Chapter 6.6, Problem 23PPE

To determine

To write: a system of inequalities that represented by the given graph.

Answer to Problem 23PPE

  $\begin{array}{l}x+3y<0\\ y-x\le 2\end{array}$

Explanation of Solution

Given information: a figure is given as follows-

  

Since the dark line passes through the points (-1, 1) and (0, 2). So, its equation is

  $\begin{array}{l}\frac{y-1}{x-(-1)}=\frac{2-1}{0-(-1)}=1\\ \Rightarrow \frac{y-1}{x+1}=1\\ \Rightarrow y-1=x+1\\ \Rightarrow y-x=2\end{array}$

This divides the xy-plane into two regions and two inequalities are possible such as $y-x\ge 2$ and $y-x\le 2$ .

Take any point from the colored region and put this value in the above inequalities which satisfies is the required inequality.

Such as take a point (0, 0) from the colored region, the inequality which contains this point is $y-x\le 2$ . Hence this is the required inequality.

Again the dotted linepasses through the points (0, 0) and (-3, 1). So, its equation is

  $\begin{array}{l}\frac{y-0}{x-0}=\frac{1-0}{-3-0}=\frac{1}{-3}\\ \Rightarrow \frac{y}{x}=\frac{1}{-3}\\ \Rightarrow -3y=x\\ \Rightarrow x+3y=0\end{array}$

This divides the xy-plane into two regions and two inequalities are possible such as $x+3y>0$ and $x+3y<0$ .

Take any point from the colored region and put this value in the above inequalities which satisfies is the required inequality.

Such as take a point (0, -1) from the colored region, the inequality which contains this point is $x+3y<0$ . Hence this is the required inequality.

Thus the required inequality represented by above figure is

  $\begin{array}{l}x+3y<0\\ y-x\le 2\end{array}$