Solve 4x^2+12y=16

$$4 x ^ { 2 } + 12 y = 16$$

Solve for y

$$y=\frac{4-x^{2}}{3}$$

Steps for Solving Linear Equation

Subtract 4x^{2} from both sides.

$$12y=16-4x^{2}$$

Divide both sides by 12.

$$\frac{12y}{12}=\frac{16-4x^{2}}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$y=\frac{16-4x^{2}}{12}$$

Divide 16-4x^{2} by 12.

$$y=\frac{4-x^{2}}{3}$$

Solve for x (complex solution)

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

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

Solve for x

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

$$x=-\sqrt{4-3y}\text{, }y\leq \frac{4}{3}$$