Chapter 1.6, Problem 61PPSE

Solution

To solve the inequality $\left|3z+15\right|\ge 0$ and graph the solution.

All real numbers

Given:

The inequality, $\left|3z+15\right|\ge 0$ .

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 $\left|3z+15\right|\ge 0$ , note that the absolute value inequality always produce positive number that is greater than or equal to zero.

Thus, the solution of the inequality is all real numbers.