<s>[INST] Solving the inequality $\frac{x}{x+1} - \frac{1}{x-3} - 2 \leq 0$ I tried to solve 
$$ \dfrac{x}{x+1} - \dfrac{1}{x-3} - 2 \leq  0$$
and I got 
$$S = \{x \in \mathbb{R} \mid x \leq -\sqrt{5} \vee x \geq \sqrt{5}\}$$
but I have plotted and here is the function graph. It doesn't seem to be a quadratic inequality. 
Does someone have a hint on how to solve it?
 [/INST] Consider $x-3> 0$
$$\dfrac{x}{x+1} - \dfrac{1}{x-3} - 2 \leq  0$$
$$\Leftrightarrow x-\dfrac{x+1}{x-3}-2(x+1)\leq  0 \space\space ||x+1 > 0$$
$$\Leftrightarrow \dfrac{x(x-3)}{x-3}-\dfrac{x+1}{x-3}-\dfrac{2(x+1)(x-3)}{x-3}\leq  0$$
$$\Leftrightarrow x^2-3x-x-1-2(x^2-2x-3) \leq 0 \space \space ||x -3>0$$
$$\Leftrightarrow -x^2+5 \leq 0$$
$$\Leftrightarrow x > \sqrt5$$
So, when $x > 3$, the above holds. You can do the other two cases.
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