Solve 2x-3y+12=0

$$2x-3y+12=0$$

Solve for x

$$x=\frac{3\left(y-4\right)}{2}$$

Steps for Solving Linear Equation

Add 3y to both sides. Anything plus zero gives itself.

$$2x+12=3y$$

Subtract 12 from both sides.

$$2x=3y-12$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{3y-12}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{3y-12}{2}$$

Divide -12+3y by 2.

$$x=\frac{3y}{2}-6$$

Solve for y

$$y=\frac{2\left(x+6\right)}{3}$$

Steps for Solving Linear Equation

Subtract 2x from both sides. Anything subtracted from zero gives its negation.

$$-3y+12=-2x$$

Subtract 12 from both sides.

$$-3y=-2x-12$$

Divide both sides by -3.

$$\frac{-3y}{-3}=\frac{-2x-12}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$y=\frac{-2x-12}{-3}$$

Divide -2x-12 by -3.

$$y=\frac{2x}{3}+4$$