You are given a system of inequalities:

1. $-12 < x - 9$
2. $-10 \geq x - 9$

Let's solve these inequalities step by step.

First inequality: $-12 < x - 9$

To solve for $x$, add 9 to both sides:

$-12 + 9 < x$ $-3 < x$

This simplifies to:

$x > -3$

Second inequality: $-10 \geq x - 9$

Again, add 9 to both sides to solve for $x$:

$-10 + 9 \geq x$ $-1 \geq x$

This simplifies to:

$x \leq -1$

Combine the results:

From the first inequality, we know $x > -3$, and from the second inequality, we know $x \leq -1$. So, the solution is:

$-3 < x \leq -1$

Final answer:

The solution to the system of inequalities is $x \in (-3, -1]$.

Would you like any more details or further explanations?

Here are five questions to extend your learning:

1. How do we graph the solution set on a number line?
2. Can you solve a system of inequalities involving absolute values?
3. How do compound inequalities differ from systems of inequalities?
4. What happens when inequalities have different signs or directions?
5. How can you verify a solution by substitution?

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