1. **State the problem:** We need to determine which of the given compound inequalities has no solution. 2. **Recall:** A compound inequality joined by "and" means the solution must satisfy both inequalities simultaneously. 3. **Analyze each compound inequality:** **(a) Solve** $7x - 8 > 9(x - 1)$ and $3x + 2 < 5x + 3$ - First inequality: $$7x - 8 > 9x - 9$$ $$7x - 8 - 9x + 9 > 0$$ $$-2x + 1 > 0$$ $$-2x > -1$$ $$\cancel{-2}x > \cancel{-1}$$ $$x < \frac{1}{2}$$ (note the inequality flips because we divided by a negative) - Second inequality: $$3x + 2 < 5x + 3$$ $$3x + 2 - 5x - 3 < 0$$ $$-2x - 1 < 0$$ $$-2x < 1$$ $$x > -\frac{1}{2}$$ - Combined solution: $$-\frac{1}{2} < x < \frac{1}{2}$$ (non-empty solution set) **(b) Solve** $-2x + 5 < -3x - 3$ and $-4x + 9 \geq -5x + 2$ - First inequality: $$-2x + 5 < -3x - 3$$ $$-2x + 5 + 3x + 3 < 0$$ $$x + 8 < 0$$ $$x < -8$$ - Second inequality: $$-4x + 9 \geq -5x + 2$$ $$-4x + 9 + 5x - 2 \geq 0$$ $$x + 7 \geq 0$$ $$x \geq -7$$ - Combined solution: $$x < -8$$ and $$x \geq -7$$ - These two conditions cannot be true simultaneously, so **no solution**. **(c) Solve** $3(x - 1) > 4(x - 2)$ and $2(x + 2) \leq 5(x - 1)$ - First inequality: $$3x - 3 > 4x - 8$$ $$3x - 3 - 4x + 8 > 0$$ $$-x + 5 > 0$$ $$-x > -5$$ $$x < 5$$ - Second inequality: $$2x + 4 \leq 5x - 5$$ $$2x + 4 - 5x + 5 \leq 0$$ $$-3x + 9 \leq 0$$ $$-3x \leq -9$$ $$x \geq 3$$ - Combined solution: $$3 \leq x < 5$$ (non-empty solution set) **(d) Solve** $-2x + 5 \geq -3x - 2$ and $-3x - 2 > -2x + 4$ - First inequality: $$-2x + 5 \geq -3x - 2$$ $$-2x + 5 + 3x + 2 \geq 0$$ $$x + 7 \geq 0$$ $$x \geq -7$$ - Second inequality: $$-3x - 2 > -2x + 4$$ $$-3x - 2 + 2x - 4 > 0$$ $$-x - 6 > 0$$ $$-x > 6$$ $$x < -6$$ - Combined solution: $$x \geq -7$$ and $$x < -6$$ - This is the interval $$[-7, -6)$$ which is non-empty. 4. **Conclusion:** The only compound inequality with no solution is option (b). **Final answer:** The compound inequality $$-2x + 5 < -3x - 3$$ and $$-4x + 9 \geq -5x + 2$$ has no solution.