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Solve the following system of equations:
$\frac{4}{x}\ +\ 3y\ =\ 8$
$\frac{6}{x}\ −\ 4y\ =\ −5$
Given:
The given system of equations is:
$\frac{4}{x}\ +\ 3y\ =\ 8$
$\frac{6}{x}\ −\ 4y\ =\ −5$
To do:
We have to solve the given system of equations.
Solution:
The given system of equations can be written as,
$\frac{4}{x}+3y=8$
Let $\frac{1}{x}=k$,
$\Rightarrow 4k+3y=8$---(i)
$\frac{6}{x}-4y=-5$
$\Rightarrow 6k-4y=-5$
$\Rightarrow 6k=4y-5$
$\Rightarrow k=\frac{4y-5}{6}$----(ii)
Substitute $k=\frac{4y-5}{6}$ in equation (i), we get,
$4(\frac{4y-5}{6})+3y=8$
$\frac{2(4y-5)}{3}+3y=8$ 
Multiplying by $3$ on both sides, we get,
$3(\frac{8y-10}{3})+3(3y)=3(8)$
$8y-10+9y=24$
$17y=24+10$
$17y=34$
$y=\frac{34}{17}$
$y=2$
Substituting the value of $y=2$ in equation (ii), we get,
$k=\frac{4(2)-5}{6}$
$k=\frac{8-5}{6}$
$k=\frac{3}{6}$
$k=\frac{1}{2}$
This implies,
$x=\frac{1}{k}=\frac{1}{\frac{1}{2}}$
$x=2$
Therefore, the solution of the given system of equations is $x=2$ and $y=2$.