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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.

The inequality 2x + 9 ≤ x + 3 ≤ 3x + 19 means that x is in the closed interval [A, B] where A is -6 and B is.

Learn how to solve the compound inequality 2x + 9 ≤ x + 3 ≤ 3x + 19 with a detailed step-by-step solution. This problem covers key algebraic concepts, such as solving linear inequalities and interpreting results within a closed interval. Suitable for students in Grades 9-11.

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