The problem asks us to simplify the expression $y = \frac{2}{x^2 - 9} + \frac{1}{x-3}$.

1. Problem Description

The problem asks us to simplify the expression $y = \frac{2}{x^2 - 9} + \frac{1}{x-3}$.

2. Solution Steps

First, we factor the denominator $x^2 - 9$:

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

So we have

$y = \frac{2}{(x-3)(x+3)} + \frac{1}{x-3}$

To add the fractions, we need a common denominator, which is $(x-3)(x+3)$. We multiply the second fraction by $\frac{x+3}{x+3}$:

$y = \frac{2}{(x-3)(x+3)} + \frac{1}{x-3} \cdot \frac{x+3}{x+3}$

$y = \frac{2}{(x-3)(x+3)} + \frac{x+3}{(x-3)(x+3)}$

Now we can add the numerators:

$y = \frac{2 + (x+3)}{(x-3)(x+3)}$

$y = \frac{2 + x + 3}{(x-3)(x+3)}$

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

$y = \frac{x+5}{x^2 - 9}$

3. Final Answer

$y = \frac{x+5}{x^2-9}$