Solve zx+y^2=z^2-yx

$$zx+ { y }^{ 2 } = { z }^{ 2 } -yx$$

Solve for x

$$\left\{\begin{matrix}\\x=z-y\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&z=-y\end{matrix}\right.$$

Steps for Solving Linear Equation

Add yx to both sides.

$$zx+y^{2}+yx=z^{2}$$

Subtract y^{2} from both sides.

$$zx+yx=z^{2}-y^{2}$$

Combine all terms containing x.

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

The equation is in standard form.

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

Divide both sides by z+y.

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

Dividing by z+y undoes the multiplication by z+y.

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

Divide \left(z-y\right)\left(z+y\right) by z+y.

$$x=z-y$$

Solve for y

$$y=z-x$$

$$y=-z$$