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

$$(2x-3)(x+5)=9(2x-3).$$ Since you are multiplying both sides by the same constant, namely $2x-3$, then they cancel out, thus bringing forward the following equation: $$x+5=9.\tag1$$

$$x(x+2)=x(-x+3).$$ In the same fashion as before, the $x$'s cancel out, i.e. $$x+2=-x+3.$$ Now, adding $x$ to both sides yeilds the following equation: $$2x + 2 = 3$$ $$\Downarrow$$ $$2x = 1.\tag2$$

Finally, solve for $x$ in Eq. $(1)$ and Eq. $(2)$. Also note that each of the common factors, $(2x-3)$ and $x$, can be equal to $0$, because for all values $n$, one always has that $0\cdot n = 0$. Thus, you have a bonus two other solutions.
