My textbook ends a proof with the following:
$|x-9| \over \sqrt(x) + 3$ < $\epsilon$
can be rearranged to conclude:
|$x-9 \over \sqrt(x) -3$ - 6| < $\epsilon$
However, I don't understand how the rearrangement of the absolute values works. How is this reasoned out explicitly?
Observe $|x - 9| = (\sqrt{x} + 3)(\sqrt{x} - 3)$.
Cancelling in the first inequality from your post, we have:
$|\sqrt{x} - 3 |< \epsilon$.
Note the latter inequality in your post says, after cancellation, that
$|\sqrt{x} + 3 - 6| < \epsilon$, i.e., $|\sqrt{x} - 3 |< \epsilon$.
This gives all the necessary pieces.
We can now alter the order slightly to make it a cleaner proof.
Proof:
Cancelling in the initial inequality, we have
$|\sqrt{x} - 3 |< \epsilon$, i.e., $|(\sqrt{x} + 3) - 6 |< \epsilon$, which is equivalent to the latter inequality. QED.