Determine if Linear x/4-y/3=1

$\frac{x}{4} - \frac{y}{3} = 1$

Step 1

Solve the equation for $y$ .

Subtract $\frac{x}{4}$ from both sides of the equation.

$- \frac{y}{3} = 1 - \frac{x}{4}$

Multiply both sides of the equation by $- 3$ .

$- 3 (- \frac{y}{3}) = - 3 (1 - \frac{x}{4})$

Simplify both sides of the equation.

Simplify the left side.

Simplify $- 3 (- \frac{y}{3})$ .

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{y}{3}$ into the numerator.

$- 3 \frac{- y}{3} = - 3 (1 - \frac{x}{4})$

Factor $3$ out of $- 3$ .

$3 (- 1) \frac{- y}{3} = - 3 (1 - \frac{x}{4})$

Cancel the common factor.

$3 \cdot - 1 \frac{- y}{3} = - 3 (1 - \frac{x}{4})$

Rewrite the expression.

$- - y = - 3 (1 - \frac{x}{4})$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 y = - 3 (1 - \frac{x}{4})$

Multiply $y$ by $1$ .

$y = - 3 (1 - \frac{x}{4})$

Simplify the right side.

Simplify $- 3 (1 - \frac{x}{4})$ .

Apply the distributive property.

$y = - 3 \cdot 1 - 3 (- \frac{x}{4})$

Multiply $- 3$ by $1$ .

$y = - 3 - 3 (- \frac{x}{4})$

Multiply $- 3 (- \frac{x}{4})$ .

Multiply $- 1$ by $- 3$ .

$y = - 3 + 3 \frac{x}{4}$

Combine $3$ and $\frac{x}{4}$ .

$y = - 3 + \frac{3 x}{4}$

Reorder $- 3$ and $\frac{3 x}{4}$ .

$y = \frac{3 x}{4} - 3$

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

Determine if Linear x/4-y/3=1

Related Questions

Support LCMGCF.com by disabling your adblocker.