Solve the following system of equations:
$\frac{5}{x+1} -\frac{2}{y-1}=\frac{1}{2}$
$\frac{10}{x+1}+\frac{2}{y-1}=\frac{5}{2}$, where $x≠-1$ and $y≠1$

Given:

The given system of equations is:

$\frac{5}{x+1} -\frac{2}{y-1}=\frac{1}{2}$

$\frac{10}{x+1}+\frac{2}{y-1}=\frac{5}{2}$, where $x≠-1$ and $y≠1$

To do:

We have to solve the given system of equations.

Solution:

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

This implies, the given system of equations can be written as,

$\frac{5}{x+1} -\frac{2}{y-1}=\frac{1}{2}$

$5u-2v=\frac{1}{2}$

$2(5u-2v)=2(\frac{1}{2})$   (Multiplying by 2 on both sides)

$10u-4v=1$

$10u-4v-1=0$---(i)

$\frac{10}{x+1}+\frac{2}{y-1}=\frac{5}{2}$

$10u+2v=\frac{5}{2}$

$2(10u+2v)=2(\frac{5}{2})$   (Multiplying by 2 on both sides)

$20u+4v=5$

$20u=5-4v$

$u=\frac{5-4v}{20}$---(ii)

Substituting $u=\frac{5-4v}{20}$ in equation (i), we get,

$10(\frac{5-4v}{20})-4v-1=0$

$\frac{5-4v}{2}=4v+1$

$5-4v=2(4v+1)$

$5-4v=8v+2$

$8v+4v=5-2$

$12v=3$

$v=\frac{3}{12}$

$v=\frac{1}{4}$

Using $v=\frac{1}{4}$ in equation (i), we get,

$10u-4(\frac{1}{4})-1=0$

$10u-1-1=0$

$10u-2=0$

$10u=2$

$u=\frac{2}{10}$

$u=\frac{1}{5}$

Using the values of $u$ and $v$, we get,

$\frac{1}{x+1}=\frac{1}{5}$

$\Rightarrow x+1=5$

$\Rightarrow x=5-1$

$\Rightarrow x=4$

$\frac{1}{y-1}=\frac{1}{4}$

$\Rightarrow y-1=4$

$\Rightarrow y=4+1$

$\Rightarrow y=5$

Therefore, the solution of the given system of equations is $x=4$ and $y=5$.