Precalculus Examples

$x = 5 y$ , $y = \frac{1}{3}$ , $y = 2$

Step 1

When two variable quantities have a constant ratio, their relationship is called a direct variation. It is said that one variable varies directly as the other. The formula for direct variation is $y = k x$ , where $k$ is the constant of variation.

$y = k x$

Step 2

Solve the equation for $k$ , the constant of variation.

$k = \frac{y}{x}$

Step 3

Replace the variables $x$ and $y$ with the actual values.

$k = \frac{\frac{1}{3}}{5 y}$

Step 4

Multiply the numerator by the reciprocal of the denominator.

$k = \frac{1}{3} \cdot \frac{1}{5 y}$

Step 5

Multiply $\frac{1}{3} \cdot \frac{1}{5 y}$ .

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Step 5.1

Multiply $\frac{1}{3}$ by $\frac{1}{5 y}$ .

$k = \frac{1}{3 (5 y)}$

Step 5.2

Multiply $5$ by $3$ .

$k = \frac{1}{15 y}$

Step 6

Use the formula $x = k y$ to substitute $\frac{1}{15 y}$ for $k$ and $2$ for $y$ .

$x = (\frac{1}{15 (2)}) \cdot (2)$

Step 7

Solve for .

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Step 7.1

Multiply $\frac{1}{15 (2)}$ by $2$ .

$x = \frac{1}{15 (2)} \cdot (2)$

Step 7.2

Multiply $\frac{1}{15 (2)}$ by $2$ .

$x = \frac{1}{15 (2)} \cdot 2$

Step 7.3

Remove parentheses.

$x = (\frac{1}{15 (2)}) \cdot (2)$

Step 7.4

Simplify $(\frac{1}{15 (2)}) \cdot (2)$ .

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Step 7.4.1

Multiply $15$ by $2$ .

$x = \frac{1}{30} \cdot 2$

Step 7.4.2

Cancel the common factor of $2$ .

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Step 7.4.2.1

Factor $2$ out of $30$ .

$x = \frac{1}{2 (15)} \cdot 2$

Step 7.4.2.2

Cancel the common factor.

$x = \frac{1}{2 \cdot 15} \cdot 2$

Step 7.4.2.3

Rewrite the expression.

$x = \frac{1}{15}$

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