Question

$$\frac{x}{x+2}-\frac{1}{b^{2}y}=\frac{8}{6-2}$$

Solve for x (complex solution)

$x=-\frac{2\left(2yb^{2}+1\right)}{yb^{2}+1}$
$b\neq 0\text{ and }y\neq 0\text{ and }b\neq -iy^{-\frac{1}{2}}\text{ and }b\neq iy^{-\frac{1}{2}}$

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Solve for x

$x=-\frac{2\left(2yb^{2}+1\right)}{yb^{2}+1}$
$b\neq 0\text{ and }y\neq 0\text{ and }\left(y>0\text{ or }|b|\neq \sqrt{-\frac{1}{y}}\right)$

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Solve for b (complex solution)

$b=-iy^{-\frac{1}{2}}\left(x+4\right)^{-\frac{1}{2}}\sqrt{x+2}$
$b=iy^{-\frac{1}{2}}\left(x+4\right)^{-\frac{1}{2}}\sqrt{x+2}\text{, }x\neq -4\text{ and }x\neq -2\text{ and }y\neq 0$

Solve for b

$b=\sqrt{-\frac{x+2}{y\left(x+4\right)}}$
$b=-\sqrt{-\frac{x+2}{y\left(x+4\right)}}\text{, }\left(x>-2\text{ and }y<0\right)\text{ or }\left(y>0\text{ and }x>-4\text{ and }x<-2\right)\text{ or }\left(x<-4\text{ and }y<0\right)$