Solve the following system of equations:
$\frac{10}{x+y} +\frac{2}{x-y}=4$
$\frac{15}{x+y}-\frac{9}{x-y}=-2$

Given:

The given system of equations is:

$\frac{10}{x+y} +\frac{2}{x-y}=4$

$\frac{15}{x+y}-\frac{9}{x-y}=-2$

To do:

We have to solve the given system of equations.

Solution:

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

This implies, the given system of equations can be written as,

$\frac{10}{x+y} +\frac{2}{x-y}=4$

$10u+2v=4$

$10u+2v-4=0$---(i)

$\frac{15}{x+y}-\frac{9}{x-y}=-2$

$15u-9v=-2$

$15u=9v-2$

$u=\frac{9v-2}{15}$---(ii)

Substituting $u=\frac{9v-2}{15}$ in equation (i), we get,

$10(\frac{9v-2}{15})+2v-4=0$

$\frac{2(9v-2)}{3}=4-2v$

$18v-4=3(4-2v)$

$18v-4=12-6v$

$18v+6v=12+4$

$24v=16$

$v=\frac{16}{24}$

$v=\frac{2}{3}$

Using $v=\frac{2}{3}$ in equation (i), we get,

$10u+2(\frac{2}{3})-4=0$

$10u+\frac{4}{3}-4=0$

$10u+\frac{4-4\times3}{3}=0$

$10u+\frac{-8}{3}=0$

$10u=\frac{8}{3}$

$u=\frac{8}{3\times10}$

$u=\frac{4}{15}$

Using the values of $u$ and $v$, we get,

$\frac{1}{x+y}=\frac{4}{15}$

$\Rightarrow x+y=\frac{15}{4}$....(iii)

$\frac{1}{x-y}=\frac{2}{3}$

$\Rightarrow x-y=\frac{3}{2}$.....(iv)

Adding  equations (iii) and (iv), we get,

$x+y+x-y=\frac{15}{4}+\frac{3}{2}$

$\Rightarrow 2x=\frac{15+2\times3}{4}$

$\Rightarrow 2x=\frac{21}{4}$

$\Rightarrow x=\frac{21}{8}$

Substituting the value of $x$ in (iii), we get,

$\frac{21}{8}+y=\frac{15}{4}$

$\Rightarrow y=\frac{15}{4}-\frac{21}{8}$

$\Rightarrow y=\frac{15\times2-21}{8}$

$\Rightarrow y=\frac{9}{8}$

Therefore, the solution of the given system of equations is $x=\frac{21}{8}$ and $y=\frac{9}{8}$.