Finite Math Examples

$2 x = \frac{5}{4} y + 1$

Step 1

Solve the equation for $y$ .

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Step 1.1

Rewrite the equation as $\frac{5}{4} y + 1 = 2 x$ .

$\frac{5}{4} y + 1 = 2 x$

Step 1.2

Combine $\frac{5}{4}$ and $y$ .

$\frac{5 y}{4} + 1 = 2 x$

Step 1.3

Subtract $1$ from both sides of the equation.

$\frac{5 y}{4} = 2 x - 1$

Step 1.4

Multiply both sides of the equation by $\frac{4}{5}$ .

$\frac{4}{5} \cdot \frac{5 y}{4} = \frac{4}{5} (2 x - 1)$

Step 1.5

Simplify both sides of the equation.

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Step 1.5.1

Simplify the left side.

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Step 1.5.1.1

Simplify $\frac{4}{5} \cdot \frac{5 y}{4}$ .

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Step 1.5.1.1.1

Cancel the common factor of $4$ .

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Step 1.5.1.1.1.1

Cancel the common factor.

$\frac{4}{5} \cdot \frac{5 y}{4} = \frac{4}{5} (2 x - 1)$

Step 1.5.1.1.1.2

Rewrite the expression.

$\frac{1}{5} (5 y) = \frac{4}{5} (2 x - 1)$

Step 1.5.1.1.2

Cancel the common factor of $5$ .

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Step 1.5.1.1.2.1

Factor $5$ out of $5 y$ .

$\frac{1}{5} (5 (y)) = \frac{4}{5} (2 x - 1)$

Step 1.5.1.1.2.2

Cancel the common factor.

$\frac{1}{5} (5 y) = \frac{4}{5} (2 x - 1)$

Step 1.5.1.1.2.3

Rewrite the expression.

$y = \frac{4}{5} (2 x - 1)$

Step 1.5.2

Simplify the right side.

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Step 1.5.2.1

Simplify $\frac{4}{5} (2 x - 1)$ .

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Step 1.5.2.1.1

Apply the distributive property.

$y = \frac{4}{5} (2 x) + \frac{4}{5} \cdot - 1$

Step 1.5.2.1.2

Multiply $\frac{4}{5} (2 x)$ .

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Step 1.5.2.1.2.1

Combine $2$ and $\frac{4}{5}$ .

$y = \frac{2 \cdot 4}{5} x + \frac{4}{5} \cdot - 1$

Step 1.5.2.1.2.2

Multiply $2$ by $4$ .

$y = \frac{8}{5} x + \frac{4}{5} \cdot - 1$

Step 1.5.2.1.2.3

Combine $\frac{8}{5}$ and $x$ .

$y = \frac{8 x}{5} + \frac{4}{5} \cdot - 1$

Step 1.5.2.1.3

Multiply $\frac{4}{5} \cdot - 1$ .

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Step 1.5.2.1.3.1

Combine $\frac{4}{5}$ and $- 1$ .

$y = \frac{8 x}{5} + \frac{4 \cdot - 1}{5}$

Step 1.5.2.1.3.2

Multiply $4$ by $- 1$ .

$y = \frac{8 x}{5} + \frac{- 4}{5}$

Step 1.5.2.1.4

Move the negative in front of the fraction.

$y = \frac{8 x}{5} - \frac{4}{5}$

Step 2

A linear equation is an equation of a straight line, which means that the degree of a linear equation must be $0$ or $1$ for each of its variables. In this case, the degree of variable $y$ is $1$ and the degree of variable $x$ is $1$ .

Linear