Chapter 1.6, Problem 63PPSE

Solution

To solve the inequality $\frac{1}{9}\left|5x-3\right|-3\ge 2$ and graph the solution.

  $x\ge \frac{48}{5}\text{or }x\le -\frac{42}{5}$

The graph of solution is shown below:

  

Given:

The inequality, $\frac{1}{9}\left|5x-3\right|-3\ge 2$ .

Concept Used:

• If there is the strict inequality symbol $<\text{ or}>$ , then use a parenthesis $\left(\right)$ on the interval
• If there is the inequality symbol $\le \text{ or}\ge$ , then use a bracket $\left[\right]$ on the interval
• 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.
• Multiplying or dividing an inequality by a negative number reverse the inequality sign.
• The absolute value property: $\text{if }\left|x\right|<a,\text{then}-a<x<a$
• The absolute value property: $\text{if }\left|x\right|>a,\text{then}x>a\text{ or }x<-a$

Calculation:

In order to solve the given inequality $\frac{1}{9}\left|5x-3\right|-3\ge 2$ , use the absolute value property and isolate the variable on one side by performing the basic algebraic operations as shown below

  $\begin{array}{c}\frac{1}{9}\left|5x-3\right|-3\ge 2\\ \frac{1}{9}\left|5x-3\right|-3+3\ge 2+3\\ 9\cdot \frac{1}{9}\left|5x-3\right|\ge 9\cdot 5\\ \left|5x-3\right|\ge 45\\ 5x-3\ge 45\text{or }5x-3\le -45\\ 5x-3+3\ge 45+3\text{or }5x-3+3\le -45+3\\ 5x\ge 48\text{or }5x\le -42\\ x\ge \frac{48}{5}\text{or }x\le -\frac{42}{5}\end{array}$

Thus, the solution of the inequality is $x\ge \frac{48}{5}\text{or }x\le -\frac{42}{5}$ .

Since the inequality sign is not strict with both endpoints, use the solid circles at both endpoints and shade the region as shown below