Chapter 6.6, Problem 34PPE

To determine

Towrite:The system of linear inequalities for given graph with given characteristics.

Answer to Problem 34PPE

The system of inequalities with required characteristics is

  $y\le -\frac{3}{2}x-2$ , $y\ge 3x+2$ .

Explanation of Solution

Given information:

The graph of system of inequalities is shown in Figure-1 here.

The characteristic of inequalities that their all solutions lie in Quadrant III.

Method used:

Find equations of lines (boundaries of shaded portions) and write inequalities corresponding to obtained lines.

Calculations:

The line AB shown in Figure-2 passes through points $\text{A}\left(0,-2\right)$ and $\text{B}\left(-2,1\right)$ . The equation of line AB from formula two points $\left(x_1,y_1\right)$ , $\left(x_2,y_2\right)$ formula $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$ is

  $y-(-2)=\frac{1-(-2)}{-2-0}\left(x-0\right)$

  $\Rightarrow y+2=-\frac{3}{2}x$

  $\Rightarrow y=-\frac{3}{2}x-2$

… (1)

The inequalities with boundary line given by equation (1) may be either $y\le -\frac{3}{2}x-2$ or $y\ge -\frac{3}{2}x-2$ . To find the correct one ,we chose a test point that lie in the solution region Quadrant III say $\left(-2,-2\right)$ and verify that it is solution by substituting in the inequalities.

Since, $-2\le -\frac{3}{2}\left(-2\right)-2\Rightarrow -2\le 3-2=1$ (true) and $-2\ge -\frac{3}{2}\left(-2\right)-2\Rightarrow -2\ge 3-2=1$ (false), therefore, the inequality whose solution is Quadrant III is $y\le -\frac{3}{2}x-2$ .

The line CD shown in Figure-2 passes through points $\text{C}\left(0,2\right)$ and $\text{B}\left(-1,-1\right)$ . The equation of line AB from formula two points $\left(x_1,y_1\right)$ , $\left(x_2,y_2\right)$ formula $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$ is

  $y-2=\frac{-1-2}{-1-0}\left(x-0\right)$

  $\Rightarrow y-2=3x$

  $\Rightarrow y=3x+2$

The inequalities with boundary line given by equation (1) may be either $y\le 3x+2$ or $y\ge 3x+2$ . To find the correct one, we chose a test point that lie in the solution region Quadrant III say $\left(-2,-2\right)$ and verify that it is the solution by substituting in the inequalities.

Since, $-2\le 3(-2)+2\Rightarrow -2\le -6+2=-4$ (false) and $-2\ge 3(-2)+2\Rightarrow -2\ge -6+2=-4$ (true), therefore, the inequality whose solution is Quadrant III is $y\ge 3x+2$ .

Verification:

The graph of obtained system of inequalities $y\le -\frac{3}{2}x-2$ and $y\ge 3x+2$ are shown in Figure-3.

Hence, the system of inequalities that has Quadrant III its solution is

  $y\le -\frac{3}{2}x-2$ , $y\ge 3x+2$

Conclusion:

The system of inequalities with required characteristics are:

  $y\le -\frac{3}{2}x-2$ , $y\ge 3x+2$ .