Question

graph the following system of inequalities by plotting all lines in the system and then shading the solution region.
$x \leq 0$
$y > 4$
$y > -x - 2$

Explanation:

Step1: Graph \( x \leq 0 \)

The inequality \( x \leq 0 \) represents all points to the left of and including the y - axis (the line \( x = 0 \)). We draw a solid vertical line at \( x = 0 \) (since the inequality is "less than or equal to") and shade the region to the left of this line.

Step2: Graph \( y>4 \)

The inequality \( y > 4 \) represents all points above the line \( y = 4 \). We draw a dashed horizontal line at \( y = 4 \) (since the inequality is "greater than", not "greater than or equal to") and shade the region above this line.

Step3: Graph \( y>-x - 2 \)

First, we consider the line \( y=-x - 2 \). The slope of this line is \( - 1 \) and the y - intercept is \( - 2 \). We draw a dashed line for \( y=-x - 2 \) (because the inequality is "greater than", not "greater than or equal to"). To determine which side to shade, we can use a test point. Let's use the point \( (0,0) \). Substituting into the inequality \( y>-x - 2 \), we get \( 0>-0 - 2\), which simplifies to \( 0 > - 2 \), a true statement. But we have to be careful with the other inequalities. However, for the line \( y=-x - 2 \), the region above the line (since \( y\) is greater than \( -x - 2 \)) is the solution for this inequality.

Step4: Find the intersection region

The solution to the system of inequalities is the region that is common to all three shaded regions. So we look for the area that is to the left of \( x = 0 \), above \( y = 4 \), and above \( y=-x - 2 \).

Answer:

To graph the system:

1. Draw a solid vertical line at \( x = 0 \) and shade left of it.
2. Draw a dashed horizontal line at \( y = 4 \) and shade above it.
3. Draw a dashed line \( y=-x - 2 \) (slope - 1, y - intercept - 2) and shade above it.

The solution region is the overlapping area of these three shaded regions (left of \( x = 0 \), above \( y = 4 \), and above \( y=-x - 2 \)).