Solve (z-1)*x^2-2zx+z+1=0

$$(z-1) \times { x }^{ 2 } -2zx+z+1=0$$

Solve for z

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

Steps for Solving Linear Equation

Use the distributive property to multiply z-1 by x^{2}.

$$zx^{2}-x^{2}-2zx+z+1=0$$

Add x^{2} to both sides. Anything plus zero gives itself.

$$zx^{2}-2zx+z+1=x^{2}$$

Subtract 1 from both sides.

$$zx^{2}-2zx+z=x^{2}-1$$

Combine all terms containing z.

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

Divide both sides by x^{2}-2x+1.

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

Dividing by x^{2}-2x+1 undoes the multiplication by x^{2}-2x+1.

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

Divide x^{2}-1 by x^{2}-2x+1.

$$z=\frac{x+1}{x-1}$$

Solve for x

$$\left\{\begin{matrix}\\x=1\text{, }&\text{unconditionally}\\x=\frac{z+1}{z-1}\text{, }&z\neq 1\end{matrix}\right.$$