Chapter 5.4, Problem 1CYU

To determine

To solve the compound inequality:

  $4\le p-8,\text{ and }p-14\le 2$

And to graph the solution on a number line.

Answer to Problem 1CYU

  $\left(-\infty ,2\right]$

  

Explanation of Solution

Given:

  $4\le p-8,\text{ and }p-14\le 2$

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:

  $4\le p-8,\text{ and }p-14\le 2$

First, solve the inequality

  $4\le p-8$

By adding 8 both sides, as

  $\begin{array}{c}4+8\le p-8+8\\ \Rightarrow 12\le p\end{array}$

Now, solve the inequality

  $p-14\le 2$

By again adding 14 both sides, as

  $\begin{array}{c}p-14+14\le 2+14\\ \Rightarrow p\le 16\end{array}$

Combining both of solutions:

  $p\le 2,y\le 16\Rightarrow p\le 2$

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

  $\left(-\infty ,2\right]$

Now, in order to sketch the graph of the solution of the inequality, since the interval of solution at right end point 2 is not strict, so use a solid circle at 2. And the solution interval has $-\infty$ at left endpoint which means every number less than and equals to 2 is in the solution interval of given inequality. The graph of the solution of inequality is shown below: