Chapter 1.6, Problem 64PPSE

Solution

To solve the inequality $\frac{1}{11}\left|2x-4\right|+10\le 11$ and graph the solution.

  $-\frac{7}{2}\le x\le \frac{15}{2}$

The graph of solution is shown below:

  

Given:

The inequality, $\frac{1}{11}\left|2x-4\right|+10\le 11$ .

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 $\frac{1}{11}\left|2x-4\right|+10\le 11$ , use the absolute value property and isolate the variable on one side by performing the basic algebraic operations as shown below

  $\begin{array}{c}\frac{1}{11}\left|2x-4\right|+10\le 11\\ \frac{1}{11}\left|2x-4\right|+10-10\le 11-10\\ \frac{1}{11}\left|2x-4\right|\le 1\\ 11\cdot \frac{1}{11}\left|2x-4\right|\le 11\cdot 1\\ \left|2x-4\right|\le 11\\ -11\le 2x-4\le 11\\ -11+4\le 2x-4+4\le 11+4\\ -7\le 2x\le 15\\ -\frac{7}{2}\le x\le \frac{15}{2}\end{array}$

Thus, the solution of the inequality is $-\frac{7}{2}\le x\le \frac{15}{2}$ .

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