\( 2 \left\{ \begin{array} { l } { x + 3 y = 5 } \\ { 2 x - y = - 4 } \end{array} \right. \)

To solve the system of equations given by \[ \begin{cases} x + 3y = 5 \quad (1) \\ 2x - y = -4 \quad (2) \end{cases} \] we can use the substitution or elimination method. Here, I'll use the substitution method. First, we can solve equation (1) for \(x\): \[ x = 5 - 3y \quad (3) \] Next, we substitute equation (3) into equation (2): \[ 2(5 - 3y) - y = -4 \] Expanding this gives: \[ 10 - 6y - y = -4 \] Combining like terms: \[ 10 - 7y = -4 \] Now, we isolate \(y\): \[ -7y = -4 - 10 \] \[ -7y = -14 \] \[ y = 2 \] Now that we have \(y\), we can substitute it back into equation (3) to find \(x\): \[ x = 5 - 3(2) \] \[ x = 5 - 6 \] \[ x = -1 \] Thus, the solution to the system of equations is: \[ (x, y) = (-1, 2) \] To verify, we can substitute \(x = -1\) and \(y = 2\) back into the original equations: 1. For \(x + 3y = 5\): \[ -1 + 3(2) = -1 + 6 = 5 \quad \text{(True)} \] 2. For \(2x - y = -4\): \[ 2(-1) - 2 = -2 - 2 = -4 \quad \text{(True)} \] Both equations are satisfied, confirming that the solution is correct. The final answer is: \[ \boxed{(-1, 2)} \]