Solve the following system of equations:
$\frac{1}{(5x)}\ +\ \frac{1}{(6y)}\ =\ 12$
$\frac{1}{(3x)}\ –\ \frac{3}{(7y)}\ =\ 8$

Given:

The given system of equations is:

$\frac{1}{(5x)}\ +\ \frac{1}{(6y)}\ =\ 12$

$\frac{1}{(3x)}\ –\ \frac{3}{(7y)}\ =\ 8$

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,

$\frac{1}{5x}\ +\ \frac{1}{6y}\ =\ 12$

$\frac{u}{5}+\frac{v}{6}=12$

$\frac{6u+5v}{30}=12$

$6u+5v=12(30)$

$6u+5v=360$-----(i)

$\frac{1}{3x}\ -\ \frac{3}{7y}\ =\ 8$

$\frac{u}{3}-\frac{3v}{7}=8$

$\frac{7u-3(3v)}{21}=8$

$7u-9v=21(8)$

$7u-9v=168$

$7u=9v+168$

$u=\frac{9v+168}{7}$

Substitute $u=\frac{9v+168}{7}$ in equation (i), we get,

$6(\frac{9v+168}{7})+5v=360$

Multiplying both sides by $7$, we get,

$7(\frac{6(9v+168)}{7}+7(5v)=7(360)$

$54v+1008+35v=2520$ 

$89v=2520-1008$

$89v=1512$ 

$v=\frac{1512}{89}$

Substituting the value of $v$ in equation (i), we get,

$6u+5(\frac{1512}{89})=360$

$6u+\frac{7560}{89}=360$

$6u=360-\frac{7560}{89}$

$6u=\frac{360(89)-7560}{89}$

$6u=\frac{32040-7560}{89}$

$6u=\frac{24480}{89}$

$u=\frac{4080}{89}$

$x=\frac{1}{u}=\frac{1}{\frac{4080}{89}}=\frac{89}{4080}$

$y=\frac{1}{v}=\frac{1}{\frac{1512}{89}}=\frac{89}{1512}$ 

Therefore, the solution of the given system of equations is $x=\frac{89}{4080}$ and $y=\frac{89}{1512}$.