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.