We are given the inequality:

$15 \leq 3(a + 3) \leq 1$

This appears to involve two inequalities combined, so we will solve it step by step. First, let's break this into two parts:

Part 1: Solve $15 \leq 3(a + 3)$

1. Start by expanding the right-hand side: $15 \leq 3a + 9$

2. Subtract 9 from both sides: $6 \leq 3a$

3. Divide both sides by 3: $2 \leq a$ or equivalently: $a \geq 2$

Part 2: Solve $3(a + 3) \leq 1$

1. Expand the left-hand side: $3a + 9 \leq 1$

2. Subtract 9 from both sides: $3a \leq -8$

3. Divide both sides by 3: $a \leq -\frac{8}{3}$

Conclusion

Now, we combine the two results from both parts:

• From Part 1: $a \geq 2$
• From Part 2: $a \leq -\frac{8}{3}$

Since there is no value of $a$ that satisfies both conditions at the same time, the inequality has no solution.

Would you like more details or have any questions?

Related questions:

1. What happens if we change the constants in the inequality?
2. How would the solution change if we only had one inequality?
3. Can this inequality system have a graphical interpretation?
4. How do we solve compound inequalities in general?
5. What would the solution look like if we changed the inequality signs?

Tip:

When solving compound inequalities, always solve each part separately and then combine the results carefully to find the correct solution set.