Direct Variation Lesson

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:

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

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

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

Let's look at a few examples.
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:
At 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:
At 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 = kxⁿ
Let's look at an example.
Example 3: Solve each direct variation problem.
If y varies directly with x² 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 = kx²
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:
At 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 = 5x²
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
A $5000 investment would earn $600 if invested at 12%.

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