Resolve $ \frac{120}{x+y} + \frac{60}{x-y} = 6;\,\frac{80}{x+y} + \frac{100}{x-y} = 7$ I want to resolve this system of equations:
$$\begin{cases} \frac{120}{x+y} + \frac{60}{x-y} = 6 \\\frac{80}{x+y} + \frac{100}{x-y} = 7\end{cases}$$
I came to equations like
 $$x - \frac{10x}{x-y} + y - \frac{10y}{x-y} = 20$$
and
$$-2xy - y^2 - 10y = 20 - x^2 -10x$$
I need to leave $x$ or $y$ alone and didn't succeed. Any help?

Let $u=\frac{1}{x+y}$ and $v=\frac{1}{x-y}$. Solve first for $u,v$, then you'll get equations $x+y=\frac{1}{u}$ and $x-y=\frac{1}{v}$ to solve for $x,y$.
(It's slightly easier to use $u=\frac{20}{x+y}$ and $v=\frac{20}{x-y}$ and then solve $x+y=\frac{20}{u}$ and $x-y=\frac{20}{v}$.)
