Solve 2x^2+2xy+y=4

$$2 { x }^{ 2 } +2xy+y=4$$

Solve for y

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

$$x\neq -\frac{1}{2}$$

Steps for Solving Linear Equation

Subtract 2x^{2} from both sides.

$$2xy+y=4-2x^{2}$$

Combine all terms containing y.

$$\left(2x+1\right)y=4-2x^{2}$$

Divide both sides by 2x+1.

$$\frac{\left(2x+1\right)y}{2x+1}=\frac{4-2x^{2}}{2x+1}$$

Dividing by 2x+1 undoes the multiplication by 2x+1.

$$y=\frac{4-2x^{2}}{2x+1}$$

Divide 4-2x^{2} by 2x+1.

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

Solve for x

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

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