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

Observe that the solution set of $x^2=x$ is $\{0,1\}$ whereas the solution set of $x=1$ is $\{1\}$.  Hence dividing by $x$ is not an equivalence transformation.  Instead of dividing by $x$ factor it out: $x^2-x=0\iff x(x-1)=0$ to get $x=0$ or $x=1$.
In your first case we will get $(2x-3)(x+5-9)=0\iff 2x-3=0$ or $x-4=0$.
