The inequality given is:

$20 > a + 9 \geq 15$

To solve for $a$, let's break this down into two separate inequalities:

First inequality:

$20 > a + 9$

Subtract 9 from both sides: $20 - 9 > a$ $11 > a \quad \text{or} \quad a < 11$

Second inequality:

$a + 9 \geq 15$

Subtract 9 from both sides: $a \geq 15 - 9$ $a \geq 6$

Final compound inequality:

$6 \leq a < 11$

So, the solution for $a$ is the compound inequality:

$6 \leq a < 11$

Would you like further details or have any questions?

Here are 5 related questions to explore:

1. How do you solve compound inequalities involving "or" conditions?
2. What are the key differences between strict inequalities and inclusive inequalities?
3. Can you solve a compound inequality where both inequalities are reversed?
4. How does solving a system of inequalities graphically work?
5. What happens if we multiply or divide both sides of an inequality by a negative number?

Tip: Always reverse the inequality sign when multiplying or dividing both sides by a negative number.