Solve the following pairs of equations:
$ \frac{2 x y}{x+y}=\frac{3}{2} $
$ \frac{x y}{2 x-y}=\frac{-3}{10}, x+y ≠ 0,2 x-y ≠ 0 $

Given:

\( \frac{2 x y}{x+y}=\frac{3}{2} \)

\( \frac{x y}{2 x-y}=\frac{-3}{10}, x+y ≠ 0,2 x-y ≠ 0 \)

To do:

We have to solve the given pairs of equations.

Solution:

$\frac{2 x y}{x+y}=\frac{3}{2}$

This implies,

$\frac{x+y}{2 x y}=\frac{2}{3}$

$\frac{x}{x y}+\frac{y}{x y}=\frac{2\times2}{3}$

$\frac{1}{y}+\frac{1}{x}=\frac{4}{3}$.......(i)

$\frac{x y}{2 x-y}=\frac{-3}{10}$

$\frac{2 x-y}{x y}=\frac{-10}{3}$

$\frac{2 x}{x y}-\frac{y}{x y}=\frac{-10}{3}$

$\frac{2}{y}-\frac{1}{x}=\frac{-10}{3}$...........(ii)

Let $\frac{1}{x}=u$ and $\frac{1}{y}=v$

This implies,

$v+u=\frac{4}{3}$.........(iii)

$2 v-u=\frac{-10}{3}$........(iv)

Adding (iii) and (iv), we get,

$3 v=\frac{4}{3}-\frac{10}{3}$

$3v=\frac{4-10}{3}$

$3v=\frac{-6}{3}$

$3 v=-2$

$v=\frac{-2}{3}$

Substituting the value of $v$ in (iii), we get,

$\frac{-2}{3}+u =\frac{4}{3}$

$u=\frac{4}{3}+\frac{2}{3}$

$u=\frac{6}{3}$

$u=2$

This implies,

$x=\frac{1}{u}=\frac{1}{2}$

$y=\frac{1}{v}=\frac{1}{\frac{-2}{3}}$

$y=\frac{-3}{2}$