1. **State the problem:** Add the two fractions $$\frac{3}{b-8} + \frac{7}{b+3}$$. 2. **Formula and rules:** To add fractions, find a common denominator, which is the product of the two denominators since they are different: $$ (b-8)(b+3) $$. 3. **Rewrite each fraction with the common denominator:** $$\frac{3}{b-8} = \frac{3(b+3)}{(b-8)(b+3)}$$ $$\frac{7}{b+3} = \frac{7(b-8)}{(b+3)(b-8)}$$ 4. **Add the numerators:** $$\frac{3(b+3)}{(b-8)(b+3)} + \frac{7(b-8)}{(b+3)(b-8)} = \frac{3(b+3) + 7(b-8)}{(b-8)(b+3)}$$ 5. **Expand the numerators:** $$3(b+3) = 3b + 9$$ $$7(b-8) = 7b - 56$$ 6. **Combine like terms:** $$3b + 9 + 7b - 56 = (3b + 7b) + (9 - 56) = 10b - 47$$ 7. **Final expression:** $$\frac{10b - 47}{(b-8)(b+3)}$$ This is the simplified sum of the two fractions. **Answer:** $$\frac{10b - 47}{(b-8)(b+3)}$$