To solve the expression given in the image:

$\frac{16-p}{(p+4)(p-6)} + \frac{2}{p+4}$

We first look for a common denominator, which in this case is $(p+4)(p-6)$. We need to rewrite each fraction with this common denominator:

1. The first fraction is already expressed with the denominator $(p+4)(p-6)$.
2. For the second fraction, multiply the numerator and the denominator by $p-6$:

$\frac{2}{p+4} = \frac{2(p-6)}{(p+4)(p-6)}$

Now, add the two fractions:

$\frac{16-p}{(p+4)(p-6)} + \frac{2(p-6)}{(p+4)(p-6)} = \frac{16-p + 2(p-6)}{(p+4)(p-6)}$

Simplify the numerator:

$16-p + 2p - 12 = p + 4$

Therefore, the expression simplifies to:

$\frac{p+4}{(p+4)(p-6)}$

Assuming $p \neq -4$, we can cancel $p+4$ from the numerator and denominator:

$\frac{1}{p-6}$

Thus, the sum of the expressions in simplest terms is:

$\frac{1}{p-6}$

If you have any questions or need further explanations on how I derived this, feel free to ask! Here are some related questions you might consider:

1. What would the result be if $p = -6$?
2. How would you evaluate this expression for specific values of $p$?
3. What happens to the expression as $p$ approaches -4?
4. Could you graph this expression to understand its behavior better?
5. What are the restrictions on the value of $p$ in this expression?

Tip: When adding fractions with algebraic expressions, always look for a common denominator first to simplify the process.