rewrite the variation equation: y = k/x with the known value of k (10):
$$y = \frac{10}{x}$$ Step 4) Substitute the remaining values and find the unknown: $$y = \frac{10}{30} = \frac{1}{3}$$ y is 1/3 when x is 30.
Example 2: Solve each inverse variation problem
y varies inversely with x and y = 4 when x = (-5). Find y when x = 20
Step 1) Write the variation equation: y = k/x or k = xy:
k = xy
Step 2) Substitute in for the given values of x and y:
k = (4)(-5) = -20
Step 3) Rewrite the variation equation: y = k/x with the known value of k (-20):
$$y = \frac{-20}{x}$$ Step 4) Substitute the remaining values and find the unknown: $$y = \frac{-20}{20} = -1$$ y is -1 when x is 20.

Inverse Variation as a Power

In some cases, we will encounter inverse variation as a power. We will utilize the same strategy to solve this type of problem. We will see inverse variation as a power presented as:
y = k/xn
Let's look at an example.
Example 3: Solve each inverse variation problem
y varies inversely with x2 and y = 5/3 when x = 3. Find y when x = 12
Step 1) Write the variation equation: y = k/x2 or k = x2y:
k = x2y
Step 2) Substitute in for the given values of x and y:
k = (3)2(5/3) = (9)(5/3) = 15
Step 3) Rewrite the variation equation: y = k/x2 with the known value of k (15):
$$y = \frac{15}{x^2}$$ Step 4) Substitute the remaining values and find the unknown: $$y = \frac{15}{12^2} = \frac{15}{144} = \frac{5}{48}$$ y is 5/48 when x is 12.

Inverse Variation Word Problems

We may also encounter a word problem that involves inverse variation. Most of these problems are fairly simple. Let's look at an example.
Example 4: Solve each inverse variation word problem
Over a given distance, speed varies inversely with time. If a race car goes a certain distance in 3 hours at 190 miles per hour, what speed is needed to go the same distance in 2 hours?
First and foremost, let's think about the distance formula. This is used with all motion word problems:
d = r • t
In our formula, d is the distance, r is the rate, and t is the time. In this case, we are saying the speed or rate varies inversely with time. We can set this equation up by solving for r: $$r = \frac{d}{t}$$ Since the distance in our problem doesn't change, it is our constant of variation. The problem tells us the race car goes a certain distance in 3 hours at 190 miles per hour. We can find distance by plugging in for r and t: $$d = (190)(3) = 570$$ Our distance or constant of variation is 570. Now we can use this to solve our problem. We are asked what speed is needed to go the same distance (570 miles) in 2 hours. Let's plug into our equation, and solve for rate: $$r = \frac{d}{t}$$ $$r = \frac{570}{2} = 285$$ This tells us the race car would need to drive at a speed of 285 miles per hour to go the same distance (570 miles) in 2 hours.