Geometry worksheet with four problems involving parallel lines and transversals, requiring students to solve for x and find angle measures using algebra.

To solve the problems, we need to use the properties of angles formed by parallel lines and a transversal. Specifically, we will use the following angle relationships:

1. Corresponding Angles: Angles in matching corners are equal.
2. Alternate Interior Angles: Angles on opposite sides of the transversal and inside the parallel lines are equal.
3. Same-Side Interior Angles: Angles on the same side of the transversal and inside the parallel lines are supplementary (sum to 180°).
4. Vertical Angles: Opposite angles formed by intersecting lines are equal.

Let's solve each problem step by step.

---

Problem 1:

Given:
- \( \angle 2 = (4x + 25)^\circ \)
- \( \angle 6 = (8x - 15)^\circ \)

Since lines \( A \) and \( B \) are parallel, and \( C \) is a transversal, \( \angle 2 \) and \( \angle 6 \) are corresponding angles. Therefore:
\[
\angle 2 = \angle 6
\]
\[
4x + 25 = 8x - 15
\]

Solve for \( x \):
\[
4x + 25 = 8x - 15
\]
\[
25 + 15 = 8x - 4x
\]
\[
40 = 4x
\]
\[
x = 10
\]

Now, find \( \angle 7 \):
- \( \angle 7 \) is the vertical angle to \( \angle 6 \), so \( \angle 7 = \angle 6 \).
- Substitute \( x = 10 \) into \( \angle 6 \):
\[
\angle 6 = 8x - 15 = 8(10) - 15 = 80 - 15 = 65^\circ
\]
Thus, \( \angle 7 = 65^\circ \).

Answer:
\[
x = 10, \quad \text{Angle 7} = 65^\circ
\]

---

Problem 2:

Given:
- \( \angle 2 = (8x + 100)^\circ \)
- \( \angle 6 = (5x + 12)^\circ \)

Since lines \( A \) and \( B \) are parallel, and \( C \) is a transversal, \( \angle 2 \) and \( \angle 6 \) are corresponding angles. Therefore:
\[
\angle 2 = \angle 6
\]
\[
8x + 100 = 5x + 12
\]

Solve for \( x \):
\[
8x + 100 = 5x + 12
\]
\[
100 - 12 = 5x - 8x
\]
\[
88 = -3x
\]
\[
x = -\frac{88}{3}
\]

Now, find \( \angle 5 \):
- \( \angle 5 \) is the vertical angle to \( \angle 6 \), so \( \angle 5 = \angle 6 \).
- Substitute \( x = -\frac{88}{3} \) into \( \angle 6 \):
\[
\angle 6 = 5x + 12 = 5\left(-\frac{88}{3}\right) + 12 = -\frac{440}{3} + 12 = -\frac{440}{3} + \frac{36}{3} = -\frac{404}{3}^\circ
\]
Thus, \( \angle 5 = -\frac{404}{3}^\circ \).

Answer:
\[
x = -\frac{88}{3}, \quad \text{Angle 5} = -\frac{404}{3}^\circ
\]

---

Problem 3:

Given:
- \( \angle 2 = (2x + 12)^\circ \)
- \( \angle 4 = (6x + 104)^\circ \)

Since lines \( A \) and \( B \) are parallel, and \( C \) is a transversal, \( \angle 2 \) and \( \angle 4 \) are alternate interior angles. Therefore:
\[
\angle 2 = \angle 4
\]
\[
2x + 12 = 6x + 104
\]

Solve for \( x \):
\[
2x + 12 = 6x + 104
\]
\[
12 - 104 = 6x - 2x
\]
\[
-92 = 4x
\]
\[
x = -23
\]

Now, find \( \angle 1 \):
- \( \angle 1 \) is the vertical angle to \( \angle 2 \), so \( \angle 1 = \angle 2 \).
- Substitute \( x = -23 \) into \( \angle 2 \):
\[
\angle 2 = 2x + 12 = 2(-23) + 12 = -46 + 12 = -34^\circ
\]
Thus, \( \angle 1 = -34^\circ \).

Answer:
\[
x = -23, \quad \text{Angle 1} = -34^\circ
\]

---

Problem 4:

Given:
- \( \angle 2 = (9x - 21)^\circ \)
- \( \angle 6 = (7x - 5)^\circ \)

Since lines \( A \) and \( B \) are parallel, and \( C \) is a transversal, \( \angle 2 \) and \( \angle 6 \) are corresponding angles. Therefore:
\[
\angle 2 = \angle 6
\]
\[
9x - 21 = 7x - 5
\]

Solve for \( x \):
\[
9x - 21 = 7x - 5
\]
\[
9x - 7x = -5 + 21
\]
\[
2x = 16
\]
\[
x = 8
\]

Now, find \( \angle 3 \):
- \( \angle 3 \) is the vertical angle to \( \angle 2 \), so \( \angle 3 = \angle 2 \).
- Substitute \( x = 8 \) into \( \angle 2 \):
\[
\angle 2 = 9x - 21 = 9(8) - 21 = 72 - 21 = 51^\circ
\]
Thus, \( \angle 3 = 51^\circ \).

Answer:
\[
x = 8, \quad \text{Angle 3} = 51^\circ
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
&x = 10, \quad \text{Angle 7} = 65^\circ \\
&x = -\frac{88}{3}, \quad \text{Angle 5} = -\frac{404}{3}^\circ \\
&x = -23, \quad \text{Angle 1} = -34^\circ \\
&x = 8, \quad \text{Angle 3} = 51^\circ
\end{aligned}
}
\]