Question

$$\frac{a}{b-x}=\frac{b}{a+2}$$

Solve for a (complex solution)

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

Solve for b (complex solution)

$\left\{\begin{matrix}b=\frac{\sqrt{x^{2}+4a^{2}+8a}+x}{2}\text{, }&\left(a\neq -2\text{ and }x\neq 0\text{ and }arg(x)\geq \pi \right)\text{ or }\left(a\neq -2\text{ and }a\neq 0\right)\\b=\frac{-\sqrt{x^{2}+4a^{2}+8a}+x}{2}\text{, }&\left(a\neq -2\text{ and }x\neq 0\text{ and }arg(x)<\pi \right)\text{ or }\left(a\neq -2\text{ and }a\neq 0\right)\end{matrix}\right.$

Solve for a

$\left\{\begin{matrix}a=-\sqrt{1+b^{2}-bx}-1\text{, }&b\neq x\text{ and }b\neq 0\text{ and }\left(b\geq \frac{\sqrt{x^{2}-4}+x}{2}\text{ or }b\leq \frac{-\sqrt{x^{2}-4}+x}{2}\text{ or }|x|\leq 2\right)\\a=\sqrt{1+b^{2}-bx}-1\text{, }&\left(b\geq \frac{\sqrt{x^{2}-4}+x}{2}\text{ or }b\leq \frac{-\sqrt{x^{2}-4}+x}{2}\text{ or }|x|\leq 2\right)\text{ and }b\neq x\end{matrix}\right.$