To determine how many solutions the given system of equations has, we can analyze the equations to see if they represent lines that are parallel, intersecting, or identical.
The system of equations is:
- \(-2x = -6y - 2\)
- \(x = 12y - 15\)
Let's first rearrange these equations into the slope-intercept form \(y = mx + b\).
For the first equation:
\[ -2x = -6y - 2 \]
Add \(6y\) to both sides:
\[ -2x + 6y = -2 \]
Now, isolate \(y\):
\[ 6y = 2x - 2 \] \[ y = \frac{1}{3}x - \frac{1}{3} \]
So, the slope (\(m_1\)) of the first equation is \(\frac{1}{3}\).
For the second equation:
\[ x = 12y - 15 \]
Rearranging this equation gives:
\[ 12y = x + 15 \] \[ y = \frac{1}{12}x + \frac{15}{12} \]
So, the slope (\(m_2\)) of the second equation is \(\frac{1}{12}\).
Comparing the slopes:
- The slope of the first equation is \(\frac{1}{3}\).
- The slope of the second equation is \(\frac{1}{12}\).
Since \(\frac{1}{3} \neq \frac{1}{12}\), the lines are not parallel, meaning they will intersect at exactly one point.
Conclusion:
The system of equations has exactly one solution.