{31 – [17 x (23 – 20) – 150 ÷ 6] + 9 x 2} ÷ 23


To solve the expression \(\{31 – [17 \times (23 – 20) – 150 ÷ 6] + 9 \times 2\} ÷ 23\), we need to follow the order of operations, commonly known as PEMDAS (Parentheses, Exponents, Multiplication and Division (left to right), Addition and Subtraction (left to right)).

1. **Parentheses**: Start with the innermost parentheses.
   \[
   (23 – 20) = 3
   \]
   Substitute back into the expression:
   \[
   31 – [17 \times 3 – 150 ÷ 6] + 9 \times 2
   \]

2. **Division inside brackets**:
   \[
   150 ÷ 6 = 25
   \]
   Substitute back into the expression:
   \[
   31 – [17 \times 3 – 25] + 9 \times 2
   \]

3. **Multiplication inside brackets**:
   \[
   17 \times 3 = 51
   \]
   Substitute back into the expression:
   \[
   31 – [51 – 25] + 9 \times 2
   \]

4. **Subtraction inside brackets**:
   \[
   51 – 25 = 26
   \]
   Substitute back into the expression:
   \[
   31 – 26 + 9 \times 2
   \]

5. **Multiplication**:
   \[
   9 \times 2 = 18
   \]
   Substitute back into the expression:
   \[
   31 – 26 + 18
   \]

6. **Subtraction and Addition**:
   \[
   31 – 26 = 5
   \]
   \[
   5 + 18 = 23
   \]

7. **Final Division**:
   \[
   23 ÷ 23 = 1
   \]

So, the value of the expression is \(1\).

Would you like more details or have any questions?

Here are 5 questions you can ask:
1. How do you solve equations with multiple parentheses?
2. Can you explain the order of operations in more detail?
3. What are some common mistakes to avoid in solving complex expressions?
4. How do you handle exponents in order of operations?
5. Can you show another example of a similar problem?

Tip: Always double-check each step when solving complex expressions to ensure accuracy.

Learn how to solve the expression {31 – [17 x (23 – 20) – 150 ÷ 6] + 9 x 2} ÷ 23 with a detailed step-by-step solution. This problem covers the order of operations (PEMDAS), including parentheses, multiplication, division, addition, and subtraction. Suitable for students in Grades 6-8.

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