# 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?

• Bring everything to a common denominator for a start. Also the $S$ you wrote equals the entire $\mathbb{R}$ which is quite clearly not the correct result.
– dxiv
Jul 21, 2017 at 22:46
• $\{x \in \mathbb{R} \mid x \leq \sqrt{5} \vee x \geq \sqrt{5}\}=\mathbb{R}$, so your result looks like strange. you should publish the derivation of your result. Jul 21, 2017 at 23:19

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}

• thanks. My problem was transforming all in a quadratic equation. I have to pay attention to not lose information. Jul 22, 2017 at 15:00
• @CarlosOliveira Thanks. I multiplied both member by $\left[\,\left(\,x + 1\,\right)\left(\,x - 3\,\right)\,\right]_{\ x\ \not\in\ \left\{-1,3\right\}}^{\,2}$ because it preserves the inequality sign. I guess it's the "straightforward way" to do it. $\substack{@\ \bullet\quad\bullet\ @\\ \smile\\ \smile}$. Jul 22, 2017 at 18:59
• I think the "straightforward way" to handle problems where the sign matters like the multiplication of an inequality by a term that contains the variable x is to split the problem into two (or more) cases: the case where the term is negativ and the inequality relation is flipped after multiplication and the case where the term is negative and the inequality relation remains the same after multiplication. But your methods using squares is very clever and may shift the split into several cases to a later point of the calculation. Jul 22, 2017 at 22:03
• @miracle173 Thanks for your remark. Jul 22, 2017 at 22:05

$$\frac{x}{x+1}-\frac1{x-3}-2\le0\iff\frac{x^2-4x-1}{(x+1)(x-3)}\le2$$ 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 $$x\in\left(-\infty,-\sqrt5\right]\cup\left(-1,\sqrt5\right]\cup(3,\infty)$$

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 $x > 3.$

• No, I think he shouldn't. If he wants to sove an inequality he should split it in different cases if this is necessary, and solve each case. Jul 22, 2017 at 6:12

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]


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

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.

• @miracle173 That stands for "in the condition that" usually denoted |. I just added the other line to make it look nicer.
– Tony
Jul 21, 2017 at 23:32
• I thought you wanted to use the ||-operator of the C lanaguage. I think you refer to the symbol | used in expressions like $\{x|\cdots\}.$ I never saw the | symbol with this meaning in another type of expression. So I think you cannot use it here. Doubling the symbol also makes no sense. $\Leftrightarrow$ is transitive and all in all you end up with $\dfrac{x}{x+1} - \dfrac{1}{x-3} - 2 \leq 0 \Leftrightarrow x > \sqrt5$ which is wrong. So you should change your notation to a correct one. Jul 22, 2017 at 5:25