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To solve this problem, we have the compound inequality:

\[
2x + 3 \leq x + 10 \leq 3x + 0
\]

This inequality can be split into two separate inequalities:

1. \(2x + 3 \leq x + 10\)
2. \(x + 10 \leq 3x + 0\)

We will solve each of these inequalities for \(x\) individually and then determine the intersection of the solutions to find the closed interval \([A, B]\).

### Step 1: Solve \(2x + 3 \leq x + 10\)

1. Subtract \(x\) from both sides:
   \[
   2x - x + 3 \leq 10
   \]
   \[
   x + 3 \leq 10
   \]

2. Subtract 3 from both sides:
   \[
   x \leq 7
   \]

So, the solution to the first inequality is \(x \leq 7\).

### Step 2: Solve \(x + 10 \leq 3x + 0\)

1. Subtract \(x\) from both sides:
   \[
   10 \leq 2x
   \]

2. Divide both sides by 2:
   \[
   5 \leq x \quad \text{or} \quad x \geq 5
   \]

So, the solution to the second inequality is \(x \geq 5\).

### Step 3: Combine the Solutions

From the two inequalities, we have:
- \(x \leq 7\)
- \(x \geq 5\)

The intersection of these two conditions is:
\[
5 \leq x \leq 7
\]

### Conclusion

The closed interval \([A, B]\) is \([5, 7]\), so:
- \(A = 5\)
- \(B = 7\)

Would you like more details or have any questions?

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**Related Questions:**

1. What would the interval be if the inequalities had strict inequalities (e.g., \(2x + 3 < x + 10\))?
2. How would the solution change if the inequalities were flipped in direction?
3. What happens if there’s no overlap between the two inequalities?
4. How can you verify the solution by testing values within the interval?
5. How can this type of problem be represented on a number line?

**Tip:** When dealing with compound inequalities, always solve each part separately before combining results.

The inequality 2x + 3 ≤ x + 10 ≤ 3x + 0 means that x is in the closed interval [A, B] where A is and B is.

Learn to solve the compound inequality 2x + 3 ≤ x + 10 ≤ 3x + 0. This problem involves solving each inequality separately, finding the intersection, and representing the solution in interval notation. Suitable for Grades 9-11.

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