Chapter 5.3, Problem 30PPS

To determine

To solve the inequality:

  $21\ge 3\left(a-7\right)+9$

And to graph the solution on a number line.

Answer to Problem 30PPS

  $\left(-\infty ,11\right]$

  

Explanation of Solution

Given:

  $21\ge 3\left(a-7\right)+9$

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. $b\le c\Rightarrow b\pm a\le c\pm a$
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 inequality

  $21\ge 3\left(a-7\right)+9$

First, simplify this inequality, as

  $\begin{array}{c}21\ge 3\left(a-7\right)+9\\ \Rightarrow 21\ge 3a-21+9\\ \Rightarrow 21\ge 3a-12\end{array}$

Now to simplify further add 12 both side of the inequality and use properties of inequality to simplify further, as

  $\begin{array}{c}21\ge 3a-12\\ \Rightarrow 21+12\ge 3a-12+12\\ \Rightarrow 33\ge 3a\\ \Rightarrow \frac{33}{3}\ge \frac{3a}{3}\\ \Rightarrow 11\ge a\end{array}$

Thus, the solution of given inequality is every number less than and equals to 11, this can be represented in interval form, as

  $\left(-\infty ,11\right]$

Now, in order to sketch the graph of the solution of the inequality, since the interval of solution at left end point 11 is not strict, so use a solid circle at 11. And the solution interval has $-\infty$ at right endpoint which means every number less than and equals to 11 is in the solution interval of given inequality. The graph of the solution of inequality is shown below: