Question

what is the sum?
$\frac{3}{x^{2}-9}+\frac{5}{x + 3}$
$\frac{8}{x^{2}+x - 6}$
$\frac{5x-12}{x - 3}$
$\frac{-5x}{(x + 3)(x - 3)}$
$\frac{5x-12}{(x + 3)(x - 3)}$

Explanation:

Step1: Factor the denominator

Factor $x^{2}-9$ using the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$. So, $x^{2}-9=(x + 3)(x - 3)$. The given expression is $\frac{3}{(x + 3)(x - 3)}+\frac{5}{x + 3}$.

Step2: Find a common denominator

The common denominator of the two fractions is $(x + 3)(x - 3)$. Rewrite $\frac{5}{x + 3}$ with the common denominator: $\frac{5}{x + 3}\times\frac{x - 3}{x - 3}=\frac{5(x - 3)}{(x + 3)(x - 3)}$.

Step3: Add the fractions

$\frac{3}{(x + 3)(x - 3)}+\frac{5(x - 3)}{(x + 3)(x - 3)}=\frac{3+5(x - 3)}{(x + 3)(x - 3)}$.

Step4: Expand and simplify the numerator

Expand $3+5(x - 3)$: $3+5x-15 = 5x-12$. So the sum is $\frac{5x - 12}{(x + 3)(x - 3)}$.

Answer:

$\frac{5x - 12}{(x + 3)(x - 3)}$