If WZ has collinear points W, X, Y, and Z in that order.

WX = YZ, XY = (4x - 2) cm, YZ = (2x - 1) cm, and WZ = 20 cm, then x =    cm, WX =    cm, and XY =    cm.

We are given the following:

- The points \( W \), \( X \), \( Y \), and \( Z \) are collinear, meaning they lie on the same straight line in that order.
- The distances between the points are given as:
  - \( WX = YZ \)
  - \( XY = 4x - 2 \, \text{cm} \)
  - \( YZ = 2x - 1 \, \text{cm} \)
  - \( WZ = 20 \, \text{cm} \)

### Step 1: Use the condition \( WZ = WX + XY + YZ \)
Since the points are collinear, the total distance \( WZ \) is the sum of the individual segments:

\[
WZ = WX + XY + YZ
\]

Substitute the expressions for \( WX \), \( XY \), and \( YZ \):

\[
20 = WX + (4x - 2) + (2x - 1)
\]

But since \( WX = YZ \), we replace \( WX \) with \( 2x - 1 \):

\[
20 = (2x - 1) + (4x - 2) + (2x - 1)
\]

### Step 2: Simplify the equation
Combine like terms:

\[
20 = 2x - 1 + 4x - 2 + 2x - 1
\]
\[
20 = 8x - 4
\]

### Step 3: Solve for \( x \)
Add 4 to both sides of the equation:

\[
24 = 8x
\]

Now divide by 8:

\[
x = 3
\]

### Step 4: Find the lengths of \( WX \), \( XY \), and \( YZ \)
Now that we know \( x = 3 \), we can substitute this value into the expressions for \( WX \), \( XY \), and \( YZ \).

- \( YZ = 2x - 1 = 2(3) - 1 = 6 - 1 = 5 \, \text{cm} \)
- \( WX = YZ = 5 \, \text{cm} \) (since \( WX = YZ \))
- \( XY = 4x - 2 = 4(3) - 2 = 12 - 2 = 10 \, \text{cm} \)

### Final answers:
- \( x = 3 \, \text{cm} \)
- \( WX = 5 \, \text{cm} \)
- \( XY = 10 \, \text{cm} \)

Would you like more details or have any questions about this solution?

Here are 5 related questions you might explore:
1. What happens if \( WX \neq YZ \)?
2. How would the result change if \( WZ \) were different?
3. Can you find \( XZ \) if the points were not collinear?
4. What if the expression for \( XY \) were quadratic instead of linear?
5. How would the problem change if the points were on a circle rather than a straight line?

**Tip:** When solving for variables, always simplify the equation before trying to isolate the variable to reduce errors.

Learn how to solve a geometry problem involving collinear points W, X, Y, and Z, with distances given as linear equations. The problem involves solving for x using the equations WX = YZ, XY = (4x - 2) cm, and YZ = (2x - 1) cm, with WZ = 20 cm. Suitable for students in Grades 7-9.

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