Solve 2x^2+xy=-16

$$2 x ^ { 2 } + x y = - 16$$

Solve for y

$$y=-2x-\frac{16}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Subtract 2x^{2} from both sides.

$$xy=-16-2x^{2}$$

The equation is in standard form.

$$xy=-2x^{2}-16$$

Divide both sides by x.

$$\frac{xy}{x}=\frac{-2x^{2}-16}{x}$$

Dividing by x undoes the multiplication by x.

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

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

$$y=-2x-\frac{16}{x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{y^{2}-128}-y}{4}$$

$$x=\frac{-\sqrt{y^{2}-128}-y}{4}$$

Solve for x

$$x=\frac{\sqrt{y^{2}-128}-y}{4}$$

$$x=\frac{-\sqrt{y^{2}-128}-y}{4}\text{, }|y|\geq 8\sqrt{2}$$