1. **State the problem:** Solve the compound inequality $$4 - 3x > x + 12 < \frac{3x + 29}{2}$$ and find the integral values of $x$ that satisfy it. 2. **Rewrite the compound inequality:** The inequality means both $$4 - 3x > x + 12$$ and $$x + 12 < \frac{3x + 29}{2}$$ must hold true simultaneously. 3. **Solve the first inequality:** $$4 - 3x > x + 12$$ Subtract $x$ from both sides: $$4 - 4x > 12$$ Subtract 4 from both sides: $$-4x > 8$$ Divide both sides by $-4$ (remember to reverse inequality sign when dividing by negative): $$x < -2$$ 4. **Solve the second inequality:** $$x + 12 < \frac{3x + 29}{2}$$ Multiply both sides by 2 to clear denominator: $$2x + 24 < 3x + 29$$ Subtract $2x$ from both sides: $$24 < x + 29$$ Subtract 29 from both sides: $$-5 < x$$ or equivalently $$x > -5$$ 5. **Combine the two results:** $$x < -2$$ and $$x > -5$$ So the solution set is $$-5 < x < -2$$ 6. **Find integral values:** The integers strictly between $-5$ and $-2$ are $$-4, -3$$ **Final answer:** The integral values of $x$ satisfying the inequality are $-4$ and $-3$.