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To solve this problem, we need to find an equation of a line that is parallel to \( y = 5x + 2 \) and passes through the point \( (-6, -1) \).

### Step-by-Step Solution:
1. **Identify the slope of the given line**: The slope-intercept form of a line is \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. From the given equation \( y = 5x + 2 \), the slope \( m \) is 5.

2. **Use the fact that parallel lines have the same slope**: Since we are looking for a line that is parallel to \( y = 5x + 2 \), the slope of the new line must also be \( m = 5 \).

3. **Substitute the slope into the slope-intercept form**: We now have the equation in the form \( y = 5x + b \). The only unknown left is the y-intercept \( b \).

4. **Use the point (-6, -1) to solve for \( b \)**: 
   - Substitute \( x = -6 \) and \( y = -1 \) into the equation:
   \[
   -1 = 5(-6) + b
   \]
   \[
   -1 = -30 + b
   \]
   \[
   b = 29
   \]

5. **Write the equation of the line**: Now that we have \( b = 29 \), the equation of the line is:
   \[
   y = 5x + 29
   \]

### Conclusion:
The correct answer is **C. \( y = 5x + 29 \)**.

Would you like more details on any step? Here are 5 questions to expand your understanding:

1. Why do parallel lines have the same slope?
2. How would you find the equation if the line were perpendicular instead of parallel?
3. How can we use point-slope form to find the equation of a line?
4. What happens if the line passed through a different point, like \( (-2, 3) \)?
5. How would you graph this equation to verify its correctness?

**Tip**: Always double-check your work by substituting the point back into the final equation to ensure it's correct!

Identify an equation in slope-intercept form for the line parallel to y = 5x + 2 that passes through (-6, -1).

Solve a problem involving finding the equation of a line parallel to y = 5x + 2 that passes through the point (-6, -1). Learn how to use slope-intercept form and the concept of parallel lines in this step-by-step solution. Suitable for students in Grades 8-10.

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