Question

systems of linear inequalities
\

$$\begin{cases} y \\geq 2x + 4 \\\\ y > -x - 2 \\end{cases}$$

drag the graph corresponding to the system of inequalities and all points that are part of the solution set into the box.
a b c
d e f
(-2, 0) (0, 4) (-4, 3) (0, 0)

Explanation:

Step1: Analyze \( y \geq 2x + 4 \)

The line \( y = 2x + 4 \) has a slope of \( 2 \) and y - intercept \( 4 \). The inequality \( y \geq 2x + 4 \) means we shade above the line (including the line, so the line is solid).

Step2: Analyze \( y > -x - 2 \)

The line \( y=-x - 2\) has a slope of \(- 1\) and y - intercept \(-2\). The inequality \( y > -x - 2\) means we shade above the line (the line is dashed since it's a strict inequality).

Step3: Find the intersection and solution region

First, find the intersection of \( y = 2x+4\) and \( y=-x - 2\). Set \( 2x + 4=-x-2\), then \( 3x=-6\), so \( x = - 2\) and \( y=0\). The solution region is where both shadings overlap. The line \( y = 2x + 4\) is solid and \( y=-x - 2\) is dashed. The overlapping region should be above \( y = 2x+4\) (solid line) and above \( y=-x - 2\) (dashed line). Looking at the graphs, graph E has the solid line \( y = 2x + 4\) (going up with slope 2, y - intercept 4) and dashed line \( y=-x - 2\) (going down with slope - 1, y - intercept - 2) with the overlapping region shaded. Now check the points:

• For \( (-2,0)\): Plug into \( y\geq2x + 4\): \( 0\geq2\times(-2)+4=0\), true. Plug into \( y >-x - 2\): \( 0>-(-2)-2=0\)? No, \( 0 = 0\), so not in \( y >-x - 2\). Wait, maybe miscalculation. Wait \( y >-x - 2\) at \( x=-2\), \( y >-(-2)-2=0\), so \( y>0\) at \( x = - 2\), so \( (-2,0)\) is not in the solution.
• For \( (0,4)\): Plug into \( y\geq2x + 4\): \( 4\geq0 + 4\), true. Plug into \( y >-x - 2\): \( 4>0 - 2= - 2\), true.
• For \( (-4,3)\): Plug into \( y\geq2x + 4\): \( 3\geq2\times(-4)+4=-4\), true. Plug into \( y >-x - 2\): \( 3>-(-4)-2=2\), true.
• For \( (0,0)\): Plug into \( y\geq2x + 4\): \( 0\geq0 + 4\)? No. So the correct graph is E and the points \( (0,4)\), \( (-4,3)\) (and the graph E). Wait, maybe the initial analysis of the graph: Let's re - check the lines. The line \( y = 2x+4\) passes through \( (0,4)\) and \( (-2,0)\). The line \( y=-x - 2\) passes through \( (0,-2)\) and \( (-2,0)\). The solution region is above \( y = 2x + 4\) (since \( y\geq2x + 4\)) and above \( y=-x - 2\) (since \( y >-x - 2\)). So the area above \( y = 2x+4\) (solid) and above \( y=-x - 2\) (dashed). The graph E has the solid line \( y = 2x + 4\) (going from \( (0,4)\) with slope 2) and dashed line \( y=-x - 2\) (going from \( (0,-2)\) with slope - 1) and the overlapping region is to the right - upper of the intersection. The points \( (0,4)\) is on \( y = 2x + 4\) (since \( 4=2\times0 + 4\)) and \( 4>-0 - 2 = 2\), true. \( (-4,3)\): \( 3\geq2\times(-4)+4=-4\), true. \( 3>-(-4)-2=2\), true. \( (0,0)\): \( 0\geq4\)? No. \( (-2,0)\): \( 0\geq0\) (for \( y\geq2x + 4\)) and \( 0>0\)? No. So the graph is E and points \( (0,4)\), \( (-4,3)\). But the problem says "Drag the graph corresponding to the system of inequalities and all points that are part of the solution set into the box." So the graph is E, and points \( (0,4)\), \( (-4,3)\). Wait, maybe I made a mistake with \( (-2,0)\). Let's re - check \( y >-x - 2\) at \( x=-2\): \( y >-(-2)-2=0\), so \( y>0\), but \( (-2,0)\) has \( y = 0\), so it's not in \( y >-x - 2\). So the correct graph is E, and points \( (0,4)\), \( (-4,3)\).

Answer:

Graph: E, Points: \( (0,4) \), \( (-4,3) \)