Question\begin{tabular}{c|l} \hline System A & The system has no solution. \\ \begin{tabular}{c} $-x-4 y=-8$ \\ $x+4 y=8$ \end{tabular} & \begin{tabular}{l} $(x, y)=(\square, \square)$ \end{tabular} \\ The system has infinitely many solutions. \\ They must satisfy the following equation: \\ $y=\square$ \end{tabular}

Studdy Solution
Since the equations are dependent, the system has infinitely many solutions. The solutions must satisfy the equation of the line represented by either equation.
Let's solve one of the equations for $y$:
Using Equation 2: $x + 4y = 8$
Rearrange to solve for $y$:
$4y = 8 - x$
$y = \frac{8 - x}{4}$
$y = 2 - \frac{x}{4}$
Thus, the solutions satisfy the equation:
$$y = 2 - \frac{x}{4}$$
The system has infinitely many solutions, and they satisfy the equation:
$$y = 2 - \frac{x}{4}$$