# Solve $\frac{4}{x+1}+\frac{2}{x-2} \leq 3$

Solve

$$\frac{4}{x+1}+\frac{2}{x-2} \leq 3$$

I must be making a very stupid mistake somewhere ... been stuck on this for 1hr+ or even more ...

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$$\begin{eqnarray} \frac{2(x-1)}{(x+1)(x-2)} &\leq& 1 \implies \\ \frac{2(x-1)}{(x+1)(x-2)} - 1 &\leq& 0 \\ \frac{(2x-2)-(x+1)(x-2)}{(x+1)(x-2)} &\leq& 0 \\ \frac{(2x-2)-(x^2-x-2)}{(x+1)(x-2)} &\leq& 0 \\ \frac{(2x-2)-(x^2-x-2)}{(x+1)(x-2)} &\leq& 0 \\ \frac{-(x^2-3x)}{(x+1)(x-2)} &\leq& 0 \\ \frac{-(x^2-3x)}{(x+1)(x-2)} &\leq& 0 \implies \\ \frac{x(x-3)}{(x+1)(x-2)} &\geq& 0 \\ \end{eqnarray}$$ If we look at this on a number line, first note that for $x$ large enough, the LHS is positive. Next, note that the LHS changes sign each time $x$ crosses a root of one of the terms in the top or bottom, i.e. when $x \in \{-1,0,2,3\}$. Thus the LHS is nonnegative for $x \in (-\infty,-1) \cup [0,2) \cup [3,\infty)$.

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You made no mistake. Note however that you need to exclude the cases $x=-1, x=2$ by hand because your initial inequality is not defined for such $x$.

Then you have to see that

$$x^4 - 4 x^3 + x^2 + 6=(x-3) (x-2) x (x+1)$$

Now look at the plot and you should know what to do:

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@jiewmeng: As Listing pointed the graph tells you what you want, however, the below image would assist you. I made it as you posted your question:

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Nice table! What did you use to make it? – amWhy Mar 3 '13 at 18:05
@amWhy: Office word. – Babak S. Mar 3 '13 at 18:07
Did you save as an image and import? Or did you scan, and import the scanned image? – amWhy Mar 3 '13 at 18:08
@amWhy: Image and then import. I convert it to pdf and then import. – Babak S. Mar 3 '13 at 18:16

One solution is given below. Another one, very close in spirit to your attempt, is given in a comment at the end.

It turns out that the numbers fit together perfectly! Rewrite our inequality as $$\frac{4}{x+1}-4 +\frac{2}{x-2}-1 \le 0.$$ This simplifies to $$-\frac{4x}{x+1}+\frac{x}{x-2} \le 0.$$ The problem now breaks up into two natural cases.

Case (i): $x\ge 0$. Here our inequality is equivalent to $$\frac{1}{x-2}\le \frac{4}{x+1}.$$ For $x >2$ this can be rewritten as $4(x-2)\ge x+1$, giving $x\ge 3$. For $0\le x<2$, we get in a similar way $x \le 3$, which holds automatically. We have obtained the intervals $[3,\infty)$ and $[0,2)$.

Case (ii): $x<0$. Here our inequality is equivalent to $$\frac{1}{x-2}\ge \frac{4}{x+1}.$$ This is automatically false if $-1<x<0$, since the left side is negative and the right side is not. When $x<-1$, the inequality is equivalent to $x+1 \ge 4(x-2)$, which holds. This gives us the additional interval $(-\infty, -1)$.

Comment: In order to solve the inequality, you used the nice strategy of multiplying by $(x+1)^2(x-2)^2$. When we do that, we have to note that $x\ne -1, 2$. Minor point. The major point is that you should "multiply out" only when you have to. Multiplying out tends to create a mess. If you keep the common factors of the two sides separated out, your calculation pushes through smoothly. Starting from your $$2(x-1)(x+1)(x-2)\le (x+1)^2(x-2)^2,$$ we obtain $$(x+1)(x-2)\left[(x+1)(x-2)-2(x-1)\right]\ge 0,$$ which simplifies to $$(x+1)(x-2)(x^2-3x) \ge 0.$$ Since $x^2-3x=x(x-3)$, the analysis becomes routine.

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