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

Remember the one rule when it comes to division:
Never divide by zero
If you want to remove $(2x-3)$ you must check that it is not equal to $0$ that is, you have to assume that $x\neq \frac{3}{2}$. Same goes with division by $x$.
In your first example, if you are looking for $x$, dividing by $(2x-3)$ without notice means that you lost a solution $\left(x=\frac{3}{2}\right)$.
