Answer
$\frac{2x-8}{(x-3)(x-5)}$.
Work Step by Step
The given expression is
$=\frac{x-3}{x^2-8x+15}+\frac{x+2}{x^2-x-6}$
Factor $x^2-8x+15$.
Rewrite the middle term $-5x$ as $-5x-3x$
$=x^2-5x-3x+15$
Group terms.
$=(x^2-5x)+(-3x+15)$
Factor each term.
$=x(x-5)-3(x-5)$
Factor out $(x-5)$.
$=(x-5)(x-3)$.
Factor $x^2-x-6$.
Rewrite the middle term $-5x$ as $-3x+2x$
$=x^2-3x+2x-6$
Group terms.
$=(x^2-3x)+(2x-6)$
Factor each term.
$=x(x-3)+2(x-3)$
Factor out $(x-3)$.
$=(x-3)(x+2)$.
Substitute all factors into the given expression.
$=\frac{x-3}{(x-5)(x-3)}+\frac{x+2}{(x-3)(x+2)}$
Cancel common terms.
$=\frac{1}{x-5}+\frac{1}{x-3}$
LCM $=$ greatest power of all the prime factors.
LCM $=(x-3)(x-5)$
$=\frac{(x-3)}{(x-3)}\cdot \frac{1}{x-5}+\frac{(x-5)}{(x-5)}\cdot \frac{1}{x-3}$
Simplify.
$=\frac{(x-3)}{(x-3)(x-5)}+\frac{(x-5)}{(x-5)(x-3)}$
Add both numerators because both denominators are equal.
$=\frac{(x-3)+(x-5)}{(x-3)(x-5)}$
Simplify.
$=\frac{x-3+x-5}{(x-3)(x-5)}$
$=\frac{2x-8}{(x-3)(x-5)}$.
