1. **State the problem:** Solve the compound inequality $$-3x + 4 < x + 7 < 2x + 5$$. 2. **Understand the compound inequality:** This means both inequalities must be true simultaneously: - $$-3x + 4 < x + 7$$ - $$x + 7 < 2x + 5$$ 3. **Solve the first inequality:** $$-3x + 4 < x + 7$$ Subtract $$x$$ from both sides: $$-3x - x + 4 < 7$$ $$-4x + 4 < 7$$ Subtract 4 from both sides: $$-4x + \cancel{4} - \cancel{4} < 7 - 4$$ $$-4x < 3$$ Divide both sides by $$-4$$ (remember to reverse inequality sign when dividing by negative): $$\frac{-4x}{-4} > \frac{3}{-4}$$ $$x > -\frac{3}{4}$$ 4. **Solve the second inequality:** $$x + 7 < 2x + 5$$ Subtract $$x$$ from both sides: $$\cancel{x} + 7 < 2x - \cancel{x} + 5$$ $$7 < x + 5$$ Subtract 5 from both sides: $$7 - 5 < x + \cancel{5} - \cancel{5}$$ $$2 < x$$ 5. **Combine the results:** From the first inequality: $$x > -\frac{3}{4}$$ From the second inequality: $$x > 2$$ Since $$x$$ must satisfy both, the solution is the intersection: $$x > 2$$ **Final answer:** $$\boxed{x > 2}$$