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?

You could also solve by eliminating one of the variables:
Multiply the first equation by 100 and the second by 60:
$$ \frac{12000}{x+y} + \frac{6000}{x-y}= 600$$
$$  \frac{4800}{x+y} + \frac{6000}{x-y}= 420$$
Subtract them from each other and get rid of $\frac{6000}{x-y}$:
$$ x+y = \frac{720}{180} = 40$$
Substitute $x+y$ into either of your original equations:
$$ \frac{120}{x+y} + \frac{60}{x-y}= 6 \implies \frac{120}{40} + \frac{60}{x-y}= 6 $$
$$ x-y = 20$$
Now solve these two simultaneous equations:
$$ x+y = 40, \ x-y = 20$$
Trust you can finish this off
