Select atl the palrs of \( z \) and \( y \) that are solutions to the system of inequalities: \( \left\{\begin{array}{l}y>3 x+2 \\ 2 x+6 y \leq 12\end{array}\right. \)

To solve the system of inequalities, let's analyze each one: 1. For the inequality \( y > 3x + 2 \): - This represents a region above the line \( y = 3x + 2 \). The line has a slope of 3 and a y-intercept of 2. 2. For the inequality \( 2x + 6y \leq 12 \): - First, simplify it to find its boundary: \( 6y \leq -2x + 12 \) or \( y \leq -\frac{1}{3}x + 2 \). This line has a slope of -1/3 and a y-intercept of 2. The region is below or on this line. Now, to find the solution set, we can graph both lines: - For \( y = 3x + 2 \), plot the point (0, 2) for the y-intercept and use the slope to find another point. For \( x = 1 \), \( y = 5 \); hence, two points are (0, 2) and (1, 5). - For \( y = -\frac{1}{3}x + 2 \), begin at (0, 2) and for \( x = 3 \), \( y = 1 \) gives the points (0, 2) and (3, 1). Next, we determine the intersection point of these two lines: Set \( 3x + 2 = -\frac{1}{3}x + 2 \): Combine and solve: \[ 3x + \frac{1}{3}x = 0 \quad \Rightarrow \quad \frac{10}{3}x = 0 \quad \Rightarrow \quad x = 0. \] Substituting \( x = 0 \) back into the first equation, \( y = 3(0) + 2 = 2 \). Therefore, the intersection point is \( (0, 2) \). Now, we need to find out which regions satisfy both inequalities: 1. Points above the line \( y = 3x + 2 \) (greater than inequality) and below the line \( y = -\frac{1}{3}x + 2 \) (less than or equal to inequality). The solutions are therefore all points in the region above \( y = 3x + 2 \) and below \( y = -\frac{1}{3}x + 2 \). In conclusion, the pairs \( (x, y) \) that satisfy both inequalities can be represented graphically, leading to a region that can be characterized by the inequality system itself.