Answer
$ \displaystyle \frac{1}{x-3},\qquad x\neq\pm 3$
Work Step by Step
When rational expressions have the same denominator, $\displaystyle \frac{A}{C}-\frac{B}{C}=\frac{A-B}{C}$.
$\displaystyle \frac{2x-7}{x^{2}-9}-\frac{x-10}{x^{2}-9}=\frac{2x-7-(x-10)}{x^{2}-9}$
Simplify the numerator: recognize the difference of squares in the denominator.
$=\displaystyle \frac{x+3}{(x-3)(x+3)},\qquad x\neq\pm 3$
All numbers producing zero in the denominator are excluded from the domain. Common factors cancel.
$= \displaystyle \frac{1}{x-3},\qquad x\neq\pm 3$
