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Solve the following system of equations:
$\frac{3}{x}\ –\ \frac{1}{y}\ =\ −9$
$\frac{2}{x}\ +\ \frac{3}{y}\ =\ 5$
Given:
The given system of equations is:
$\frac{3}{x}\ –\ \frac{1}{y}\ =\ −9$
$\frac{2}{x}\ +\ \frac{3}{y}\ =\ 5$
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,
$\frac{3}{x}\ -\ \frac{1}{y}\ =\ -9$
$3u-v=-9$-----(i)
$\frac{2}{x}\ +\ \frac{3}{y}\ =\ 5$
$2u+3v=5$
$2u=5-3v$
$u=\frac{5-3v}{2}$
Substitute $u=\frac{5-3v}{2}$ in equation (i), we get,
$3(\frac{5-3v}{2})-v=-9$
Multiplying both sides by $2$, we get,
$2(\frac{3(5-3v)}{2})-2(v)=2(-9)$ 
$15-9v-2v=-18$ 
$-11v=-18-15$
$-11v=-33$ 
$v=\frac{-33}{-11}$
$v=3$
This implies,
$u=\frac{5-3(3)}{2}$
$u=\frac{5-9}{2}$
$u=\frac{-4}{2}$
$u=-2$
$x=\frac{1}{u}=\frac{1}{-2}=-\frac{1}{2}$
$y=\frac{1}{v}=\frac{1}{3}$ 
Therefore, the solution of the given system of equations is $x=-\frac{1}{2}$ and $y=\frac{1}{3}$.