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 ?

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 $(-24, -2) \cup (9, \infty)$.
