Chapter 5.2, Problem 64S

To determine

To solve the inequality:

  $-\frac{2}{3}h\le 9$

And to graph the solution on a number line.

Answer to Problem 64S

The solution of the inequality is $\left(-\frac{27}{2},\infty \right)\text{ or }h>-13.5$ and corresponding correct option is b.

  

Explanation of Solution

Given:

  $-\frac{2}{3}h\le 9$

Concept Used:

• To graph the endpoint with strict inequality $<\text{ or}>$, use the open parenthesis or a hollow circle at that endpoint.
• To graph the endpoint with inequality symbol $\le \text{ or}\ge$, use the bracket or a solid circle at that endpoint.
• Properties of inequality:
1. $b\le c\Rightarrow ab\le ac,\text{ if }a\ge 0$
2. $b\le c\Rightarrow ab\ge ac,\text{ if }a\le 0$

Calculation:

To solve the inequality:

  $-\frac{2}{3}h\le 9$

Multiply both sides by -1 to reverse the inequality:

  $\begin{array}{c}-\frac{2}{3}h<9\\ \left(-\frac{2}{3}h\right)\left(-1\right)>9\left(-1\right)\\ \frac{2}{3}h>-9\end{array}$

Multiplyboth sides by 3

  $\begin{array}{c}\frac{2}{3}h>-9\\ 3\cdot \frac{2}{3}h>3\left(-9\right)\\ 2h>-27\end{array}$

Now, divideboth sides by 2

  $\begin{array}{c}\frac{2h}{2}>\frac{-27}{2}\\ h>-\frac{27}{2}\\ h>-13.5\end{array}$

Thus, the inequality is true for hgreater than $-13.5$ , thus the solution of given inequality in interval form is:

  $\left(-13.5,\infty \right)$

Now in order to sketch the graph of the solution of the inequality, since the interval of solution at left end point $-13.5$ is strict, so use a blank circle at $-13.5$ . And the solution interval has $\infty$ at right endpoint which means every number greater than from $-13.5$ to infinity is in the solution interval and does not contain the left end point in the number line. The graph of the inequality is shown below:

  

Hence, option b is correct.