Solve the following system of equations:
$\frac{1}{(2x)}\ +\ \frac{1}{(3y)}\ =\ 2$
$\frac{1}{(3x)}\ +\ \frac{1}{(2y)}\ =\ \frac{13}{6}$

Given:

The given system of equations is:

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

$\frac{1}{(3x)}\ +\ \frac{1}{(2y)}\ =\ \frac{13}{6}$

To do:

We have to solve the given system of equations.

Solution:

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

This implies,

The given system of equations can be written as,

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

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

$\frac{3u+2v}{6}=2$

$3u+2v=6(2)$

$3u+2v=12$-----(i)

$\frac{1}{(3x)}\ +\ \frac{1}{(2y)}\ =\ \frac{13}{6}$

$\frac{u}{3}+\frac{v}{2}=\frac{13}{6}$

$\frac{2u+3v}{6}=\frac{13}{6}$

$2u+3v=13$

$2u=13-3v$

$u=\frac{13-3v}{2}$

Substitute $u=\frac{13-3v}{2}$ in equation (i), we get,

$3(\frac{13-3v}{2})+2v=12$

Multiplying both sides by $2$, we get,

$2(\frac{3(13-3v)}{2})+2(2v)=2(12)$ 

$39-9v+4v=24$ 

$-5v=24-39$

$-5v=-15$ 

$v=\frac{-15}{-5}$

$v=3$

This implies,

$u=\frac{13-3(3)}{2}$

$u=\frac{13-9}{2}$

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

$u=2$

$x=\frac{1}{u}=\frac{1}{2}$

$y=\frac{1}{v}=\frac{1}{3}$ 

Therefore, the solution of the given system of equations is $x=\frac{1}{2}$ and $y=\frac{1}{3}$.