2. \( \left\{\begin{array}{l}-3 x+6 y=-18 \\ y=\frac{1}{2} x-3\end{array}\right. \)

To solve this system of equations, we can first simplify the first equation \(-3x + 6y = -18\). Dividing the entire equation by -3 gives us \(x - 2y = 6\) or rearranging it to \(x = 2y + 6\). Now, we can substitute the expression for \(y\) from the second equation \(y = \frac{1}{2}x - 3\) into the rearranged first equation. Setting the two equations equal gives us: \[ 2\left(\frac{1}{2}x - 3\right) + 6 = x \] This simplifies to \(x - 6 + 6 = x\), confirming that the second equation is a representation of points that lie along a line parallel to the first. To find a solution, we can choose any value for \(x\). For \(x = 6\): \[ y = \frac{1}{2}(6) - 3 = 0 \] So the point \( (6, 0) \) is a solution to the system. Therefore, the two equations represent the same line, indicating an infinite number of solutions along that line! But if you start with the original equations, you’ll discover this is a classic case of dependent equations!