What is the Mathematical Property that justifies equating coefficients while solving partial fractions? The McGraw Hill PreCaculus Textbook gives several good examples of solving partial fractions, and they justify all but one step with established mathematical properties.
In the 4th step of Example 1, when going from:
$$1x + 13 = (A+B)x+(4A-5B)$$
they say to "equate the coefficients", writing the linear system
$$A+B = 1$$
$$4A-5B=13$$
It is a simple step, color coded in the textbook for easy understanding, but McGraw Hill does not justify it with any mathematical property, postulate or theorem.  Addition and/or multiplication properties of equality don't seem to apply directly.
Can someone help me justify this step?!

Presumably it is stated that this equation must hold for all $x$. That means that in particular it must hold for $x=0$. If we set $x=0$ in your equation, we find the equation
$13=4A-5B$
which is the second equation you wrote. Let us now look at $x=1$, for which we find
$1+13=(A+B)+(4A-5B)$
We already know that $4A-5B=13$ from our previous step, which we can thus put into the equation, leading to $1+13=(A+B)+13$. Subtracting $13$ from either side leaves us with
$1=A+B$
which is your second equation.
Note that this can be done more generally, i.e. even if the equation only has to hold for a few $x$, not including $0$ or $1$. If you want a proof that does not use $x=0$ and $x=1$ let me know.
