Solve z=xy+2x^2+y

$$z = x y + 2 x ^ { 2 } + y$$

Solve for y (complex solution)

$$\left\{\begin{matrix}y=-\frac{2x^{2}-z}{x+1}\text{, }&x\neq -1\\y\in \mathrm{C}\text{, }&z=2\text{ and }x=-1\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

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

Subtract 2x^{2} from both sides.

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

Combine all terms containing y.

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

Divide both sides by x+1.

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

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

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

Solve for y

$$\left\{\begin{matrix}y=-\frac{2x^{2}-z}{x+1}\text{, }&x\neq -1\\y\in \mathrm{R}\text{, }&z=2\text{ and }x=-1\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

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

Subtract 2x^{2} from both sides.

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

Combine all terms containing y.

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

Divide both sides by x+1.

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

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

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

Solve for x (complex solution)

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

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

Solve for x

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

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