Question

which graph shows the solution to the system of linear inequalities?
y ≥ \\(\frac{2}{3}x + 3\\)
y < -\\(\frac{1}{3}x + 2\\)

Explanation:

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

The inequality \( y \geq \frac{2}{3}x + 3 \) has a solid line (because of \( \geq \)) with slope \( \frac{2}{3} \) and y - intercept 3. The region above the line (including the line) is shaded.

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

The inequality \( y < -\frac{1}{3}x + 2 \) has a dashed line (because of \( < \)) with slope \( -\frac{1}{3} \) and y - intercept 2. The region below the line (not including the line) is shaded.

Step3: Find the overlapping region

We need to find the graph where the shading from \( y \geq \frac{2}{3}x + 3 \) (above the solid line) and the shading from \( y < -\frac{1}{3}x + 2 \) (below the dashed line) overlap.

Looking at the four graphs:

• For the first inequality \( y \geq \frac{2}{3}x + 3 \), the solid line is \( y=\frac{2}{3}x + 3 \). For the second inequality \( y < -\frac{1}{3}x + 2 \), the dashed line is \( y = -\frac{1}{3}x+2 \).

The fourth graph (the last one in the given set) has the region above the solid line \( y=\frac{2}{3}x + 3 \) (shaded purple) and below the dashed line \( y=-\frac{1}{3}x + 2 \) (shaded light blue - the overlapping region is the area that satisfies both inequalities). Wait, actually, let's re - check:

Wait, the first inequality \( y\geq\frac{2}{3}x + 3 \): when we graph \( y = \frac{2}{3}x+3 \), for \( x = 0,y = 3 \), for \( x=-3,y = 1 \). The region above this line (including the line) is shaded. The second inequality \( y<-\frac{1}{3}x + 2 \): when \( x = 0,y = 2 \), for \( x = 6,y=- 2 \). The region below this line (dashed) is shaded.

The correct graph should have the solid line \( y=\frac{2}{3}x + 3 \) with the area above it shaded and the dashed line \( y = -\frac{1}{3}x+2 \) with the area below it shaded, and the overlapping region is the solution.

Looking at the four graphs, the fourth graph (the one with the purple - like shading above the solid line and light blue shading below the dashed line, with the overlapping region being the area that is both above the solid line and below the dashed line) is the correct one. But actually, let's check the intercepts and slopes again.

Wait, maybe I made a mistake in the initial graph identification. Let's list the characteristics:

• Solid line: \( y=\frac{2}{3}x + 3 \) (slope \( \frac{2}{3} \), y - intercept 3)
• Dashed line: \( y=-\frac{1}{3}x + 2 \) (slope \( -\frac{1}{3} \), y - intercept 2)

The correct graph is the one where:

• The solid line \( y = \frac{2}{3}x+3 \) has the region above it shaded.
• The dashed line \( y=-\frac{1}{3}x + 2 \) has the region below it shaded.

The fourth graph (the last one) has the area above the solid line (shaded purple) and below the dashed line (shaded light blue), and the overlapping area is the solution.

Answer:

The fourth graph (the last graph in the given set of four graphs) shows the solution to the system of linear inequalities. (If we assume the graphs are labeled as 1,2,3,4 from top - left to bottom - right, the answer is the 4th graph)