1. **State the problem:** Simplify the expression $$\frac{n - 4}{7n^2 - 28n} + \frac{3n}{n - 2}$$ and identify which of the given options it equals. 2. **Factor the denominators:** - Factor the quadratic in the first denominator: $$7n^2 - 28n = 7n(n - 4)$$ 3. **Rewrite the expression with factored denominators:** $$\frac{n - 4}{7n(n - 4)} + \frac{3n}{n - 2}$$ 4. **Simplify the first fraction:** Since $n - 4$ appears in numerator and denominator, cancel it: $$\frac{\cancel{n - 4}}{7n\cancel{(n - 4)}} = \frac{1}{7n}$$ 5. **Rewrite the expression:** $$\frac{1}{7n} + \frac{3n}{n - 2}$$ 6. **Find common denominator:** The denominators are $7n$ and $n - 2$, so common denominator is $7n(n - 2)$. 7. **Rewrite each fraction with common denominator:** $$\frac{1}{7n} = \frac{n - 2}{7n(n - 2)}$$ $$\frac{3n}{n - 2} = \frac{3n \cdot 7n}{7n(n - 2)} = \frac{21n^2}{7n(n - 2)}$$ 8. **Add the fractions:** $$\frac{n - 2}{7n(n - 2)} + \frac{21n^2}{7n(n - 2)} = \frac{n - 2 + 21n^2}{7n(n - 2)}$$ 9. **Simplify numerator:** $$21n^2 + n - 2$$ 10. **Factor numerator if possible:** Try to factor $21n^2 + n - 2$: Find two numbers that multiply to $21 \times (-2) = -42$ and add to $1$. These are $7$ and $-6$. Rewrite: $$21n^2 + 7n - 6n - 2 = 7n(3n + 1) - 2(3n + 1) = (7n - 2)(3n + 1)$$ 11. **Rewrite the expression:** $$\frac{(7n - 2)(3n + 1)}{7n(n - 2)}$$ 12. **Check for common factors to cancel:** No common factors between numerator and denominator. 13. **Final simplified expression:** $$\frac{(7n - 2)(3n + 1)}{7n(n - 2)}$$ 14. **Compare with options:** None of the options match this expression directly. 15. **Re-examine the problem:** The original problem asks to divide the rational expression, but the user input is an addition. If the problem is to divide $$\frac{n - 4}{7n^2 - 28n}$$ by $$\frac{3n}{n - 2}$$, then: 16. **Rewrite division as multiplication by reciprocal:** $$\frac{n - 4}{7n^2 - 28n} \div \frac{3n}{n - 2} = \frac{n - 4}{7n^2 - 28n} \times \frac{n - 2}{3n}$$ 17. **Factor denominator:** $$7n^2 - 28n = 7n(n - 4)$$ 18. **Substitute:** $$\frac{n - 4}{7n(n - 4)} \times \frac{n - 2}{3n}$$ 19. **Cancel $n - 4$:** $$\frac{\cancel{n - 4}}{7n\cancel{(n - 4)}} \times \frac{n - 2}{3n} = \frac{1}{7n} \times \frac{n - 2}{3n} = \frac{n - 2}{21n^2}$$ 20. **Final answer:** $$\frac{n - 2}{21n^2}$$ which matches option B. **Answer: Option B) $\frac{n - 2}{21n^2}$**