1. The problem is to simplify the expression $$\frac{6}{c+5} + \frac{7}{c+2}$$. 2. To add these fractions, we need a common denominator. The common denominator is $$(c+5)(c+2)$$. 3. Rewrite each fraction with the common denominator: $$\frac{6}{c+5} = \frac{6(c+2)}{(c+5)(c+2)}$$ $$\frac{7}{c+2} = \frac{7(c+5)}{(c+5)(c+2)}$$ 4. Now add the numerators: $$\frac{6(c+2)}{(c+5)(c+2)} + \frac{7(c+5)}{(c+5)(c+2)} = \frac{6(c+2) + 7(c+5)}{(c+5)(c+2)}$$ 5. Expand the numerators: $$6(c+2) = 6c + 12$$ $$7(c+5) = 7c + 35$$ 6. Add the expanded numerators: $$6c + 12 + 7c + 35 = (6c + 7c) + (12 + 35) = 13c + 47$$ 7. So the expression becomes: $$\frac{13c + 47}{(c+5)(c+2)}$$ 8. This is the simplified form since the numerator cannot be factored further to cancel with the denominator. Final answer: $$\frac{13c + 47}{(c+5)(c+2)}$$