Chapter 6, Problem 56SGR

To determine

To graph : To solve the below mentioned inequalities by graphing:

  $y\le -x-3$

And $y\ge 3x-2$

Explanation of Solution

Given information :

Given the two inequalities:

  $y\le -x-3$

  $y\ge 3x-2$

Graph : The graph plotted below using the inequalities:

  $y\le -x-3$

  $y\ge 3x-2$

Equation $y\le -x-3$ is represented by light red area in above graph.

And $y\ge 3x-2$ is represented by light blue area in above graph.

Interpretation : >From the above graph, we see that $y\le -x-3$ is the bottom area of line $y=-x-3$ (including line $y=-x-3$ ) which has been represented in light red. And we see that $y\ge 3x-2$ is the left area of the line $y=3x-2$ (including the value of $y=3x-2$ ) which has been represented in the light blue area.

So final solution is common color area which is in dark blue color. It covers the area bottom to $y=-x-3$ and right to $y=3x-2$ including the points on lines of common color and including point of intersection of the lines.