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:
$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$

Reducing to the same denominator, we get
$$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}\iff -\frac{4x+3}{(x+2)(x-1)(x-3)}>0 $$
Now the sign of $-\dfrac{4x+3}{(x+2)(x-1)(x-3)}$, when defined, is the sign of the product $p(x)=-(4x+3)(x+2)(x-1)(x-3)$, which by Bolzano's theorem can change sign only at $-2, -\dfrac34,1$ and $3$. As $p(0)=-\frac12<0$, we only have to alternate signs to obtain the following table:
$$\begin{array}{*{9}{c}}
&-2&&-\frac34&&1&&2& \\
\hline
-&\Vert&  + &0&\color{red}{\mathbf-}&\Vert&+&\Vert&-\end{array}, $$
whence the solutions:
$$(-2,-\frac34)\cup(1,2).$$
