a) \( \left\{\begin{array}{c}2 x+3 y=-1 \\ 4 x-y=3\end{array}\right. \)

To solve the system of equations, we can use the substitution or elimination method. Let’s go with the elimination method! First, we can multiply the first equation by 2 to align the coefficients of \(x\): \[ \begin{cases} 4x + 6y = -2 \\ 4x - y = 3 \end{cases} \] Now, we can subtract the second equation from the first to eliminate \(x\): \[ (4x + 6y) - (4x - y) = -2 - 3 \] Which simplifies to: \[ 7y = -5 \quad \Rightarrow \quad y = -\frac{5}{7} \] Now substitute \(y\) back into one of the original equations to find \(x\). Let's use the first equation: \[ 2x + 3\left(-\frac{5}{7}\right) = -1 \] This gives us: \[ 2x - \frac{15}{7} = -1 \quad \Rightarrow \quad 2x = -1 + \frac{15}{7} = \frac{-7 + 15}{7} = \frac{8}{7} \] Thus, \[ x = \frac{8}{14} = \frac{4}{7}. \] So the solution is: \[ \left(x, y\right) = \left(\frac{4}{7}, -\frac{5}{7}\right). \]