Solution for $\frac{3}{x-9} \gt \frac{2}{x+2}$ What I did:
$\frac{3}{x-9} \gt \frac{2}{x+2}$
$3(x+2) \gt 2(x-9)$
$3x+6 \gt 2x-18$
$x \gt -24$
When typing this in in symbolab, it showed me that the solution is $-24 \lt x \lt -2$ or $x \gt 9$
What did i do wrong ?
How come i didnt get the correct solution ?
 A: HINT
The first step
$$\frac{3}{x-9} \gt \frac{2}{x+2}\iff 3(x+2) \gt 2(x-9)$$
is wrong since $x-9$ and $x+2$ are not both positive (or both negative) in general.
To solve properly use that
$$\frac{3}{x-9} \gt \frac{2}{x+2}\iff \frac{3}{x-9} - \frac{2}{x+2}>0\iff\frac{3(x+2)-2(x-9)}{(x-9)(x+2)}>0$$$$\iff\frac{x+24}{(x-9)(x+2)}>0$$
and study separetely the sign for numerator and denominator.
A: $\frac{3}{x-9} -\frac{2}{x+2}$ may only change sign where it’s zero (at $x=-24)$ or where it’s not defined (at $x=9$ or $ x=-2$).  Now determine the sign on each of the four intervals given by those three numbers.
A: A shorter way
The domain of the inequality is $\mathbf R\smallsetminus\{-2,9\}$.
You can multiply both sides of the inequality by a positive number. To remove the denominators, we'll multiply by $(x+2)^2(x-9)^2$ and obtain
\begin{align}
3(x+2)^2(x-9)>2(x+2)(x-9)^2&\iff(x+2)(x-9)\bigl(3(x+2)-2(x-9)\bigr) >0\\
&\iff (x+2)(x-9)(x+24)>0.
\end{align}
Now , a cubic polynomial with simple roots has alternating signs on the successive intervals determined by the roots, and this particular polynomial tends to $+\infty$ when $x$ tends to   $+\infty$, so the solutions are
$$-24<x<-2\enspace\text{ or }\enspace x>9.$$
A: Consider the function $f(x) = \dfrac{3}{x-9} - \dfrac{2}{x+2}$.
The function $f(x)$ has a zero at $x=-24$ and poles at $x=-2$ and at $x=9$. The two poles and the one zero divide the $x-$axis into four intervals
$$(-\infty, -24), \ (-24, -2), \ (-2, 9), (9, \infty) $$
In each of those intervals, $f(x)$ is either entirely positive or entirely negative.
We note that 
$\text{$-100 \in (-\infty, -24)$ and $f(-100) < 0$}$
$\text{$-13 \in (-24, -2)$ and $f(-13) = \dfrac{1}{22}$} > 0$
$\text{$0 \in (-2, 9)$ and $f(0) = -\dfrac 43 < 0$}$
$\text{$100 \in (9, \infty)$ and $f(100) > 0$}$
So the solution set is $(-24, -2) \cup (9, \infty)$.
