Question

which graph shows the solution to the system of linear inequalities?
y ≥ 2x + 1
y ≤ 2x − 2

Explanation:

Step1: Analyze the slopes and y-intercepts

The two inequalities are \( y \geq 2x + 1 \) and \( y \leq 2x - 2 \). Both have a slope of \( 2 \), so the lines are parallel. The first inequality \( y \geq 2x + 1 \) has a y-intercept at \( (0, 1) \) and should be a solid line (since it's \( \geq \)) with shading above the line. The second inequality \( y \leq 2x - 2 \) has a y-intercept at \( (0, -2) \) and should be a solid line (since it's \( \leq \)) with shading below the line.

Step2: Check for overlapping region

Since the lines are parallel (same slope) and \( 2x + 1 \) is always greater than \( 2x - 2 \) (because \( 1 > -2 \)), there is no overlapping region between the two shadings. So we need to find the graph where the first line (with y-intercept 1) has shading above it, the second line (with y-intercept -2) has shading below it, and there's no overlap.

Looking at the options, the last graph (with red line \( y = 2x + 1 \) shaded above and blue line \( y = 2x - 2 \) shaded below, no overlap) matches this. Wait, actually, let's re - check. Wait, the two lines \( y = 2x+1 \) and \( y = 2x - 2 \) are parallel. The region \( y\geq2x + 1 \) is above \( y = 2x + 1 \), and \( y\leq2x - 2 \) is below \( y = 2x - 2 \). Since \( 2x+1>2x - 2 \) for all x, there is no solution (no overlapping shaded area). So we need to find the graph where the shading for \( y\geq2x + 1 \) and \( y\leq2x - 2 \) do not overlap.

Looking at the four graphs:

• First graph: Shading between the lines? No, because the lines are parallel and \( 2x + 1 \) is above \( 2x - 2 \).
• Second graph: Also shading between? No.
• Third graph: Shading in the same area? No.
• Fourth graph: The red line ( \( y = 2x + 1 \)) has shading above (orange) and the blue line ( \( y = 2x - 2 \)) has shading below (blue), and there's no overlap, which is correct because there's no solution.

Answer:

The fourth graph (with red line \( y = 2x + 1 \) shaded above and blue line \( y = 2x - 2 \) shaded below, no overlapping region) shows the solution (which is no solution, as the regions don't overlap). If we assume the fourth graph is the one with the red line at y - intercept 1 and blue at y - intercept - 2, with respective shadings, that's the correct one.