Chapter 2.6, Problem 18E

To determine

To solve: the inequality with its graph and graph the solution.

Answer to Problem 18E

The required result is $v\ge -6\text{ and }v\le 3$

Its graph should have a closed point at -6 and a shaded line going to the right until the closed point at 3.

Explanation of Solution

Given:

  $\frac{2}{3}\left|4v+6\right|-2\le 10$

Calculation:

  $\begin{array}{l}\frac{2}{3}\left|4v+6\right|-2\le 10\\ \frac{2}{3}\left|4v+6\right|-2+2\le 10+2\text{ Add both sides by 2}\\ \left(\frac{3}{2}\right)\left(\frac{2}{3}\right)\left|4v+6\right|\le 12\left(\frac{3}{2}\right)\text{ Multiply both sides by }\frac{3}{2}\\ \left|4v+6\right|\le 18\text{ Simplify}\\ 4v+6\ge -18\text{ and }4v+6\le 18\text{ Write a compound inequality}\\ 4v+6-6\ge -18-6\text{ and }4v+6-6\le 18-6\text{ Subtract both sides of both inequalities by 6}\\ \frac{4v}{4}\ge \frac{-24}{4}\text{ and }\frac{4v}{4}\le \frac{12}{4}\text{ Divide both sides of both inequalities by 4}\\ v\ge -6\text{ and }v\le 3\text{ Simplify}\end{array}$

Since the inequalities are joined by “and”, we should take the intersection of the inequalities as its solution set.

Its graph should have a closed point at -6 and a shaded line going to the right until the closed point at 3, as in:

  

Conclusion:

Therefore, its graph should have a closed point at -6 and a shaded line going to the right until the closed point at 3.

The required result is $v\ge -6\text{ and }v\le 3$