Solve the following systems of equations:

The given equations are:

$\frac{4}{x}+3 y=8$$\ldots(i)$

$\frac{6}{x}-4 y=-5$$\cdots(i i)$

Multiply equation (i) by 4 and equation (ii) by 3 and add both equations we get

$\frac{16}{x}+12 y=32$

Put the value of $x$ in equation $(i)$ we get

$\frac{4}{2}+3 y=8$

$\Rightarrow 3 y=6$

$\Rightarrow y=2$

Hence the value of $x=2$ and $y=2$.