Solve the following system of equations:

$\frac{1}{3x+y} +\frac{1}{3x-y}=\frac{3}{4}$
$\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=-\frac{1}{8}$

Given:

The given system of equations is:

$\frac{1}{3x+y} +\frac{1}{3x-y}=\frac{3}{4}$

$\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=-\frac{1}{8}$

To do:

We have to solve the given system of equations.

Solution:

Let $\frac{1}{3x+y}=u$ and $\frac{1}{3x-y}=v$

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

$\frac{1}{3x+y} +\frac{1}{3x-y}=\frac{3}{4}$

$u+v=\frac{3}{4}$

$4(u+v)=3$

$4u+4v=3$

$4u+4v-3=0$---(i)

$\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=-\frac{1}{8}$

$\frac{u}{2}-\frac{v}{2}=-\frac{1}{8}$

$8(\frac{u}{2}-\frac{v}{2})=-8(\frac{1}{8})$    (Multiplying both sides by 8)

$4u-4v=-1$

$4u-4v+1=0$---(ii)

Adding equations (i) and (ii), we get,

$4u+4v-3+4u-4v+1=0$

$8u-2=0$

$8u=2$

$u=\frac{2}{8}$

$u=\frac{1}{4}$

Using $u=\frac{1}{4}$ in equation (i), we get,

$4(\frac{1}{4})+4v-3=0$

$1+4v-3=0$

$4v-2=0$

$4v=2$

$v=\frac{2}{4}$

$v=\frac{1}{2}$

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

$\frac{1}{3x+y}=\frac{1}{4}$

$\Rightarrow 3x+y=4$.....(iii)

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

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

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

$3x+y+3x-y=4+2$

$6x=6$

$x=\frac{6}{6}$

$x=1$

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

$3(1)-y=2$

$y=3-2$

$y=1$

Therefore, the solution of the given system of equations is $x=1$ and $y=1$.