Chapter 1.6, Problem 26PPSE

Solution

To solve the inequality $\left|6y-2\right|+4<22$ and graph the solution.

  $-\frac{8}{3}<y<\frac{10}{3}$

The graph of solution is shown below:

  

Given:

The inequality, $\left|6y-2\right|+4<22$ .

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: $\text{if }\left|x\right|<a,\text{then}-a<x<a$

Calculation:

In order to solve the given inequality $\left|6y-2\right|+4<22$ , use the absolute value property and isolate the variable on one side by performing the basic algebraic operations as shown below

  $\begin{array}{c}\left|6y-2\right|+4<22\\ \left|6y-2\right|+4-4<22-4\\ \left|6y-2\right|<18\\ -18<6y-2<18\\ -18+2<6y-2+2<18+2\\ -16<6y<20\\ \frac{-16}{6}<\frac{6y}{6}<\frac{20}{6}\\ -\frac{8}{3}<y<\frac{10}{3}\end{array}$

Thus, the solution of the inequality is $-\frac{8}{3}<y<\frac{10}{3}$ .

Since the inequality sign is strict with both endpoints, use the hollow circles at both endpoints and shade the region between them as shown below