Solve 4x-3y+12=0

$$4x-3y+12=0$$

Solve for x

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

Steps for Solving Linear Equation

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

$$4x+12=3y$$

Subtract 12 from both sides.

$$4x=3y-12$$

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide -12+3y by 4.

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

Solve for y

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

Steps for Solving Linear Equation

Subtract 4x from both sides. Anything subtracted from zero gives its negation.

$$-3y+12=-4x$$

Subtract 12 from both sides.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide -4x-12 by -3.

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