Lesson Objectives

- Demonstrate an understanding of linear equations in two variables
- Learn how to solve a direct variation problem
- Learn how to solve a direct variation as a power problem
- Learn how to solve a word problem that involves direct variation

## How to Solve a Direct Variation Problem

In this lesson, we will learn about direct variation and direct variation as a power. When we work with a direct variation problem, we have a variable k,
which is known as the constant of variation. We may also refer to k as the constant of proportionality. Our direct variation equation is given as:

y = kx

When we studied linear equations in two variables, we worked with a similar equation form. Recall that our slope-intercept form of a linear equation in two variables was given as:

y = mx + b

If we say our y-intercept occurs at (0,0) and replace the slope (m) with (k), we obtain:

y = kx

k is just the slope or the change in y per 1 unit change in x:

If k > 0, then as x ↑ by 1 unit, y ↑ k units.

If k > 0, then as x ↓ by 1 unit, y ↓ k units.

Let's suppose we saw the following direct variation equation:

y = 4x

We can see that our constant of variation is 4. This means each time we increase x by 1, y will increase by 4:

We often need to find the value for k. We can do this by solving our equation for k:
$$y = kx$$
$$k = \frac{y}{x}$$

Example 1: Solve each direct variation problem

If y varies directly with x and y = 12 when x = 4, find y when x = 7.

Step 1) Write the variation equation: k = y/x

$$k = \frac{y}{x}$$ Step 2) Substitute in for the given values and find the value of k:

In the beginning of the problem, we are told that y = 12, when x = 4. We will plug in for x and y and find the value of k: $$k = \frac{y}{x}$$ $$k = \frac{12}{4} = 3$$ k = 3

Step 3) Rewrite the variation equation with the known value of k:

y = 3x

Step 4) Substitute the remaining values and find the unknown:

We are told to find y when x is 7. Since we know k is 3, we can plug in a 3 for k and a 7 for x and find y: $$y = 3x$$ $$y = 3(7)$$ $$y = 21$$ y is 21 when x is 7.

Example 2: Solve each direct variation problem

If y varies directly with x and y = 6 when x = 12, find y when x = 18.

Step 1) Write the variation equation: k = y/x

$$k = \frac{y}{x}$$ Step 2) Substitute in for the given values and find the value of k:

In the beginning of the problem, we are told that y = 6, when x = 12. We will plug in for x and y and find the value of k: $$k = \frac{y}{x}$$ $$k = \frac{6}{12} = \frac{1}{2}$$ k = 1/2

Step 3) Rewrite the variation equation with the known value of k:

y = (1/2)x

Step 4) Substitute the remaining values and find the unknown:

We are told to find y when x is 18. Since we know k is 1/2, we can plug in a 1/2 for k and an 18 for x and find y: $$y = \frac{1}{2}x$$ $$y = \frac{1}{2} \cdot 18$$ $$y = 9$$ y is 9 when x is 18.

y = kx

Let's look at an example.

Example 3: Solve each direct variation problem

If y varies directly with x

Step 1) Write the variation equation:

In this case, we have x raised to the second power. This means our equation will change to:

y = kx

We can also solve for k and obtain:

$$k = \frac{y}{x^2}$$ Step 2) Substitute in for the given values and find the value of k:

In the beginning of the problem, we are told that y = 45, when x = 3. We will plug in for x and y and find the value of k: $$k = \frac{y}{x^2}$$ $$k = \frac{45}{(3)^2} = \frac{45}{9} = 5$$ k = 5

Step 3) Rewrite the variation equation with the known value of k:

y = 5x

Step 4) Substitute the remaining values and find the unknown:

We are told to find y when x is (-4). Since we know k is 5, we can plug in a 5 for k and a (-4) for x and find y: $$y = 5x^2$$ $$y = 5(-4)^2$$ $$y = 5(16) = 80$$ y is 80 when x is (-4).

Example 4: Solve each direct variation word problem

The yearly simple interest earned on an investment is given with the formula: I = prt, where I is the amount of simple interest earned, p is the principal (amount invested), r is the rate (interest rate as a decimal), and t is the time (given in years). If we discuss a scenario where our time period was 1 year, we could rewrite our formula as: I = pr(1) or I = pr. For a given principal and time period of 1 year, we can say that the simple interest earned varies directly with the rate of interest. Suppose a one-year investment at 7% annual simple interest yields $350. How much would the same investment earn at 12% annual simple interest?

Let's start with what is known from the first scenario. Setup the simple interest formula and fill in for time (1), Interest (350), and rate (.07 for 7%):

I = prt

Since time is 1, we can rewrite our formula as:

I = pr

350 = p(.07)

We can solve for p:

$$p = \frac{350}{.07} = 5000$$ Our principal is $5000. This is our constant of variation and will not change throughout the problem. We want to know how much the same $5000 investment would earn if the rate was 12%. Let's plug in and find our answer:

I = pr

I = 5000(.12) = 600

So our $5000 investment would earn $600 if invested at 12%.

y = kx

When we studied linear equations in two variables, we worked with a similar equation form. Recall that our slope-intercept form of a linear equation in two variables was given as:

y = mx + b

If we say our y-intercept occurs at (0,0) and replace the slope (m) with (k), we obtain:

y = kx

k is just the slope or the change in y per 1 unit change in x:

If k > 0, then as x ↑ by 1 unit, y ↑ k units.

If k > 0, then as x ↓ by 1 unit, y ↓ k units.

Let's suppose we saw the following direct variation equation:

y = 4x

We can see that our constant of variation is 4. This means each time we increase x by 1, y will increase by 4:

y = 4x | |
---|---|

x = -2 | y = -8 |

x = -1 | y = -4 |

x = 0 | y = 0 |

x = 1 | y = 4 |

x = 2 | y = 8 |

x = 3 | y = 12 |

### Solving a Direct Variation Problem

- Write the variation equation: y = kx or k = y/x
- Substitute in for the given values and find the value of k
- Rewrite the variation equation: y = kx with the known value of k
- Substitute the remaining values and find the unknown

Example 1: Solve each direct variation problem

If y varies directly with x and y = 12 when x = 4, find y when x = 7.

Step 1) Write the variation equation: k = y/x

$$k = \frac{y}{x}$$ Step 2) Substitute in for the given values and find the value of k:

In the beginning of the problem, we are told that y = 12, when x = 4. We will plug in for x and y and find the value of k: $$k = \frac{y}{x}$$ $$k = \frac{12}{4} = 3$$ k = 3

Step 3) Rewrite the variation equation with the known value of k:

y = 3x

Step 4) Substitute the remaining values and find the unknown:

We are told to find y when x is 7. Since we know k is 3, we can plug in a 3 for k and a 7 for x and find y: $$y = 3x$$ $$y = 3(7)$$ $$y = 21$$ y is 21 when x is 7.

Example 2: Solve each direct variation problem

If y varies directly with x and y = 6 when x = 12, find y when x = 18.

Step 1) Write the variation equation: k = y/x

$$k = \frac{y}{x}$$ Step 2) Substitute in for the given values and find the value of k:

In the beginning of the problem, we are told that y = 6, when x = 12. We will plug in for x and y and find the value of k: $$k = \frac{y}{x}$$ $$k = \frac{6}{12} = \frac{1}{2}$$ k = 1/2

Step 3) Rewrite the variation equation with the known value of k:

y = (1/2)x

Step 4) Substitute the remaining values and find the unknown:

We are told to find y when x is 18. Since we know k is 1/2, we can plug in a 1/2 for k and an 18 for x and find y: $$y = \frac{1}{2}x$$ $$y = \frac{1}{2} \cdot 18$$ $$y = 9$$ y is 9 when x is 18.

### Direct Variation as a Power

In some cases, we will encounter direct variation as a power. We will utilize the same strategy to solve this type of problem. We will see direct variation as a power presented as:y = kx

^{n}Let's look at an example.

Example 3: Solve each direct variation problem

If y varies directly with x

^{2}and y = 45 when x = 3, find y when x = (-4).Step 1) Write the variation equation:

In this case, we have x raised to the second power. This means our equation will change to:

y = kx

^{2}We can also solve for k and obtain:

$$k = \frac{y}{x^2}$$ Step 2) Substitute in for the given values and find the value of k:

In the beginning of the problem, we are told that y = 45, when x = 3. We will plug in for x and y and find the value of k: $$k = \frac{y}{x^2}$$ $$k = \frac{45}{(3)^2} = \frac{45}{9} = 5$$ k = 5

Step 3) Rewrite the variation equation with the known value of k:

y = 5x

^{2}Step 4) Substitute the remaining values and find the unknown:

We are told to find y when x is (-4). Since we know k is 5, we can plug in a 5 for k and a (-4) for x and find y: $$y = 5x^2$$ $$y = 5(-4)^2$$ $$y = 5(16) = 80$$ y is 80 when x is (-4).

### Direct Variation Word Problems

We may also encounter a word problem that involves direct variation. Most of these problems are fairly simple. Let's look at an example.Example 4: Solve each direct variation word problem

The yearly simple interest earned on an investment is given with the formula: I = prt, where I is the amount of simple interest earned, p is the principal (amount invested), r is the rate (interest rate as a decimal), and t is the time (given in years). If we discuss a scenario where our time period was 1 year, we could rewrite our formula as: I = pr(1) or I = pr. For a given principal and time period of 1 year, we can say that the simple interest earned varies directly with the rate of interest. Suppose a one-year investment at 7% annual simple interest yields $350. How much would the same investment earn at 12% annual simple interest?

Let's start with what is known from the first scenario. Setup the simple interest formula and fill in for time (1), Interest (350), and rate (.07 for 7%):

I = prt

Since time is 1, we can rewrite our formula as:

I = pr

350 = p(.07)

We can solve for p:

$$p = \frac{350}{.07} = 5000$$ Our principal is $5000. This is our constant of variation and will not change throughout the problem. We want to know how much the same $5000 investment would earn if the rate was 12%. Let's plug in and find our answer:

I = pr

I = 5000(.12) = 600

So our $5000 investment would earn $600 if invested at 12%.

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