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Solve the following system of equations:
$\frac{1}{(7x)}\ +\ \frac{1}{(6y)}\ =\ 3$
$\frac{1}{(2x)}\ –\ \frac{1}{(3y)}\ =\ 5$
Given:
The given system of equations is:
$\frac{1}{(7x)}\ +\ \frac{1}{(6y)}\ =\ 3$
$\frac{1}{(2x)}\ –\ \frac{1}{(3y)}\ =\ 5$
To do:
We have to solve the given system of equations.
Solution:
The given system of equations can be written as,
$\frac{1}{7x}+\frac{1}{6y}=3$
Multiplying by $2$ on both sides, we get,
$\frac{2}{7x}+\frac{2}{6y}=2(3)$
$\frac{2}{7x}+\frac{1}{3y}=6$-----(i)
$\frac{1}{2x}-\frac{1}{3y}=5$----(ii)
Adding equations (i) and (ii), we get,
$\frac{2}{7x}+\frac{1}{3y}+\frac{1}{2x}-\frac{1}{3y}=6+5$
$\frac{2}{7x}+\frac{1}{2x}=11$
$\frac{2(2)+7(1)}{14x}=11$
$\frac{4+7}{14x}=11$
$11=11(14x)$
$14x=1$
$x=\frac{1}{14}$
Substitute $x=\frac{1}{14}$ in equation (i), we get,
$\frac{2}{7(\frac{1}{14})}+\frac{1}{3y}=6$
$2\times2+\frac{1}{3y}=6$ 
$4+\frac{1}{3y}=6$ 
$\frac{1}{3y}=6-4$
$\frac{1}{3y}=2$ 
$1=2(3y)$
$6y=1$
$y=\frac{1}{6}$ 
Therefore, the solution of the given system of equations is $x=\frac{1}{14}$ and $y=\frac{1}{6}$.