Question: Solve the rational inequality and graph the solution set on a real number line. Express the solution set in interval notation.
$$\frac{x-8}{x+3} > 0$$
Solve the inequality. What is the solution set? Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. The solution set is ____
(Simplify your answer. Type your answer in interval notation. Type an exact answer, using radicals as needed. Use integers or fractions for any numbers in the expression.)
B. The solution set is the empty set.
1. **State the problem:** Solve the inequality $$\frac{x-8}{x+3} > 0$$ and express the solution set in interval notation.
2. **Identify critical points:** The numerator $$x-8=0$$ when $$x=8$$ and the denominator $$x+3=0$$ when $$x=-3$$. These points divide the number line into intervals.
3. **Intervals to test:**
- $$(-\infty, -3)$$
- $$(-3, 8)$$
- $$(8, \infty)$$
4. **Check signs on each interval:**
- For $$x < -3$$, pick $$x=-4$$: numerator $$-4-8=-12$$ (negative), denominator $$-4+3=-1$$ (negative), fraction is $$\frac{negative}{negative} = positive$$, so inequality holds.
- For $$-3 < x < 8$$, pick $$x=0$$: numerator $$0-8=-8$$ (negative), denominator $$0+3=3$$ (positive), fraction is $$\frac{negative}{positive} = negative$$, inequality does not hold.
- For $$x > 8$$, pick $$x=9$$: numerator $$9-8=1$$ (positive), denominator $$9+3=12$$ (positive), fraction is $$\frac{positive}{positive} = positive$$, inequality holds.
5. **Check domain restrictions:** $$x \neq -3$$ because denominator cannot be zero.
6. **Write solution set:** The inequality holds for $$x \in (-\infty, -3) \cup (8, \infty)$$.
7. **Final answer:**
$$\boxed{(-\infty, -3) \cup (8, \infty)}$$