1. **State the problem:** We need to find the value of $k$ such that the compound inequality $$4 - \frac{x}{3} > 3x - k > -13$$ holds true. 2. **Rewrite the compound inequality as two separate inequalities:** - $$4 - \frac{x}{3} > 3x - k$$ - $$3x - k > -13$$ 3. **Solve the first inequality for $k$:** $$4 - \frac{x}{3} > 3x - k \implies k > 3x - \left(4 - \frac{x}{3}\right) = 3x - 4 + \frac{x}{3} = \frac{9x}{3} + \frac{x}{3} - 4 = \frac{10x}{3} - 4$$ 4. **Solve the second inequality for $k$:** $$3x - k > -13 \implies -k > -13 - 3x \implies k < 3x + 13$$ 5. **Combine the inequalities for $k$:** $$\frac{10x}{3} - 4 < k < 3x + 13$$ 6. **Interpretation:** For the compound inequality to hold for all $x$, the interval for $k$ must be valid for all $x$. However, since $k$ depends on $x$ here, to find a fixed $k$ that satisfies the inequality for all $x$, the inequalities must overlap for all $x$. 7. **Find $k$ such that $$\frac{10x}{3} - 4 < 3x + 13$$ for all $x$:** $$\frac{10x}{3} - 4 < 3x + 13 \implies \frac{10x}{3} - 3x < 13 + 4 \implies \frac{10x - 9x}{3} < 17 \implies \frac{x}{3} < 17 \implies x < 51$$ 8. **Since $x$ must be less than 51 for the inequality to hold, the compound inequality holds for $x < 51$. To find $k$ independent of $x$, consider the boundary at $x=51$:** $$k > \frac{10(51)}{3} - 4 = \frac{510}{3} - 4 = 170 - 4 = 166$$ $$k < 3(51) + 13 = 153 + 13 = 166$$ 9. **Therefore, $k$ must satisfy:** $$166 < k < 166$$ which is impossible unless $k=166$. 10. **Check if $k=166$ satisfies the original inequality:** At $x=51$, both inequalities become equalities, so $k=166$ is the value that makes the compound inequality hold. **Final answer:** $$k = 166$$