1. **State the problem:** Simplify the expression $$\frac{-6 - \frac{6}{3^2} - (-3)}{-6 - \frac{-7 - \frac{5}{7^1} - 6}{3}}$$ using order of operations. 2. **Recall order of operations:** Parentheses, Exponents, Multiplication/Division (left to right), Addition/Subtraction (left to right). 3. **Simplify inside the numerator:** - Calculate $3^2 = 9$. - Calculate $\frac{6}{9} = \frac{2}{3}$. - The numerator becomes $-6 - \frac{2}{3} - (-3) = -6 - \frac{2}{3} + 3$. 4. **Combine terms in numerator:** - Convert $-6$ and $3$ to fractions with denominator 3: $-6 = -\frac{18}{3}$, $3 = \frac{9}{3}$. - Sum: $-\frac{18}{3} - \frac{2}{3} + \frac{9}{3} = \frac{-18 - 2 + 9}{3} = \frac{-11}{3}$. 5. **Simplify inside the denominator:** - Calculate $7^1 = 7$. - Calculate $\frac{5}{7} = \frac{5}{7}$. - Calculate $-7 - \frac{5}{7} - 6$. 6. **Combine terms in denominator numerator:** - Convert $-7$ and $-6$ to fractions with denominator 7: $-7 = -\frac{49}{7}$, $-6 = -\frac{42}{7}$. - Sum: $-\frac{49}{7} - \frac{5}{7} - \frac{42}{7} = \frac{-49 - 5 - 42}{7} = \frac{-96}{7}$. 7. **Divide by 3 in denominator:** - $\frac{-96}{7} \div 3 = \frac{-96}{7} \times \frac{1}{3} = \frac{-96}{21}$. - Simplify $\frac{-96}{21}$ by dividing numerator and denominator by 3: $$\frac{\cancel{-96}^{32}}{\cancel{21}^7} = \frac{-32}{7}$$ 8. **Denominator becomes:** $-6 - \left(-\frac{32}{7}\right) = -6 + \frac{32}{7}$. 9. **Convert $-6$ to fraction with denominator 7:** $-6 = -\frac{42}{7}$. 10. **Sum denominator:** $-\frac{42}{7} + \frac{32}{7} = \frac{-42 + 32}{7} = \frac{-10}{7}$. 11. **Final expression:** $$\frac{\frac{-11}{3}}{\frac{-10}{7}} = \frac{-11}{3} \times \frac{7}{-10}$$ 12. **Multiply fractions:** $$\frac{-11 \times 7}{3 \times -10} = \frac{-77}{-30}$$ 13. **Simplify signs:** $$\frac{-77}{-30} = \frac{77}{30}$$ 14. **Final answer:** $$\boxed{\frac{77}{30}}$$