Chapter 1.6, Problem 30PPSE

Solution

To solve the inequality $3\left|5t-1\right|+9\le 23$ and graph the solution.

  $-\frac{11}{15}\le t\le \frac{17}{15}$

The graph of solution is shown below:

  

Given:

The inequality, $3\left|5t-1\right|+9\le 23$ .

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 $3\left|5t-1\right|+9\le 23$ , use the absolute value property and isolate the variable on one side by performing the basic algebraic operations as shown below

  $\begin{array}{c}3\left|5t-1\right|+9\le 23\\ 3\left|5t-1\right|+9-9\le 23-9\\ 3\left|5t-1\right|\le 14\\ \frac{3\left|5t-1\right|}{3}\le \frac{14}{3}\\ \left|5t-1\right|\le \frac{14}{3}\\ -\frac{14}{3}\le 5t-1\le \frac{14}{3}\\ -\frac{14}{3}+1\le 5t-1+1\le \frac{14}{3}+1\\ \frac{-14+3}{3}\le 5t\le \frac{14+3}{3}\\ \frac{\frac{-11}{3}}{5}\le \frac{5t}{5}\le \frac{\frac{17}{3}}{5}\\ -\frac{11}{15}\le t\le \frac{17}{15}\end{array}$

Thus, the solution of the inequality is $-\frac{11}{15}\le t\le \frac{17}{15}$ .

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