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Solve the following system of equations:
$\frac{2}{x}\ +\ \frac{3}{y}\ =\ 13$
$\frac{5}{x}\ –\ \frac{4}{y}\ =\ -2$
Given:
The given system of equations is:
$\frac{2}{x}\ +\ \frac{3}{y}\ =\ 13$
$\frac{5}{x}\ –\ \frac{4}{y}\ =\ -2$
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,
$\frac{2}{x}\ +\ \frac{3}{y}\ =\ 13$
$2u+3v=13$-----(i)
$\frac{5}{x}\ -\ \frac{4}{y}\ =\ -2$
$5u-4v=-2$
$5u=4v-2$
$u=\frac{4v-2}{5}$
Substitute $u=\frac{4v-2}{5}$ in equation (i), we get,
$2(\frac{4v-2}{5})+3v=13$
Multiplying both sides by $5$, we get,
$5(\frac{2(4v-2)}{5})+5(3v)=5(13)$ 
$8v-4+15v=65$ 
$23v=65+4$
$23v=69$ 
$v=\frac{69}{23}$
$v=3$
This implies,
$u=\frac{4(3)-2}{5}$
$u=\frac{12-2}{5}$
$u=\frac{10}{5}$
$u=2$
$x=\frac{1}{u}=\frac{1}{2}$
$y=\frac{1}{v}=\frac{1}{3}$ 
Therefore, the solution of the given system of equations is $x=\frac{1}{2}$ and $y=\frac{1}{3}$.