Reduce to the lowest terms: 3/(x+3) + 2/(3-x) + (2x+24)/(x^2-9).

We notice that the denominator of the 3rd fraction is a
difference of squares that returns the product:

x^2 - 9 =
(x-3)(x+3)

We notice that we can factorize by -1 the
denominator of the 2nd fraction:

2/(3-x) =
-2/(x-3)

We'll re-write the  sum of
fractions:

3/(x+3) - 2/(x-3) + (2x+24)/
(x-3)(x+3)

We'll multiply each fraction by the missing
factor in order to get the LCD (x-3)(x+3)

3(x-3)/(x-3)(x+3)
- 2(x+3)/(x-3)(x+3) + (2x+24)/ (x-3)(x+3)

Since the
fractions have the same denominators, we can re-write the
sum:

[3(x-3) - 2(x+3) + (2x+24)]/
(x-3)(x+3)

(3x - 9 - 2x - 6 + 2x + 24)/
(x-3)(x+3)

We'll combine and eliminate like
terms:

(3x - 9 - 6 + 24)/
(x-3)(x+3)

(3x+9)/
(x-3)(x+3)

We'll factorize the numerator by
3:

3(x+3)/ (x-3)(x+3)

We'll
simplify and we'll
get:

3/(x-3)

The
result of the given sum, reduced to the lowest terms, is
3/(x-3).