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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 $y=\frac{89}{1512}$.