To find the solution to the given system of equations, we need to first find the equation of the line that passes through the points (-3, 0) and (0, -6), and then solve the second equation \(3y + 30 = 6x\).
Step 1: Find the equation of the first line.
We can start by calculating the slope (m) of the line that passes through the points (-3, 0) and (0, -6):
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-6 - 0}{0 - (-3)} = \frac{-6}{3} = -2 \]
Now, using the slope-intercept form of a line \(y = mx + b\), we can substitute one of the points to find b. We can use the point (0, -6):
\[ y = -2x + b \implies -6 = -2(0) + b \implies b = -6 \]
So the equation of the first line is:
\[ y = -2x - 6 \]
Step 2: Simplify the second equation.
We start with the second equation:
\[ 3y + 30 = 6x \]
Let's solve for y. First, we can isolate \(3y\):
\[ 3y = 6x - 30 \]
Now, divide by 3:
\[ y = 2x - 10 \]
Step 3: Find the intersection of the two lines.
We now have the two equations:
- \(y = -2x - 6\)
- \(y = 2x - 10\)
To find their intersection, we set the two equations equal to each other:
\[ -2x - 6 = 2x - 10 \]
Now we can combine like terms:
\[ -2x - 2x = -10 + 6 \] \[ -4x = -4 \implies x = 1 \]
Step 4: Substitute x back to find y.
We can substitute \(x = 1\) into either equation to find \(y\). Using the second equation:
\[ y = 2(1) - 10 = 2 - 10 = -8 \]
Thus, the solution to the system of equations is:
\[ (1, -8) \]
Among the provided options, this corresponds to:
\[ \boxed{(1, -8)} \]