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The problem involves three parallel lines, \( l \parallel m \parallel n \), with angles formed by a transversal cutting across them. The goal is to find the value of \( x \) using the information provided about the angles.

The given angles are:
- \( (9x - 6)^\circ \) at the top,
- \( (6x + 6)^\circ \) at the bottom.

Since \( l \parallel m \parallel n \), and a transversal intersects these parallel lines, the angles formed are either corresponding, alternate interior, or co-interior (same-side interior) angles. In this case, the two angles provided are alternate interior angles. By the property of alternate interior angles, they must be equal.

Thus, we can set up the equation:
\[
(9x - 6) = (6x + 6)
\]

### Solving for \( x \):
1. Start with the equation:
   \[
   9x - 6 = 6x + 6
   \]
2. Subtract \( 6x \) from both sides:
   \[
   3x - 6 = 6
   \]
3. Add 6 to both sides:
   \[
   3x = 12
   \]
4. Divide both sides by 3:
   \[
   x = 4
   \]

So, the value of \( x \) is \( 4 \).

Would you like more details or have any further questions?

### Follow-up questions:
1. What are alternate interior angles, and how do they behave with parallel lines?
2. How do corresponding angles relate to each other in a parallel lines scenario?
3. Could the problem have been solved using a different angle relationship?
4. What would happen if the lines weren't parallel—how would that affect the angles?
5. Can you verify the solution by substituting \( x = 4 \) into the given expressions?

### Tip:
When working with angles formed by parallel lines and a transversal, remember that alternate interior angles are always equal, while co-interior angles sum to 180°.

Given l || m || n, find the value of x. The given angles are (9x - 6)° and (6x + 6)° formed by a transversal cutting across three parallel lines.

This problem involves three parallel lines intersected by a transversal, forming angles expressed as (9x - 6)° and (6x + 6)°. Learn how to use the Alternate Interior Angles Theorem and algebraic methods to solve for x. Suitable for students in Grades 8-10.

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