We are asked to simplify the compound fraction $\frac{\frac{x-7}{x+2}}{\frac{x^2-49}{5x+10}}$.

1. Problem Description

We are asked to simplify the compound fraction $\frac{\frac{x-7}{x+2}}{\frac{x^2-49}{5x+10}}$.

2. Solution Steps

First, we rewrite the compound fraction as a division problem:

$\frac{\frac{x-7}{x+2}}{\frac{x^2-49}{5x+10}} = \frac{x-7}{x+2} \div \frac{x^2-49}{5x+10}$.

Next, we change the division to multiplication by inverting the second fraction:

$\frac{x-7}{x+2} \div \frac{x^2-49}{5x+10} = \frac{x-7}{x+2} \cdot \frac{5x+10}{x^2-49}$.

Now, we factor the expressions in the numerators and denominators:

$x^2 - 49 = (x-7)(x+7)$ (difference of squares)

$5x + 10 = 5(x+2)$

Substituting the factored expressions, we have:

$\frac{x-7}{x+2} \cdot \frac{5(x+2)}{(x-7)(x+7)}$.

Now, we cancel out the common factors: $(x-7)$ and $(x+2)$ from the numerator and the denominator.

$\frac{x-7}{x+2} \cdot \frac{5(x+2)}{(x-7)(x+7)} = \frac{1}{1} \cdot \frac{5}{x+7} = \frac{5}{x+7}$.

3. Final Answer

$\frac{5}{x+7}$