Suggestions on solving a system of equations How do you suggest solving the following sytem? 
$$\begin{cases} \dfrac{12}{x}+\dfrac{12}{y}=1 \\ \dfrac{6}{x-2}+\dfrac{8}{y-6}=\dfrac{2}{3} \end{cases}$$
After simplifying the equations, I got $xy-12y=12x$ and $xy-11y=18x-90$. 
Should I continue from here or there's a better solution?

Alternatively:
$$\begin{cases} \dfrac{12}{x}+\dfrac{12}{y}=1 \\ \dfrac{6}{x-2}+\dfrac{8}{y-6}=\dfrac{2}{3} \end{cases} \Rightarrow 
\begin{cases} \dfrac{12}{x}=1-\dfrac{12}{y} \\ \dfrac{6}{x-2}=\dfrac23-\dfrac{8}{y-6} \end{cases} \Rightarrow 
\begin{cases} x=\dfrac{12y}{y-12} \\ x=\dfrac{18(y-6)}{2y-36}+2=\dfrac{22y-180}{2y-36} \end{cases}$$
Now equate them:
$$12y(2y-36)=(y-12)(22y-180) \Rightarrow y^2+6y-1080=0 \Rightarrow \\
y_{1,2}=-36;30 \Rightarrow x_{1,2}=9;20.$$
