https://file.aiquickdraw.com/mathbot/file/1726239441027krxk3bfb.jpg

Let's solve the equations from the image.

### Problem 7:
\[
|x + 6| = 3x - 8
\]

#### Case 1: \( x + 6 = 3x - 8 \)
\[
x + 6 = 3x - 8 \implies 6 + 8 = 3x - x \implies 14 = 2x \implies x = 7
\]

#### Case 2: \( -(x + 6) = 3x - 8 \)
\[
-x - 6 = 3x - 8 \implies -6 + 8 = 3x + x \implies 2 = 4x \implies x = \frac{1}{2}
\]

#### Check for extraneous solutions:
For \(x = 7\):
\[
|7 + 6| = 3(7) - 8 \implies 13 = 13 \quad \text{(valid)}
\]

For \(x = \frac{1}{2}\):
\[
\left|\frac{1}{2} + 6\right| = 3\left(\frac{1}{2}\right) - 8 \implies \left|\frac{13}{2}\right| \neq \frac{-13}{2} \quad \text{(invalid)}
\]

Thus, the solution is \( x = 7 \).

---

### Problem 8:
\[
|2m - 3| = 4m - 1
\]

#### Case 1: \( 2m - 3 = 4m - 1 \)
\[
2m - 3 = 4m - 1 \implies -3 + 1 = 4m - 2m \implies -2 = 2m \implies m = -1
\]

#### Case 2: \( -(2m - 3) = 4m - 1 \)
\[
-2m + 3 = 4m - 1 \implies 3 + 1 = 4m + 2m \implies 4 = 6m \implies m = \frac{2}{3}
\]

#### Check for extraneous solutions:
For \(m = -1\):
\[
|2(-1) - 3| = 4(-1) - 1 \implies | -5 | = -5 \quad \text{(valid)}
\]

For \(m = \frac{2}{3}\):
\[
|2\left(\frac{2}{3}\right) - 3| = 4\left(\frac{2}{3}\right) - 1 \implies | \frac{4}{3} - 3| = \frac{8}{3} - 1 \implies | \frac{-5}{3} | \neq \frac{5}{3} \quad \text{(invalid)}
\]

Thus, the solution is \( m = -1 \).

---

### Problem 9:
\[
|2x + 5| = 3x + 4
\]

#### Case 1: \( 2x + 5 = 3x + 4 \)
\[
2x + 5 = 3x + 4 \implies 5 - 4 = 3x - 2x \implies x = 1
\]

#### Case 2: \( -(2x + 5) = 3x + 4 \)
\[
-2x - 5 = 3x + 4 \implies -5 - 4 = 3x + 2x \implies -9 = 5x \implies x = -\frac{9}{5}
\]

#### Check for extraneous solutions:
For \(x = 1\):
\[
|2(1) + 5| = 3(1) + 4 \implies |7| = 7 \quad \text{(valid)}
\]

For \(x = -\frac{9}{5}\):
\[
|2\left(-\frac{9}{5}\right) + 5| = 3\left(-\frac{9}{5}\right) + 4 \implies \left|\frac{-18}{5} + \frac{25}{5}\right| = \frac{-27}{5} + \frac{20}{5} \implies \left|\frac{7}{5}\right| \neq \frac{-7}{5} \quad \text{(invalid)}
\]

Thus, the solution is \( x = 1 \).

---

### Problem 10:
\[
|3m - 2| = 2m - 3
\]

#### Case 1: \( 3m - 2 = 2m - 3 \)
\[
3m - 2 = 2m - 3 \implies -2 + 3 = 2m - 3m \implies 1 = -m \implies m = -1
\]

#### Case 2: \( -(3m - 2) = 2m - 3 \)
\[
-3m + 2 = 2m - 3 \implies 2 + 3 = 2m + 3m \implies 5 = 5m \implies m = 1
\]

#### Check for extraneous solutions:
For \(m = -1\):
\[
|3(-1) - 2| = 2(-1) - 3 \implies |-5| = -5 \quad \text{(valid)}
\]

For \(m = 1\):
\[
|3(1) - 2| = 2(1) - 3 \implies |1| \neq -1 \quad \text{(invalid)}
\]

Thus, the solution is \( m = -1 \).

---

### Final Solutions:
7. \( x = 7 \)  
8. \( m = -1 \)  
9. \( x = 1 \)  
10. \( m = -1 \)

Would you like further explanations or details on any specific step?

### 5 Questions for Further Practice:
1. How do you solve equations involving absolute values with inequalities?
2. What is the difference between extraneous solutions and valid solutions?
3. How can graphing help visualize solutions to absolute value equations?
4. How do you handle absolute value equations when both sides are complex expressions?
5. What strategies help in identifying extraneous solutions early?

### Tip:
Always check both the positive and negative cases in absolute value equations to avoid missing any potential solutions.

Solve the absolute value equations: |x + 6| = 3x - 8, |2m - 3| = 4m - 1, |2x + 5| = 3x + 4, |3m - 2| = 2m - 3.

Learn how to solve absolute value equations such as |x + 6| = 3x - 8 and |2m - 3| = 4m - 1 with detailed step-by-step solutions. Understand the key concepts of solving absolute value and linear equations, and check for extraneous solutions. Suitable for students in Grades 8-10.

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