Solve (2-y)x^2+3(y-3)x+(3-y)=0

$$(2-y) { x }^{ 2 } +3(y-3)x+(3-y)=0$$

Solve for y

$$y=-\frac{2x^{2}-9x+3}{-x^{2}+3x-1}$$

$$x\neq \frac{\sqrt{5}+3}{2}\text{ and }x\neq \frac{3-\sqrt{5}}{2}$$

Steps for Solving Linear Equation

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

$$2x^{2}-yx^{2}+3\left(y-3\right)x+3-y=0$$

Use the distributive property to multiply 3 by y-3.

$$2x^{2}-yx^{2}+\left(3y-9\right)x+3-y=0$$

Use the distributive property to multiply 3y-9 by x.

$$2x^{2}-yx^{2}+3yx-9x+3-y=0$$

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

$$-yx^{2}+3yx-9x+3-y=-2x^{2}$$

Add 9x to both sides.

$$-yx^{2}+3yx+3-y=-2x^{2}+9x$$

Subtract 3 from both sides.

$$-yx^{2}+3yx-y=-2x^{2}+9x-3$$

Combine all terms containing y.

$$\left(-x^{2}+3x-1\right)y=-2x^{2}+9x-3$$

Divide both sides by -x^{2}+3x-1.

$$\frac{\left(-x^{2}+3x-1\right)y}{-x^{2}+3x-1}=\frac{-2x^{2}+9x-3}{-x^{2}+3x-1}$$

Dividing by -x^{2}+3x-1 undoes the multiplication by -x^{2}+3x-1.

$$y=\frac{-2x^{2}+9x-3}{-x^{2}+3x-1}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{\left(y-3\right)\left(5y-19\right)}-3y+9}{2\left(2-y\right)}\text{; }x=\frac{-\sqrt{\left(y-3\right)\left(5y-19\right)}-3y+9}{2\left(2-y\right)}\text{, }&y\neq 2\\x=\frac{1}{3}\approx 0.333333333\text{, }&y=2\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{\left(y-3\right)\left(5y-19\right)}-3y+9}{2\left(2-y\right)}\text{; }x=\frac{-\sqrt{\left(y-3\right)\left(5y-19\right)}-3y+9}{2\left(2-y\right)}\text{, }&y\geq \frac{19}{5}\text{ or }\left(y\neq 2\text{ and }y\leq 3\right)\\x=\frac{1}{3}\approx 0.333333333\text{, }&y=2\end{matrix}\right.$$