Question

which graph shows the solution to the system of linear inequalities?
y > \frac{2}{3}x + 3
y \leq -\frac{1}{3}x + 2

Explanation:

Step1: Analyze \( y > \frac{2}{3}x + 3 \)

The inequality \( y > \frac{2}{3}x + 3 \) has a slope of \( \frac{2}{3} \) and a y - intercept of 3. Since the inequality is "greater than", the line should be dashed (because the points on the line are not included in the solution set) and we shade the region above the line.

Step2: Analyze \( y \leq -\frac{1}{3}x + 2 \)

The inequality \( y \leq -\frac{1}{3}x + 2 \) has a slope of \( -\frac{1}{3} \) and a y - intercept of 2. Since the inequality is "less than or equal to", the line should be solid (because the points on the line are included in the solution set) and we shade the region below the line.

Step3: Find the intersection of the two regions

The solution to the system of inequalities is the region that is shaded by both inequalities. We need to check which graph has a dashed line for \( y=\frac{2}{3}x + 3 \) (shaded above) and a solid line for \( y = -\frac{1}{3}x+2 \) (shaded below), and the overlapping region of these two shadings.

(Note: Since the actual graphs are not fully visible in the text - based description, but the general method is as above. If we assume that the correct graph has a dashed line with slope \( \frac{2}{3} \), y - intercept 3 (shaded above) and a solid line with slope \( -\frac{1}{3} \), y - intercept 2 (shaded below), and the overlapping region of these two shadings.)

Answer:

(The answer would be the graph that has a dashed line \( y=\frac{2}{3}x + 3 \) (shaded above) and a solid line \( y = -\frac{1}{3}x + 2 \) (shaded below) with the overlapping region of the two shadings. Since the options are not fully provided in a text - based selectable format here, but the method to identify is as described above.)