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 ... 

 A: 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:

A: $$
\begin{eqnarray}
\frac{2(x-1)}{(x+1)(x-2)}             &\leq& 1 \implies \\
\frac{2(x-1)}{(x+1)(x-2)} - 1         &\leq& 0 \\
\frac{(2x-2)-(x+1)(x-2)}{(x+1)(x-2)}  &\leq& 0 \\ 
\frac{(2x-2)-(x^2-x-2)}{(x+1)(x-2)}   &\leq& 0 \\
\frac{(2x-2)-(x^2-x-2)}{(x+1)(x-2)}   &\leq& 0 \\
\frac{-(x^2-3x)}{(x+1)(x-2)}          &\leq& 0 \\
\frac{-(x^2-3x)}{(x+1)(x-2)}          &\leq& 0 \implies \\
\frac{x(x-3)}{(x+1)(x-2)}             &\geq& 0 \\
\end{eqnarray}
$$
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
$x \in (-\infty,-1) \cup [0,2) \cup [3,\infty)$.
A: @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:

A: 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.    
