<s>[INST] What would be the fastest way of solving the following inequality $\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$ What would be the fastest way of solving the following inequality:
$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$
 [/INST] $\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$
$\frac{(x+1)(x-3)}{(x-1)(x+2)(x-3)}-\frac{(x)(x+2)}{(x-1)(x+2)(x-3)}>0$
$\frac{(x+1)(x-3)-(x)(x+2)}{(x-1)(x+2)(x-3)}>0$
$\frac{4x+3}{(x-1)(x+2)(x-3)}<0$
the inequations are:


*

*$x>-2$

*$4x+3>0$ or $x>-\frac{4}{3}$

*$x>1$

*$x>3$


We need to have one or three inequations true to have the big inequality hold.
You will find that the answer is  $(-2,-\frac{3}{4})\cup(1,3)$
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