Chapter 3.3, Problem 39PPSE

Solution

To check : The ordered pair $\left(0,-2\right)$ is the solution of the given inequalities.

Given information:

The given ordered pair is $\left(0,-2\right)$ and the inequalities along with graph:

  

Concept used:

If the given ordered pair is in the solution then the ordered pair must satisfied the inequality.

Calculations:

Check $\left(0,-2\right)$ is true for $A)x+y\le 2$ :

Substitute $\left(0,-2\right)$ in to $A)x+y\le 2$ .

  $\begin{array}{c}-2+0\le 2\\ -2\le 2\end{array}$

Inequality is true for this coordinate.

So $\left(0,-2\right)$ is solution for $A)x+y\le 2$ .

Now check $\left(0,-2\right)$ is true for $B)y\le \frac{3}{2}x-1$ :

Substitute $\left(0,-2\right)$ in to $B)y\le \frac{3}{2}x-1$ .

  $-2\le -1$

Inequality is true for this coordinate.

So $\left(0,-2\right)$ is a solution for $B)y\le \frac{3}{2}x-1$ .

Now check $\left(0,-2\right)$ is true for $C)y>-\frac{1}{3}x-2$ :

Substitute $\left(0,-2\right)$ in to $C)y>-\frac{1}{3}x-2$ .

  $-2>-2$

Inequality is not true for this coordinate.

So $\left(0,-2\right)$ is not a solution for $C)y>-\frac{1}{3}x-2$ .

Thus $\left(0,-2\right)$ is a solution of the inequalities $A)x+y\le 2$ , $B)y\le \frac{3}{2}x-1$ .