Is there a contradiction is this exercise? The following exercise was a resolution to this problem

Let $\displaystyle\frac{2x+5}{(x-3)(x-7)}=\frac{A}{(x-7)}+\frac{B}{(x-3)}\space  \forall \space x \in \mathbb{R}$. Find the values for $A$ and $B$

The propose resolution was:
In order to isolate $A$ on the right side, multiply all the equation by $x-7$
$\displaystyle\frac{(2x+5)(x-7)}{(x-3)(x-7)}=\frac{A(x-7)}{(x-7)}+\frac{B(x-7)}{(x-3)}$
Now is my doubt. The resolution suggests that $x-7$ cancel out.
$\displaystyle\frac{(2x+5)}{(x-3)}=A+\frac{B(x-7)}{(x-3)}$
But, $x-7$ can be equal zero. In this situation, is allowed to perform this operation? One can say "for every $x\neq7$", but on the next step the resolution says "for $x=7$ we have".
$\displaystyle\frac{(14+5)}{(7-3)}=A+\frac{B(0)}{(7-3)} \Leftrightarrow A=\frac{19}{4}$
I think there is a contradiction is this resolution.
 A: The incorrect bit is "$\forall x\in\Bbb R$." We must indeed rule out $x=7$ and $x=3$ in order to avoid problems. What we can still do in that case is take the limit as $x$ approaches $7$ (we can't just plug $x=7$ in, if we've declared that $x\ne 7$), and get the same result.
A: If $\frac{2x+5}{x-3}=A+B\frac{x-7}{x-3}$ for all $x\ne 7$ (or $3$), then we can take the limit as $x\to 7$ on both sides to get the given result, and since both sides are continuous, it is equivalent to just plugging in $x=7$.
A: I would do it in another way. the equalation is $$\frac {2x+5}{(x-3)(x-7)}=\frac A {x-3}+\frac B {x-7}$$. what we can do is to take the common domniator on the right side getting $$\frac {2x+5}{(x-3)(x-7)}=\frac {A(x-7)+B(x-3)} {(x-7)(x-3)}$$. since the numerator in both sides is condcuted from numbers and x, we can conclude that: $$\begin{cases}
A+B=2 &  \\
-7A-3B=5 & \
\end{cases}
$$. solving this system is relatively simple and by substituting $A=2-B$  in the second equaltion you get $$B=4.75,A=-2.75$$
