Question

find the graph of this system of linear inequalities. \\(\

$$\begin{cases} y \\geq x - 2 \\\\ y > -2x - 1 \\end{cases}$$

\\)

Explanation:

Step1: Analyze \( y \geq x - 2 \)

The line \( y = x - 2 \) has a slope of \( 1 \) and a y - intercept of \( - 2 \). Since the inequality is \( y\geq x - 2 \), we draw a solid line (because the inequality includes equality) for \( y=x - 2 \) and shade the region above the line (because \( y \) is greater than or equal to \( x - 2 \)).

Step2: Analyze \( y > - 2x - 1 \)

The line \( y=-2x - 1 \) has a slope of \( - 2 \) and a y - intercept of \( - 1 \). Since the inequality is \( y > - 2x - 1 \), we draw a dashed line (because the inequality does not include equality) for \( y = - 2x-1 \) and shade the region above the line (because \( y \) is greater than \( - 2x - 1 \)).

Step3: Find the intersection of the two shaded regions

The solution to the system of inequalities is the region that is shaded for both \( y\geq x - 2 \) and \( y > - 2x - 1 \).

• For the line \( y=x - 2 \) (solid line), when \( x = 0 \), \( y=-2 \); when \( y = 0 \), \( x = 2 \).
• For the line \( y=-2x - 1 \) (dashed line), when \( x = 0 \), \( y=-1 \); when \( y = 0 \), \( x=-\frac{1}{2}\).

By looking at the three graphs:

• The first graph: The shaded region for \( y\geq x - 2 \) (above the solid line) and \( y > - 2x - 1 \) (above the dashed line) does not match as the intersection is not correct.
• The second graph: The region above the solid line \( y = x - 2 \) and above the dashed line \( y=-2x - 1 \) is the overlapping region. The solid line is \( y=x - 2 \) (passes through \( (0,-2) \) and \( (2,0) \)) and the dashed line is \( y=-2x - 1 \) (passes through \( (0, - 1) \) and \( (-\frac{1}{2},0) \)). The shaded region in the second graph is the area that is above both lines, which is the correct intersection.
• The third graph: The shaded region is not the intersection of the two required regions.

Answer:

The second graph (the middle one) is the graph of the system of linear inequalities.