https://file.aiquickdraw.com/mathbot/file/1728069863945j2hqmniw.jpg

The image you provided shows two parallel lines cut by a transversal, with two corresponding angles marked. One angle is labeled \(13x + 3\) and the other angle is labeled \(14x - 6\).

Since the angles are corresponding angles (angles in the same position formed by the transversal and parallel lines), they are congruent. Therefore, we can set up an equation to solve for \(x\):

\[
13x + 3 = 14x - 6
\]

### Step-by-step solution:
1. Subtract \(13x\) from both sides:
\[
3 = 14x - 13x - 6
\]
This simplifies to:
\[
3 = x - 6
\]

2. Add 6 to both sides to isolate \(x\):
\[
3 + 6 = x
\]
\[
x = 9
\]

Thus, \(x = 9\).

### Verification:
Substitute \(x = 9\) back into the expressions for both angles:
- The first angle: \(13x + 3 = 13(9) + 3 = 117 + 3 = 120^\circ\)
- The second angle: \(14x - 6 = 14(9) - 6 = 126 - 6 = 120^\circ\)

Both angles are \(120^\circ\), confirming that the solution is correct.

Would you like any further details on this solution?

Here are five related questions for further exploration:
1. What are alternate interior angles, and how do they compare to corresponding angles?
2. How can you solve for the value of \(x\) if the two angles were supplementary instead of congruent?
3. What other angle relationships can you explore when parallel lines are cut by a transversal?
4. What are the properties of vertically opposite angles?
5. How would the equation change if the angles given were exterior angles?

**Tip:** When working with parallel lines and a transversal, always look for angle relationships like corresponding, alternate interior, or alternate exterior to set up equations.

Solve for x in the given geometry problem where two parallel lines are cut by a transversal. The angles are expressed as 13x + 3 and 14x - 6.

Learn how to solve for x in a geometry problem involving parallel lines cut by a transversal. The angles are expressed as 13x + 3 and 14x - 6. Using algebra and the Corresponding Angles Postulate, this solution is ideal for students in Grades 7-10.

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