Chapter 5.4, Problem 12PPS

To determine

To solve the compound inequality:

  $5h-4\ge 6,\text{ and }7h+11<32$

And to graph the solution on a number line.

Answer to Problem 12PPS

  $\left[2,3\right)$

  

Explanation of Solution

Given:

  $5h-4\ge 6,\text{ and }7h+11<32$

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. $b\le c\Rightarrow b\pm a\le c\pm a$
2. $b\le c\Rightarrow ab\le ac,\text{ if }a>0$
3. $b\le c\Rightarrow ab\ge ac,\text{ if }a<0$

Calculation:

In order to solve the compound inequality:

  $5h-4\ge 6,\text{ and }7h+11<32$

First, solve the inequality

  $5h-4\ge 6$

First, by adding 14 both sides, as

  $\begin{array}{c}5h-4+4\ge 6+4\\ \Rightarrow 5h\ge 10\end{array}$

Now, divide both sides by 5

  $\begin{array}{c}\frac{5h}{5}\ge \frac{10}{5}\\ \Rightarrow h\ge 2\end{array}$

Now, solve the inequality

  $7h+11<32$

First, by subtracting 11 from both sides, as

  $\begin{array}{c}7h+11-11<32-11\\ \Rightarrow 7h+0<21\\ \Rightarrow 7h<21\end{array}$

Now, divide both sides of above inequality by 7, as

  $\begin{array}{c}7h<21\\ \Rightarrow \frac{7h}{7}<\frac{21}{7}\\ \Rightarrow h<3\end{array}$

Combining both of solutions:

  $h<3,\text{ and }2\le h\Rightarrow 2\le h<3$

Thus, the solution of given inequality is every number greater than and equals to 2 but strictly less than 3, this can be represented in interval form, as

  $\left[2,3\right)$

Now, in order to sketch the graph of the solution of the inequality, since the interval of solution at left end point, 2, is not strict, so use a solid circle at 2. And the solution interval at right endpoint, 3, is strict so use a hollow circle at 3. The graph of the solution of inequality is shown below: