\( \begin{aligned} x+3 y & =1 \\ 2 x+3 y & =5 \\ \text { 3) } \quad x-3 y & =-7\end{aligned} \)

To solve the system of equations, we can start by simplifying the first two equations. First, we'll subtract the first equation from the second: \[ (2x + 3y) - (x + 3y) = 5 - 1 \] This simplifies to: \[ x = 4 \] Now, we can substitute \(x = 4\) back into the first equation: \[ 4 + 3y = 1 \] Solving for \(y\): \[ 3y = 1 - 4 \\ 3y = -3 \\ y = -1 \] So we have \(x = 4\) and \(y = -1\). For the third equation \(x - 3y = -7\), we can substitute \(x = 4\) and \(y = -1\): \[ 4 - 3(-1) = -7 \\ 4 + 3 = 7 \neq -7 \] This means that the first two equations are consistent, but they disagree with the third. Therefore, the system has no solution.