Solve x^2-z/y-2=x

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

Solve for y

$$\left\{\begin{matrix}y=-\frac{z}{2+x-x^{2}}\text{, }&z\neq 0\text{ and }x\neq -1\text{ and }x\neq 2\\y\neq 0\text{, }&\left(x=2\text{ or }x=-1\right)\text{ and }z=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by y.

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

Subtract xy from both sides.

$$yx^{2}-z+y\left(-2\right)-xy=0$$

Add z to both sides. Anything plus zero gives itself.

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

Combine all terms containing y.

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

The equation is in standard form.

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

Divide both sides by x^{2}-2-x.

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

Dividing by x^{2}-2-x undoes the multiplication by x^{2}-2-x.

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

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

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

Variable y cannot be equal to 0.

$$y=\frac{z}{\left(x-2\right)\left(x+1\right)}\text{, }y\neq 0$$

Solve for x (complex solution)

$$x=\frac{\sqrt{y\left(9y+4z\right)}+y}{2y}$$

$$x=\frac{-\sqrt{y\left(9y+4z\right)}+y}{2y}\text{, }y\neq 0$$

Solve for x

$$x=\frac{\sqrt{y\left(9y+4z\right)}+y}{2y}$$

$$x=\frac{-\sqrt{y\left(9y+4z\right)}+y}{2y}\text{, }\left(z\leq -\frac{9y}{4}\text{ and }y<0\right)\text{ or }\left(z\geq -\frac{9y}{4}\text{ and }y>0\right)\text{ or }\left(y\neq 0\text{ and }z=-\frac{9y}{4}\right)$$