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 ... 


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