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 $y=\frac{1}{6}$.