Precalculus Examples

$y = 15$ , $x = 10$ , $x = 6$

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.

$k = \frac{15}{10}$

Step 4

Cancel the common factor of $15$ and $10$ .

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Step 4.1

Factor $5$ out of $15$ .

$k = \frac{5 (3)}{10}$

Step 4.2

Cancel the common factors.

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Step 4.2.1

Factor $5$ out of $10$ .

$k = \frac{5 \cdot 3}{5 \cdot 2}$

Step 4.2.2

Cancel the common factor.

$k = \frac{5 \cdot 3}{5 \cdot 2}$

Step 4.2.3

Rewrite the expression.

$k = \frac{3}{2}$

Step 5

Use the formula $y = k x$ to substitute $\frac{3}{2}$ for $k$ and $6$ for $x$ .

$y = (\frac{3}{2}) \cdot (6)$

Step 6

Solve for .

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Step 6.1

Multiply $\frac{3}{2}$ by $6$ .

$y = \frac{3}{2} \cdot (6)$

Step 6.2

Multiply $\frac{3}{2}$ by $6$ .

$y = \frac{3}{2} \cdot 6$

Step 6.3

Remove parentheses.

$y = (\frac{3}{2}) \cdot (6)$

Step 6.4

Simplify $(\frac{3}{2}) \cdot (6)$ .

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Step 6.4.1

Cancel the common factor of $2$ .

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Step 6.4.1.1

Factor $2$ out of $6$ .

$y = \frac{3}{2} \cdot (2 (3))$

Step 6.4.1.2

Cancel the common factor.

$y = \frac{3}{2} \cdot (2 \cdot 3)$

Step 6.4.1.3

Rewrite the expression.

$y = 3 \cdot 3$

Step 6.4.2

Multiply $3$ by $3$ .

$y = 9$

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