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Let's solve each of the problems from Exercises 5 to 8 based on the provided conditions.

### 5. Passing through (-8, -10) and parallel to the line whose equation is \( y = -4x + 3 \)
- **Step 1: Identify the slope of the given line**: The given line is \( y = -4x + 3 \), so its slope \( m \) is -4.
- **Step 2: Use the point-slope form equation**:
  \[
  y - y_1 = m(x - x_1)
  \]
  Substituting \( m = -4 \), \( x_1 = -8 \), and \( y_1 = -10 \):
  \[
  y - (-10) = -4(x - (-8))
  \]
  Simplifying:
  \[
  y + 10 = -4(x + 8)
  \]
  Expanding:
  \[
  y + 10 = -4x - 32
  \]
  Solving for \( y \):
  \[
  y = -4x - 42
  \]
  This is the slope-intercept form.

- **Point-slope form**: \( y + 10 = -4(x + 8) \)
- **Slope-intercept form**: \( y = -4x - 42 \)

---

### 6. Passing through (-8, -7) and parallel to the line whose equation is \( y = -4x + 4 \)
- **Step 1: Identify the slope of the given line**: The slope is \( m = -4 \).
- **Step 2: Use the point-slope form equation**:
  \[
  y - (-7) = -4(x - (-8))
  \]
  Simplifying:
  \[
  y + 7 = -4(x + 8)
  \]
  Expanding:
  \[
  y + 7 = -4x - 32
  \]
  Solving for \( y \):
  \[
  y = -4x - 39
  \]
  This is the slope-intercept form.

- **Point-slope form**: \( y + 7 = -4(x + 8) \)
- **Slope-intercept form**: \( y = -4x - 39 \)

---

### 7. Passing through (2, -3) and perpendicular to the line whose equation is \( y = \frac{1}{5}x + 6 \)
- **Step 1: Find the slope of the perpendicular line**: The slope of the given line is \( m = \frac{1}{5} \). For perpendicular lines, the slope is the negative reciprocal. So, the new slope \( m = -5 \).
- **Step 2: Use the point-slope form**:
  \[
  y - (-3) = -5(x - 2)
  \]
  Simplifying:
  \[
  y + 3 = -5(x - 2)
  \]
  Expanding:
  \[
  y + 3 = -5x + 10
  \]
  Solving for \( y \):
  \[
  y = -5x + 7
  \]
  This is the slope-intercept form.

- **Point-slope form**: \( y + 3 = -5(x - 2) \)
- **Slope-intercept form**: \( y = -5x + 7 \)

---

### 8. Passing through (-4, 2) and perpendicular to the line whose equation is \( y = \frac{3}{5}x + 7 \)
- **Step 1: Find the slope of the perpendicular line**: The slope of the given line is \( m = \frac{3}{5} \). The slope of the perpendicular line is the negative reciprocal, so \( m = -\frac{5}{3} \).
- **Step 2: Use the point-slope form**:
  \[
  y - 2 = -\frac{5}{3}(x - (-4))
  \]
  Simplifying:
  \[
  y - 2 = -\frac{5}{3}(x + 4)
  \]
  Expanding:
  \[
  y - 2 = -\frac{5}{3}x - \frac{20}{3}
  \]
  Adding 2 (which is \( \frac{6}{3} \)) to both sides:
  \[
  y = -\frac{5}{3}x - \frac{20}{3} + \frac{6}{3}
  \]
  Simplifying:
  \[
  y = -\frac{5}{3}x - \frac{14}{3}
  \]
  This is the slope-intercept form.

- **Point-slope form**: \( y - 2 = -\frac{5}{3}(x + 4) \)
- **Slope-intercept form**: \( y = -\frac{5}{3}x - \frac{14}{3} \)

---

Let me know if you need more details or clarifications on any step! 

### Related Questions:
1. How do you find the slope of a line perpendicular to another line?
2. What is the general process to convert from point-slope form to slope-intercept form?
3. How do you find the equation of a line passing through a point and parallel to another line?
4. Why do you use the negative reciprocal of the slope for perpendicular lines?
5. What is the relationship between slope and the steepness of a line?

### Tip: 
Always ensure that you simplify the final form of the equation, especially when working with fractions in slopes.

In Exercises 5–8, use the given conditions to write an equation for each line in point-slope form and slope-intercept form. 5. Passing through (-8, -10) and parallel to the line whose equation is y = -4x + 3. 6. Passing through (-8, -7) and parallel to the line whose equation is y = -4x + 4. 7. Passing through (2, -3) and perpendicular to the line whose equation is y = 1/5x + 6. 8. Passing through (-4, 2) and perpendicular to the line whose equation is y = 3/5x + 7.

Learn how to write equations of lines passing through given points, parallel or perpendicular to other lines. This problem set involves point-slope and slope-intercept forms of linear equations, covering essential algebraic concepts for Grades 9-12.

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