To solve the system of equations:

1. Equations given:
   • Equation 1: $-x - 3y = -6$
   • Equation 2: $y = -\frac{1}{3}x + 1$

Step 1: Rewrite the first equation in slope-intercept form ($y = mx + b$).

-x - 3y = -6 \quad \text{(Add xto both sides:)} \quad -3y = x - 6 \quad \text{(Divide by-3):} \quad y = -\frac{1}{3}x + 2

So, the first equation becomes: $y = -\frac{1}{3}x + 2$

Step 2: Compare slopes of the two lines.

The second equation is already in slope-intercept form: $y = -\frac{1}{3}x + 1$

Both equations have the same slope ($-\frac{1}{3}$) but different y-intercepts (2 and 1). This means the two lines are parallel.

Step 3: Conclusion

Since the lines are parallel and never intersect, there is no solution to this system of equations.

Would you like a graph to visualize this solution or any further explanation?

Here are 5 related questions to deepen your understanding:

1. How do you determine if two lines are parallel using their equations?
2. What happens when the slopes and y-intercepts of two lines are equal?
3. How do you find the point of intersection if two lines are not parallel?
4. Can two lines with different slopes ever be parallel?
5. What is the geometric interpretation of "no solution" in a system of linear equations?

Tip: Always rewrite equations into slope-intercept form ($y = mx + b$) for easier comparison and graphing.