Solve x^2+y^2-2x+y=x^2-y^2

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

Solve for x

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

Steps for Solving Linear Equation

Subtract x^{2} from both sides.

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

Combine x^{2} and -x^{2} to get 0.

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

Subtract y^{2} from both sides.

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

Combine -y^{2} and -y^{2} to get -2y^{2}.

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

Subtract y from both sides.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide -y\left(1+2y\right) by -2.

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

Solve for y (complex solution)

$$y=\frac{\sqrt{16x+1}-1}{4}$$

$$y=\frac{-\sqrt{16x+1}-1}{4}$$

Solve for y

$$y=\frac{\sqrt{16x+1}-1}{4}$$

$$y=\frac{-\sqrt{16x+1}-1}{4}\text{, }x\geq -\frac{1}{16}$$