Question 6 of 10 Solve the system of equations below. \[ \begin{aligned} 4 x+y=16 \\ 2 x+3 y=-2\end{aligned} \] \[ \begin{array}{l}\text { A. }(5,-4) \\ \text { B. }(4,-5) \\ \text { c. }(5,4) \\ \text { D. }(-5,4)\end{array} \]

To solve the system of equations, we will use substitution or elimination. Let's start by solving for \(y\) from the first equation: \[ y = 16 - 4x \] Now we can substitute this expression for \(y\) into the second equation: \[ 2x + 3(16 - 4x) = -2 \] Distributing the 3 gives: \[ 2x + 48 - 12x = -2 \] Combining like terms: \[ -10x + 48 = -2 \] Next, we'll isolate \(x\): \[ -10x = -2 - 48 \] \[ -10x = -50 \] \[ x = 5 \] Now that we have \(x\), we can substitute it back into the first equation to find \(y\): \[ 4(5) + y = 16 \] \[ 20 + y = 16 \] \[ y = 16 - 20 \] \[ y = -4 \] Thus, the solution to the system of equations is \((5, -4)\), which corresponds to option A.