Resolve $ \frac{120}{x+y} + \frac{60}{x-y} = 6;\,\frac{80}{x+y} + \frac{100}{x-y} = 7$

Question

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?

Answer 1

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

Answer 2

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 $x-y=\frac{20}{v}$.)