Solve f^-1y+x/x-2=x

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

Solve for f

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

Steps for Solving Linear Equation

Multiply both sides of the equation by x-2.

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

Use the distributive property to multiply f^{-1} by y+x.

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

Use the distributive property to multiply x by x-2.

$$f^{-1}y+f^{-1}x=x^{2}-2x$$

Reorder the terms.

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

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

$$1x+1y=fx^{2}-2xf$$

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

$$fx^{2}-2xf=1x+1y$$

Reorder the terms.

$$fx^{2}-2fx=x+y$$

Combine all terms containing f.

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

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

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

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

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

Divide y+x by x^{2}-2x.

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

Variable f cannot be equal to 0.

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

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{4fy+4f^{2}+4f+1}+2f+1}{2f}\text{, }&\left(f\neq 0\text{ and }y\neq -2\right)\text{ or }\left(f\neq 0\text{ and }f\neq \frac{1}{2}\text{ and }arg(2f-1)\geq \pi \right)\\x=-\frac{\sqrt{4fy+4f^{2}+4f+1}-2f-1}{2f}\text{, }&\left(arg(1-2f)\geq \pi \text{ and }f\neq \frac{1}{2}\right)\text{ or }\left(f\neq 0\text{ and }y\neq -2\right)\end{matrix}\right.$$