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

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

Solve for x

$$\left\{\begin{matrix}x=\frac{z^{2}}{2y}-\frac{y}{2}\text{, }&y\neq 0\\x\in \mathrm{R}\text{, }&y=0\text{ and }z=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Add yx to both sides.

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

Combine yx and yx to get 2yx.

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

Subtract y^{2} from both sides.

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

Divide both sides by 2y.

$$\frac{2yx}{2y}=\frac{\left(z-y\right)\left(y+z\right)}{2y}$$

Dividing by 2y undoes the multiplication by 2y.

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

Divide \left(z-y\right)\left(z+y\right) by 2y.

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

Solve for y

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

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