Chapter 6.5, Problem 1P

a.

To determine

Whether $\left(3,6\right)$ is the solution of inequality $y\le \frac{2}{3}x+4$ .

Answer to Problem 1P

The point $\left(3,6\right)$ is the solution of inequality $y\le \frac{2}{3}x+4$ .

Explanation of Solution

Given:

The point $\left(3,6\right)$ is the solution of inequality $y\le \frac{2}{3}x+4$ .

Concept used:

The given solution must satisfy the inequality.

Calculations:

To verify $\left(3,6\right)$ is the solution of inequality $y\le \frac{2}{3}x+4$ , substitute $x=3$ and $y=6$ in the given inequality.

  $6\le \frac{2}{3}\cdot 3+4$

  $\Rightarrow 6\le 2+4=6$

Since, point $\left(3,6\right)$ satisfy the inequality $y\le \frac{2}{3}x+4$ , therefore, the point $\left(3,6\right)$ is the solution of inequality $y\le \frac{2}{3}x+4$ .

Conclusion:

Thepoint $\left(3,6\right)$ is the solution of inequality $y\le \frac{2}{3}x+4$ .

b.

To determine

To explain:Suppose an ordered pair is not a solution of $y>x+10$ . Must it be the solution of $y<x+10$ .

Answer to Problem 1P

If an ordered pair is not a solution of $y>x+10$ , then surely, it is a solution of the inequality $y<x+10$ .

Explanation of Solution

Given:

An ordered pair is not a solution of $y>x+10$ . Is it a solution of the inequality $y<x+10$ .

Concept used:

An equation or inequality divides the region of coordinate plane into two parts.

Calculations:

Since, the inequality $y>x+10$ divides the region of coordinate plane into two parts. Therefore, if a point (ordered pair) does not fall within one region, i.e. is not the solution of inequality corresponding to this region, then surely that ordered pair will fall in the region of inequality $y<x+10$ , i.e., will be surely the solution of inequality $y<x+10$ .

The graphs of inequalities $y>x+10$ and $y<x+10$ are shown in Figure-1 here.

See ordered pair $\left(-9,5\right)$ is solution of $y>x+10$ but not of $y<x+10$ , and $\left(3,6\right)$ is not the solution of $y>x+10$ but it is solution of $y<x+10$ .

  

Thus, if an ordered pair is not a solution of $y>x+10$ , then surely, it is a solution of the inequality $y<x+10$ .

Conclusion:

If an ordered pair is not a solution of $y>x+10$ , then surely, it is a solution of the inequality $y<x+10$ .