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Let's solve the problems from the image step by step.

### Problem 7:
We are given three equations:
- \( BD = 7x - 10 \)
- \( BC = 4x - 29 \)
- \( CD = 5x - 9 \)

Since \( BD = BC + CD \) (because \( B \), \( C \), and \( D \) are collinear points), we can set up the following equation:

\[
BD = BC + CD
\]

Substituting the given values:

\[
7x - 10 = (4x - 29) + (5x - 9)
\]

Now, let's solve for \( x \):

\[
7x - 10 = 4x - 29 + 5x - 9
\]

Combine like terms:

\[
7x - 10 = 9x - 38
\]

Subtract \( 9x \) from both sides:

\[
7x - 9x - 10 = -38
\]

\[
-2x - 10 = -38
\]

Add 10 to both sides:

\[
-2x = -28
\]

Divide by -2:

\[
x = 14
\]

Now that we have \( x = 14 \), let's find the values for \( BC \), \( CD \), and \( BD \):

1. \( BC = 4x - 29 = 4(14) - 29 = 56 - 29 = 27 \)
2. \( CD = 5x - 9 = 5(14) - 9 = 70 - 9 = 61 \)
3. \( BD = 7x - 10 = 7(14) - 10 = 98 - 10 = 88 \)

### Final answers for Problem 7:
- \( x = 14 \)
- \( BC = 27 \)
- \( CD = 61 \)
- \( BD = 88 \)

---

### Problem 8:
We are given the following conditions:
- \( BD \equiv BC \) (i.e., \( BD = BC \))
- \( BD = 5x - 26 \)
- \( BC = 2x + 1 \)
- \( AC = 43 \)

Since \( BD = BC \), we can set the two expressions equal:

\[
5x - 26 = 2x + 1
\]

Now, solve for \( x \):

\[
5x - 2x = 1 + 26
\]

\[
3x = 27
\]

\[
x = 9
\]

Now that we have \( x = 9 \), substitute this into either \( BD \) or \( BC \) (since they are equal):

\[
BD = 5(9) - 26 = 45 - 26 = 19
\]

Next, we know that \( AB + BC = AC \), so:

\[
AB + 19 = 43
\]

Solve for \( AB \):

\[
AB = 43 - 19 = 24
\]

### Final answer for Problem 8:
- \( AB = 24 \)

---

Would you like further clarification or more details on any of these solutions?

Here are five related questions to practice with:
1. If \( BD = 6x - 12 \) and \( BC = 3x - 8 \), and \( CD = 4x - 5 \), how would you solve for \( x \)?
2. What if \( BD \neq BC \)? How would that change the solution in problem 8?
3. How can you verify that a line segment is congruent to another?
4. Can you apply these steps to solve problems involving triangles instead of line segments?
5. What are some real-life applications of these geometric principles?

**Tip:** Always set up relationships between segments or angles before solving equations to ensure accuracy.

Solve for the unknowns in the two geometry problems involving line segments BD, BC, CD, and AC.

Learn how to solve for the lengths of line segments BD, BC, and CD using algebra and geometry principles. This problem covers algebraic manipulation and the Segment Addition Postulate. Suitable for students in Grades 8-10.

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