Algebra II: Basic Adding fractions question Disclaimer: I'm new to adding fractions this way.
Does anyone understand how the author justifies the circled equation?

 A: First consider (as a concrete example) the possible fractions that have $6$ as the denominator. They can all be expressed as some combination of halves or thirds:
$\frac 16 = \frac 12 - \frac 13$
$\frac 26 = \frac 13$
$\frac 36 = \frac 12$
$\frac 46 = \frac 13 + \frac 13$
$\frac 56 = \frac 12 + \frac 13$
This observation leads us to conclude that all fractions with denominator $pq$ can be expressed as a combination of $\frac 1p$ and $\frac 1q$ (provided $p$ and $q$ have no common factors).
Moving on to algebraic fractions, all fractions with denominator $(x-p)(x-q)$ can be expressed as a combination of $\frac 1{x-p}$ and $\frac 1{x-q}$.
A: When we write out an equation such as
$$\frac{7x-1}{(x+2)(x-3)} = \frac A{x+2} + \frac B{x-3},$$
where $A$ and $B$ are unknown real numbers, the general idea is
that we want to solve the equation for $A$ and $B$.
(I assume it is stated, or at least implicit, in the preceding
discussion of the "original fraction", that the equation must be true
for all values of $x$.)
For a very general equation with two unknowns, it is possible there is a
unique solution for the two unknowns; it is also possible that
there are many solutions, or that there are none at all.
You can always write an equation with two unknowns without knowing
whether it actually is possible for those two unknowns to be given values
that make the equation true. If that is not possible, you have an
equation with no solutions.
This particular equation has properties that you (hopefully) will
come to recognize (because you will likely use equations like this to perform
partial-fraction decomposition of integrals if you study calculus),
which guarantee it has a solution; but if this has not already been shown
to you, then the justification comes later, when you actually solve for
$A$ and $B$, thereby proving that a solution exists.
