Algebra Examples

x = 5 y

$x = 5 y$ ,

y = \frac{1}{3}

$y = \frac{1}{3}$ ,

y = 2

$y = 2$

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

$y = k x$ , where

k

$k$ is the constant of variation.

y = k x

$y = k x$

Step 2

Solve the equation for

k

$k$ , the constant of variation.

k = \frac{y}{x}

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

Step 3

Replace the variables

x

$x$ and

y

$y$ with the actual values.

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

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

Step 4

Multiply the numerator by the reciprocal of the denominator.

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

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

Step 5

Multiply

\frac{1}{3} \cdot \frac{1}{5 y}

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

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Step 5.1

Multiply

\frac{1}{3}

$\frac{1}{3}$ by

\frac{1}{5 y}

$\frac{1}{5 y}$ .

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

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

Step 5.2

Multiply

5

$5$ by

3

$3$ .

k = \frac{1}{15 y}

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

k = \frac{1}{15 y}

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

Step 6

Use the formula

x = k y

$x = k y$ to substitute

\frac{1}{15 y}

$\frac{1}{15 y}$ for

k

$k$ and

2

$2$ for

y

$y$ .

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

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

Step 7

Solve for

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Step 7.1

Multiply

\frac{1}{15 (2)}

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

2

$2$ .

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

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

Step 7.2

Multiply

\frac{1}{15 (2)}

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

2

$2$ .

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

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

Step 7.3

Remove parentheses.

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

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

Step 7.4

Simplify

(\frac{1}{15 (2)}) \cdot (2)

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

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Step 7.4.1

Multiply

15

$15$ by

2

$2$ .

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

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

Step 7.4.2

Cancel the common factor of

2

$2$ .

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Step 7.4.2.1

Factor

2

$2$ out of

30

$30$ .

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

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

Step 7.4.2.2

Cancel the common factor.

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

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

Step 7.4.2.3

Rewrite the expression.

x = \frac{1}{15}

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

x = \frac{1}{15}

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

x = \frac{1}{15}

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

x = \frac{1}{15}

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

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