Solve {l}{(x=a-1/a+1)}{x(3-2b)-1=x(2-xb-b^2)}

$$\left. \begin{array} { l } { ( x = \frac { a - 1 } { a + 1 } ) } \\ { x ( 3 - 2 b ) - 1 = x ( 2 - x b - b ^ { 2 } ) } \end{array} \right.$$

Solve for x (complex solution)

$$x=-\frac{1-a}{a+1}$$

$$\left(a=0\text{ and }b=1\right)\text{ or }\left(a=\frac{\sqrt{b^{4}-4b^{3}+6b^{2}+1}+b+1}{b^{2}-b}\text{ and }b\neq 1\text{ and }b\neq 0\right)\text{ or }\left(a=\frac{-\sqrt{b^{4}-4b^{3}+6b^{2}+1}+b+1}{b^{2}-b}\text{ and }b\neq 1\text{ and }b\neq 0\right)$$

Solve for x

$$x=-\frac{1-a}{a+1}$$

$$\left(a=0\text{ and }b=1\right)\text{ or }\left(a=\frac{\sqrt{b^{4}-4b^{3}+6b^{2}+1}+b+1}{b^{2}-b}\text{ and }b\neq 1\text{ and }b\neq 0\text{ and }b^{4}-4b^{3}+6b^{2}\geq -1\text{ and }b^{4}-4b^{3}+6b^{2}+1\geq 0\right)\text{ or }\left(a=\frac{-\sqrt{b^{4}-4b^{3}+6b^{2}+1}+b+1}{b^{2}-b}\text{ and }b\neq 1\text{ and }b\neq 0\text{ and }b^{4}-4b^{3}+6b^{2}\geq -1\text{ and }b^{4}-4b^{3}+6b^{2}+1\geq 0\right)$$