The inequality provided is:

$2x + 9 \leq x + 3 \leq 3x + 19$

We need to solve this inequality step by step.

Step 1: Break down the inequality into two parts.

1. $2x + 9 \leq x + 3$
2. $x + 3 \leq 3x + 19$

Solving the first inequality:

$2x + 9 \leq x + 3$ Subtract $x$ from both sides: $x + 9 \leq 3$ Now subtract 9 from both sides: $x \leq -6$ This gives $x \leq -6$.

Solving the second inequality:

$x + 3 \leq 3x + 19$ Subtract $x$ from both sides: $3 \leq 2x + 19$ Now subtract 19 from both sides: $-16 \leq 2x$ Divide by 2: $-8 \leq x$ This gives $x \geq -8$.

Step 2: Combine the two parts:

The solution to the inequality is: $-8 \leq x \leq -6$

Step 3: Closed interval $[A, B]$:

In interval notation, $A = -8$ and $B = -6$.

Let me know if you need any further details or explanations!

Follow-up Questions:

1. How do you solve compound inequalities in general?
2. Can you explain why the solution is a closed interval in this case?
3. What is the significance of combining two inequalities like this?
4. What happens if you encounter a "greater than" rather than a "less than" inequality?
5. How can you graph the solution to this inequality on a number line?

Tip: Always isolate the variable step by step when dealing with compound inequalities to avoid mistakes.