Solve the following system of equations:
$\frac{6}{x+y} =\frac{7}{x-y}+3$
$\frac{1}{2(x+y)}=\frac{1}{3(x-y)}$

Given:

The given system of equations is:

$\frac{6}{x+y} =\frac{7}{x-y}+3$

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

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,

$\frac{6}{x+y} =\frac{7}{x-y}+3$

$6u=7v+3$

$6u-7v-3=0$---(i)

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

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

$3u=2v$

$u=\frac{2v}{3}$---(ii)

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

$6(\frac{2v}{3})-7v-3=0$

$2(2v)-7v=3$

$4v-7v=3$

$-3v=3$

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

$v=-1$

Using $v=-1$ in equation (i), we get,

$6u-7(-1)-3=0$

$6u+7-3=0$

$6u+4=0$

$6u=-4$

$u=\frac{-4}{6}$

$u=\frac{-2}{3}$

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

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

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

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

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

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

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

$2x=\frac{-3-1\times2}{2}$

$2x=\frac{-5}{2}$

$x=\frac{\frac{-5}{2}}{2}$

$x=\frac{-5}{4}$

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

$\frac{-5}{4}-y=-1$

$y=\frac{-5}{4}+1$

$y=\frac{-5+4\times1}{4}$

$y=\frac{-1}{4}$

Therefore, the solution of the given system of equations is $x=\frac{-5}{4}$ and $y=\frac{-1}{4}$.