Solve 2x^2-xy-2y+1=0

$$2 { x }^{ 2 } -xy-2y+1=0$$

Solve for y

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

$$x\neq -2$$

Steps for Solving Linear Equation

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

$$-xy-2y+1=-2x^{2}$$

Subtract 1 from both sides.

$$-xy-2y=-2x^{2}-1$$

Combine all terms containing y.

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

Divide both sides by -x-2.

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

Dividing by -x-2 undoes the multiplication by -x-2.

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

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

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

Solve for x (complex solution)

$$x=\frac{\sqrt{y^{2}+16y-8}+y}{4}$$

$$x=\frac{-\sqrt{y^{2}+16y-8}+y}{4}$$

Solve for x

$$x=\frac{\sqrt{y^{2}+16y-8}+y}{4}$$

$$x=\frac{-\sqrt{y^{2}+16y-8}+y}{4}\text{, }y\geq 6\sqrt{2}-8\text{ or }y\leq -6\sqrt{2}-8$$