Why can't I cancel $2x-3$ from $(2x-3)(x+5)=9(2x-3)$? Why are these simplifications wrong?
$$\begin{align}
(2x-3)(x+5)=9(2x-3) &\quad\to\quad \frac{(2x-3)(x+5)}{2x-3} = \frac{9(2x-3)}{2x-3} \quad\to\quad x+5 = 9\\[4pt]
x(x+2)=x(-x+3) &\quad\to\quad \frac{x(x+2)}{x} = \frac{x(-x+3)}{x} \quad\to\quad x+2=-x+3
\end{align}$$

From here
$$(2x-3)(x+5)=9(2x-3)$$
we can observe that $2x-3=0$ is a solution and for $2x-3\neq 0$ we can cancel out and obtain
$$(2x-3)(x+5)=9(2x-3)\iff x+5=9\iff x=4$$
thus the solutions for the original equation are $x=\frac32$ and $x=4$.
As an alternative note that
$$(2x-3)(x+5)=9(2x-3)\iff 2x^2+7x-15=18x-27 \iff2x^2-11x+12=0$$
and then
$$x_{1,2}=\frac{11\pm\sqrt{121-96}}{4}=\frac{11\pm5}{4}=\frac32,4$$
