Solve 2x^2-xy-y=0

$$2 { x }^{ 2 } -xy-y=0$$

Solve for y

$$y=\frac{2x^{2}}{x+1}$$

$$x\neq -1$$

Steps for Solving Linear Equation

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

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

Combine all terms containing y.

$$\left(-x-1\right)y=-2x^{2}$$

Divide both sides by -x-1.

$$\frac{\left(-x-1\right)y}{-x-1}=-\frac{2x^{2}}{-x-1}$$

Dividing by -x-1 undoes the multiplication by -x-1.

$$y=-\frac{2x^{2}}{-x-1}$$

Divide -2x^{2} by -x-1.

$$y=\frac{2x^{2}}{x+1}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{y\left(y+8\right)}+y}{4}$$

$$x=\frac{-\sqrt{y\left(y+8\right)}+y}{4}$$

Solve for x

$$x=\frac{\sqrt{y\left(y+8\right)}+y}{4}$$

$$x=\frac{-\sqrt{y\left(y+8\right)}+y}{4}\text{, }y\leq -8\text{ or }y\geq 0$$