Solve each of the following systems of equations by the method of cross-multiplication :

GIVEN: 

$\frac{5}{x+y}-\frac{2}{x-y}=-1$

$\frac{15}{x+y}+\frac{7}{x-y}=10$

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

Rewriting the equation again

$\frac{5}{x+y}-\frac{2}{x-y}+1=0$

$\frac{15}{x+y}+\frac{7}{x-y}-10=0$

Taking $u=\frac{1}{x+y}$ and $v=\frac{1}{x-y}$

$5 u-2 v+1=0$...(1)

$15 u+7 v-10=0$...$.(2)$

By cross multiplication method we get

$\frac{u}{(20)-(7)}=\frac{-v}{(-50)-(15)}=\frac{1}{(35)-(-30)}$

$\Rightarrow \frac{u}{13}=\frac{-v}{-65}=\frac{1}{65}$

$\Rightarrow \frac{u}{13}=\frac{v}{65}=\frac{1}{65}$

$\Rightarrow \frac{u}{13}=\frac{1}{65}$

$\Rightarrow u=\frac{1}{5}$

And

$\frac{v}{65}=\frac{1}{65}$

$v=1$

We know that 

$u=\frac{1}{x+y}$ and $v=\frac{1}{x-y}$

$\Rightarrow \frac{1}{5}=\frac{1}{x+y}$

$\Rightarrow x+y=5 \quad \ldots \ldots . .(3)$

and

$1=\frac{1}{x-y}$

$\Rightarrow x-y=1$ $...(4)$

Adding equation (3) and (4)

$2 x=6$

$x=3$

Substituting value of x in equation (3) we get

$\begin{aligned} y &=5-3 \\ &=2 \end{aligned}$

$=2$

Hence we get the value of $x=3$ and $y=2$