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In this problem, we are given a diagram with two parallel lines \(QP\) and \(SR\) cut by a transversal line \(NO\), and we are asked to solve for \(x\).

The angles formed by the intersections of these lines are given as:
- \( \angle QTU = (6x + 12)^\circ \)
- \( \angle TUS = (10x - 8)^\circ \)

Since \( \angle QTU \) and \( \angle TUS \) are consecutive interior angles, we know that the sum of these two angles must be 180°, according to the consecutive interior angles theorem for parallel lines.

### Equation Setup:
\[
(6x + 12) + (10x - 8) = 180
\]

### Solving for \(x\):
1. Combine like terms:
\[
6x + 10x + 12 - 8 = 180
\]
\[
16x + 4 = 180
\]

2. Subtract 4 from both sides:
\[
16x = 176
\]

3. Divide by 16:
\[
x = \frac{176}{16} = 11
\]

Thus, \(x = 11\).

Would you like further clarification or more details on any step? Here are some follow-up questions:

1. What is the relationship between parallel lines and angles formed by transversals?
2. How would you verify the solution for \(x\) using the given angles?
3. What other angle pairs could be considered in such a diagram?
4. How does the consecutive interior angles theorem work?
5. Can you think of any real-life examples where this kind of geometry applies?

**Tip:** When working with transversals and parallel lines, remember the different angle relationships such as corresponding angles, alternate interior angles, and consecutive interior angles to solve problems more effectively.

Solve for x in the given diagram with parallel lines and a transversal, where ∠QTU = (6x + 12)° and ∠TUS = (10x - 8)°.

This problem involves finding the value of x in a diagram with two parallel lines cut by a transversal. Using the consecutive interior angles theorem, we set up an equation with ∠QTU = (6x + 12)° and ∠TUS = (10x - 8)°, and solve for x. Suitable for students in Grades 8-10.

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