# Solve direct variation problems

In the example above, Nicole’s earnings can be found by multiplying her sales by her commission. The formula *e* = 0.16*s* tells us her earnings, *e*, come from the product of 0.16, her commission, and the sale price of the vehicle. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive.

s, sales prices |
e = 0.16s |
Interpretation |
---|---|---|

$4,600 | e = 0.16(4,600) = 736 |
A sale of a $4,600 vehicle results in $736 earnings. |

$9,200 | e = 0.16(9,200) = 1,472 |
A sale of a $9,200 vehicle results in $1472 earnings. |

$18,400 | e = 0.16(18,400) = 2,944 |
A sale of a $18,400 vehicle results in $2944 earnings. |

Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from $4,600 to $9,200, and we double the earnings from $736 to $1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called **direct variation**. Each variable in this type of relationship **varies directly **with the other.

The graph below represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula [latex]y=k{x}^{n}[/latex] is used for direct variation. The value *k* is a nonzero constant greater than zero and is called the **constant of variation**. In this case, *k *= 0.16 and *n *= 1.

**Figure 1**

### A General Note: Direct Variation

If *x *and *y* are related by an equation of the form

then we say that the relationship is **direct variation** and *y* **varies directly** with the *n*th power of *x*. In direct variation relationships, there is a nonzero constant ratio [latex]k=\frac{y}{{x}^{n}}[/latex], where *k* is called the **constant of variation**, which help defines the relationship between the variables.

### How To: Given a description of a direct variation problem, solve for an unknown.**
**

- Identify the input,
*x*, and the output,*y*. - Determine the constant of variation. You may need to divide
*y*by the specified power of*x*to determine the constant of variation. - Use the constant of variation to write an equation for the relationship.
- Substitute known values into the equation to find the unknown.

### Example 1: Solving a Direct Variation Problem

The quantity *y* varies directly with the cube of *x*. If *y *= 25 when *x *= 2, find *y* when *x* is 6.

### Solution

The general formula for direct variation with a cube is [latex]y=k{x}^{3}[/latex]. The constant can be found by dividing *y* by the cube of *x*.

Now use the constant to write an equation that represents this relationship.

Substitute *x* = 6 and solve for *y*.

### Q & A

**Do the graphs of all direct variation equations look like Example 1?**

*No. Direct variation equations are power functions—they may be linear, quadratic, cubic, quartic, radical, etc. But all of the graphs pass through (0, 0).*

### Try It 1

The quantity *y* varies directly with the square of *x*. If *y *= 24 when *x *= 3, find *y* when *x* is 4.

## Licenses & Attributions

### CC licensed content, Shared previously

- Precalculus.
**Provided by:**OpenStax**Authored by:**Jay Abramson, et al..**Located at:**https://openstax.org/books/precalculus/pages/1-introduction-to-functions.**License:**CC BY: Attribution.**License terms:**Download For Free at : http://cnx.org/contents/[email protected]..

## Analysis of the Solution

The graph of this equation is a simple cubic, as shown below.

Figure 2