1. **State the problem:** Simplify the expression $$\frac{8}{3v-6} + \frac{6v}{v+3}$$. 2. **Identify the denominators and factor if possible:** - The first denominator is $$3v-6$$, which can be factored as $$3(v-2)$$. - The second denominator is $$v+3$$, which is already factored. 3. **Rewrite the expression with factored denominators:** $$\frac{8}{3(v-2)} + \frac{6v}{v+3}$$ 4. **Find the least common denominator (LCD):** The LCD is $$3(v-2)(v+3)$$. 5. **Rewrite each fraction with the LCD as denominator:** $$\frac{8}{3(v-2)} = \frac{8(v+3)}{3(v-2)(v+3)}$$ $$\frac{6v}{v+3} = \frac{6v \cdot 3(v-2)}{3(v-2)(v+3)} = \frac{18v(v-2)}{3(v-2)(v+3)}$$ 6. **Add the numerators over the common denominator:** $$\frac{8(v+3) + 18v(v-2)}{3(v-2)(v+3)}$$ 7. **Expand the numerators:** $$8(v+3) = 8v + 24$$ $$18v(v-2) = 18v^2 - 36v$$ 8. **Combine the numerator terms:** $$8v + 24 + 18v^2 - 36v = 18v^2 - 28v + 24$$ 9. **Write the full expression:** $$\frac{18v^2 - 28v + 24}{3(v-2)(v+3)}$$ 10. **Factor the numerator if possible:** First, factor out the common factor 2: $$2(9v^2 - 14v + 12)$$ Try to factor $$9v^2 - 14v + 12$$: - The product of $$9 \times 12 = 108$$. - We look for two numbers that multiply to 108 and add to -14, but no such integer pair exists. Therefore, the numerator does not factor nicely over integers. 11. **Simplify the fraction by canceling common factors:** The denominator has a factor 3, numerator has factor 2, no common factors to cancel. 12. **Final simplified expression:** $$\frac{2(9v^2 - 14v + 12)}{3(v-2)(v+3)}$$ This is the simplest form. **Answer:** $$\boxed{\frac{2(9v^2 - 14v + 12)}{3(v-2)(v+3)}}$$