https://file.aiquickdraw.com/mathbot/file/1726706046748qqj7ygk9.jpg

In the given image, lines \( l \), \( m \), and \( n \) are parallel, and we are tasked with finding the value of \( x^\circ \).

### Step-by-Step Solution:

1. **Understand the setup:**
   - We are told that lines \( l \parallel m \parallel n \), meaning these lines are parallel to each other.
   - There is a transversal cutting through these parallel lines. 
   - One of the angles formed by this transversal is 124°, and the angle labeled \( x^\circ \) is adjacent to it.

2. **Use the property of supplementary angles:**
   - When a transversal crosses parallel lines, consecutive interior angles formed on the same side of the transversal are **supplementary** (i.e., their sum is 180°).
   - Here, the angle \( 124^\circ \) and \( x^\circ \) form a linear pair (adjacent angles along a straight line), so they are supplementary.

3. **Set up the equation:**
   \[
   x^\circ + 124^\circ = 180^\circ
   \]

4. **Solve for \( x \):**
   \[
   x = 180^\circ - 124^\circ = 56^\circ
   \]

### Final Answer:
\[
x = 56^\circ
\]

Let me know if you'd like further clarification or have any questions.

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Here are five related questions for practice:

1. What is the sum of the interior angles on the same side of the transversal?
2. If another angle on the transversal is given as 56°, what is the corresponding alternate interior angle?
3. How would the problem change if the lines were not parallel?
4. What is the relationship between corresponding angles when a transversal cuts through parallel lines?
5. How do you identify supplementary angles in a diagram with parallel lines and a transversal?

**Tip:** Always look for linear pairs or alternate angles when dealing with parallel lines and transversals.

Given l || m || n, find the value of x where the angle adjacent to x is 124°.

Learn how to solve for angle x given parallel lines l, m, n, and a transversal that forms a 124° angle. This problem uses the supplementary angles theorem and is suitable for students in Grades 8-10.

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