Solve for y 3x-4y=12 ; for y

$3 x - 4 y = 12$ ; for $y$

Subtract $3 x$ from both sides of the equation.

$- 4 y = 12 - 3 x$

Divide each term in $- 4 y = 12 - 3 x$ by $- 4$ and simplify.

Divide each term in $- 4 y = 12 - 3 x$ by $- 4$ .

$\frac{- 4 y}{- 4} = \frac{12}{- 4} + \frac{- 3 x}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 y}{- 4} = \frac{12}{- 4} + \frac{- 3 x}{- 4}$

Divide $y$ by $1$ .

$y = \frac{12}{- 4} + \frac{- 3 x}{- 4}$

Simplify the right side.

Simplify each term.

Divide $12$ by $- 4$ .

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

Dividing two negative values results in a positive value.

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