A) \( \begin{aligned} 4 x+y & =-16 \\ x-y & =-1\end{aligned} \)

To solve the system of equations, we can use substitution or elimination. Here, let’s use substitution. From the second equation, \( x - y = -1 \) can be rearranged to find \( x = y - 1 \). Now, substitute \( x \) into the first equation: \( 4(y - 1) + y = -16 \) \( 4y - 4 + y = -16 \) \( 5y - 4 = -16 \) Adding 4 to both sides gives: \( 5y = -12 \) Thus, \( y = -\frac{12}{5} \). Now substitute \( y \) back into the equation for \( x \): \( x = -\frac{12}{5} - 1 = -\frac{12}{5} - \frac{5}{5} = -\frac{17}{5} \). So, the solution is \( x = -\frac{17}{5} \) and \( y = -\frac{12}{5} \).