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 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
$$-24<x<-2\enspace\text{ or }\enspace x>9.$$
