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?
 A: $$
\frac{x}{x+1}-\frac1{x-3}-2\le0\iff\frac{x^2-4x-1}{(x+1)(x-3)}\le2
$$
which, for $x\in(-1,3)$ is the same as
$$
x^2-4x-1\ge2x^2-4x-6\iff x^2\le5\iff x\in\left(-1,\sqrt5\right]
$$
and which, for $x\not\in[-1,3]$ is the same as
$$
x^2-4x-1\le2x^2-4x-6\iff x^2\ge5\iff x\not\in\left(-\sqrt5,3\right]
$$
So the full solution is
$$
x\in\left(-\infty,-\sqrt5\right]\cup\left(-1,\sqrt5\right]\cup(3,\infty)
$$
A: You should learn to draw graphs of Moebius functions, it will help. The first has a horizontal asymptote of $y=1,$ the second has $y=0.$
As far as sketching by hand, which I strongly recommend, a single Moebius function $y = \frac{ax+b}{cx+d}$ (with $c \neq 0$ and $ad-bc \neq 0$ ) gives a hyperbola, with one vertical asymptote at $x = -d/c$ and horizontal at $y = a/c.$ If you add or subtract two of these, you still get horizontal asymptotes, aftre that some care is needed.



Let's see, we are asked when $y \leq 2.$ That would be when $x \leq - \sqrt 5,$ then $-1 < x \leq \sqrt 5,$ then $x > 3.$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{x \not\in \braces{-1,3}}$:

\begin{align}
&{x \over x + 1} - {1 \over x - 3} - 2 \leq  0
\\[5mm] &\ \implies
x\pars{x + 1}\pars{x - 3}^{2} - \pars{x + 1}^{2}\pars{x - 3} -
2\pars{x + 1}^{2}\pars{x - 3}^{2} \leq 0
\\[5mm] &\ \implies 0 \geq
\pars{x + 1}\pars{x - 3}\bracks{x\pars{x - 3} - \pars{x + 1} -
2\pars{x + 1}\pars{x - 3}}
\\[3mm] &\
\phantom{\implies \geq}= \pars{x + 1}\pars{x - 3}\pars{5 - x^{2}}
\\[5mm] &\ \implies 
\bbx{\pars{x + \root{5}}\pars{x + 1}\pars{x - \root{5}}\pars{x - 3} \geq 0}
\label{1}\tag{1}
\end{align}

\begin{align}
&\eqref{1} \,\,\,\stackrel{x\ \not\in\ \braces{-1,3}}{\large\implies}\,\,\,
\bbox[15px,#ffe,border:1px dotted navy]{{x \in \left(-\infty,\root{5}\right]\bigcup\left(-1,\root{5}\right]\bigcup
\left(3,+\infty\vphantom{\root{5}}\right)}}
\\[3mm] &\
\mbox{as depicted in the following picture.}
\\ &\ \mbox{}
\end{align}


A: Group the inequality, which will yield three solution regions:
$$\frac{5-x^2}{(x-3) (x+1)} \leq 0$$
To check... Mathematica gives the proper solution:
Reduce[x/(x + 1) - 1/(x - 3) - 2 <= 0, x]

$x\leq -\sqrt{5}\lor -1<x\leq \sqrt{5}\lor x>3$
A: 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.
