Solve 2xy-4x+2y-y^2=3

$$2 x y - 4 x + 2 y - y ^ { 2 } = 3$$

Solve for x

$$x=-\frac{-y^{2}+2y-3}{2\left(y-2\right)}$$

$$y\neq 2$$

Steps for Solving Linear Equation

Subtract 2y from both sides.

$$2xy-4x-y^{2}=3-2y$$

Add y^{2} to both sides.

$$2xy-4x=3-2y+y^{2}$$

Combine all terms containing x.

$$\left(2y-4\right)x=3-2y+y^{2}$$

The equation is in standard form.

$$\left(2y-4\right)x=y^{2}-2y+3$$

Divide both sides by 2y-4.

$$\frac{\left(2y-4\right)x}{2y-4}=\frac{y^{2}-2y+3}{2y-4}$$

Dividing by 2y-4 undoes the multiplication by 2y-4.

$$x=\frac{y^{2}-2y+3}{2y-4}$$

Divide 3-2y+y^{2} by 2y-4.

$$x=\frac{y^{2}-2y+3}{2\left(y-2\right)}$$

Solve for y (complex solution)

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

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

Solve for y

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

$$y=\sqrt{x^{2}-2x-2}+x+1\text{, }x\geq \sqrt{3}+1\text{ or }x\leq 1-\sqrt{3}$$