Solve -ax^2+2ax-2x+2/xleft(2-xright)=0

$$\frac{ -a { x }^{ 2 } +2ax-2x+2 }{ x \left( 2-x \right) } = 0$$

Solve for a

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

$$x\neq 0\text{ and }x\neq 2$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x\left(-x+2\right).

$$\left(-a\right)x^{2}+2ax-2x+2=0$$

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

$$\left(-a\right)x^{2}+2ax+2=2x$$

Subtract 2 from both sides.

$$\left(-a\right)x^{2}+2ax=2x-2$$

Reorder the terms.

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

Combine all terms containing a.

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

The equation is in standard form.

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

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

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

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

$$a=\frac{2x-2}{2x-x^{2}}$$

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

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

Solve for x

$$\left\{\begin{matrix}x=-\frac{\sqrt{a^{2}+1}-a+1}{a}\text{; }x=\frac{\sqrt{a^{2}+1}+a-1}{a}\text{, }&a\neq 0\\x=1\text{, }&a=0\end{matrix}\right.$$