1. 2x + 11 2. (2z + y) -5 3. |(y + z) - x| x is 11 y is 0 z is 22

Let's solve each expression using the values \( x = 11 \), \( y = 0 \), and \( z = 22 \).

1. **For \( 2x + 11 \)**:
   \[
   2(11) + 11 = 22 + 11 = 33
   \]

2. **For \( (2z + y) - 5 \)**:
   \[
   (2(22) + 0) - 5 = (44 + 0) - 5 = 44 - 5 = 39
   \]

3. **For \( |(y + z) - x| \)**:
   \[
   |(0 + 22) - 11| = |22 - 11| = |11| = 11
   \]

### Summary of Results:
1. \( 2x + 11 = 33 \)
2. \( (2z + y) - 5 = 39 \)
3. \( |(y + z) - x| = 11 \)

Would you like any further details or explanations? Here are some related questions:

1. How do you simplify expressions with absolute values?
2. What are some common applications of linear equations?
3. Can you explain the significance of variable substitution in algebra?
4. What are the properties of absolute values?
5. How would these calculations change if the values of \( x \), \( y \), and \( z \) were different?

**Tip:** Always double-check your values before substituting them into equations to avoid errors.

Explore the solutions for the expressions 2x + 11, (2z + y) - 5, and |(y + z) - x| using x = 11, y = 0, and z = 22. This guide covers key algebraic concepts and theorems, suitable for students in Grades 7-9.

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