1. **State the problem:** Solve the inequality $$-3 > \frac{6}{5}(-2x - 1) - 1$$. 2. **Distribute the fraction:** Multiply $$\frac{6}{5}$$ by each term inside the parentheses: $$-3 > \frac{6}{5} \times (-2x) + \frac{6}{5} \times (-1) - 1$$ $$-3 > -\frac{12}{5}x - \frac{6}{5} - 1$$ 3. **Combine like terms on the right side:** $$-3 > -\frac{12}{5}x - \frac{6}{5} - \frac{5}{5}$$ $$-3 > -\frac{12}{5}x - \frac{11}{5}$$ 4. **Add $$\frac{11}{5}$$ to both sides to isolate the term with $$x$$:** $$-3 + \frac{11}{5} > -\frac{12}{5}x$$ Convert $$-3$$ to $$\frac{-15}{5}$$: $$\frac{-15}{5} + \frac{11}{5} > -\frac{12}{5}x$$ $$\frac{-4}{5} > -\frac{12}{5}x$$ 5. **Divide both sides by $$-\frac{12}{5}$$ to solve for $$x$$.** Remember, dividing by a negative number reverses the inequality sign: $$\frac{\frac{-4}{5}}{\cancel{-\frac{12}{5}}} < x$$ Simplify the division: $$\frac{-4}{5} \times \frac{5}{-12} < x$$ $$\frac{-4 \times 5}{5 \times -12} < x$$ $$\frac{-20}{-60} < x$$ $$\frac{1}{3} < x$$ 6. **Final answer:** $$x > \frac{1}{3}$$ This means all values of $$x$$ greater than $$\frac{1}{3}$$ satisfy the inequality.