Solve 3x-4y+12=0

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

Solve for x

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

Steps for Solving Linear Equation

Add 4y to both sides. Anything plus zero gives itself.

$$3x+12=4y$$

Subtract 12 from both sides.

$$3x=4y-12$$

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide -12+4y by 3.

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

Solve for y

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

Steps for Solving Linear Equation

Subtract 3x from both sides. Anything subtracted from zero gives its negation.

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

Subtract 12 from both sides.

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

Divide both sides by -4.

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

Dividing by -4 undoes the multiplication by -4.

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

Divide -3x-12 by -4.

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