<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] $$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$$
$$\frac{(x+1)}{(x-1)(x+2)}-\frac{(x)}{(x-1)(x-3)}>0$$
$$\frac{1}{(x-1)}\left(\frac{(x+1)}{(x+2)}-\frac{(x)}{(x-3)}\right)>0$$
$$\frac{1}{(x-1)}\left(\frac{(x+1)(x-3)-x(x+2)}{(x+2)(x-3)}\right)>0$$
$$\frac{x^2-2x-3-x^2-2x}{(x-1)(x+2)(x-3)}>0$$
$$\frac{4x+3}{(x-1)(x+2)(x-3)}<0$$
The above inequality implies the following:
$$x\not = 1,x\not = -2,x\not = 3,x\not = -\frac{3}{4}$$
Now check the values of $x$ in the following intervals: $(-\infty,-2),(-2,-\frac{3}{4}),(-\frac{3}{4},1),(1,3),(3,+\infty)$ 
You will find that the inequality holds in the interval $(-2,-\frac{3}{4})\cup(1,3)$
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