Solve x^2-2yx-y+1=0

$${ x }^{ 2 } -2yx-y+1=0$$

Solve for y

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

$$x\neq -\frac{1}{2}$$

Steps for Solving Linear Equation

Subtract x^{2} from both sides. Anything subtracted from zero gives its negation.

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

Subtract 1 from both sides.

$$-2yx-y=-x^{2}-1$$

Combine all terms containing y.

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

Divide both sides by -2x-1.

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

Dividing by -2x-1 undoes the multiplication by -2x-1.

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

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

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

Solve for x (complex solution)

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

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

Solve for x

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

$$x=-\sqrt{y^{2}+y-1}+y\text{, }y\geq \frac{\sqrt{5}-1}{2}\text{ or }y\leq \frac{-\sqrt{5}-1}{2}$$