Finding all possible complex solutions to the system $xy = \sqrt 2(x + y)$, $yz = \sqrt 3(y + z)$, $zx = \sqrt 5(z + x)$ 
Find all possible complex solutions to:
$$\begin{cases}xy = \sqrt 2(x + y) \\
yz = \sqrt 3(y + z)\\
zx = \sqrt 5(z + x)\end{cases}$$

I have some ideas:
$$xy-\sqrt 2y=\sqrt 2x\implies y=\frac{\sqrt 2x}{x-\sqrt 2} $$
and
$$yz-\sqrt 3y=\sqrt 3z\implies y=\frac{\sqrt 3z}{z-\sqrt 3} $$
and
$$xz-\sqrt 5x=\sqrt 5z\implies x=\frac{\sqrt 5z}{z-\sqrt 5} $$
So I got
$$y=\frac{\sqrt 2x}{x-\sqrt 2}=\frac{\sqrt 3z}{z-\sqrt 3}=\frac{\sqrt 2×\frac{\sqrt 5z}{z-\sqrt 5}}{\frac{\sqrt 5z}{z-\sqrt 5}-\sqrt 2}$$
From here we can go to a solution, but that brings up really scary equations.  Is there any clever way to solve this?  That is a minor Olympiad problem.

The problem is not so bad.
From the first equation $y=\frac{\sqrt{2} x}{x-\sqrt{2}}$; from the third $z=\frac{\sqrt{5} x}{x-\sqrt{5}}$
Plug in the second
$$\frac{x \left(\left(\sqrt{6}-\sqrt{10}+\sqrt{15}\right) x-2
   \sqrt{30}\right)}{\left(x-\sqrt{2}\right) \left(x-\sqrt{5}\right)}=0$$ So, if $x\neq 0$, then
$$x=\frac{2 \sqrt{30}}{\sqrt{6}-\sqrt{10}+\sqrt{15}}\qquad y=\frac{2 \sqrt{30}}{-\sqrt{6}+\sqrt{10}+\sqrt{15}}\qquad z=\frac{2 \sqrt{30}}{\sqrt{6}+\sqrt{10}-\sqrt{15}}$$
