Chapter 2.6, Problem 16E

To determine

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

Answer to Problem 16E

The required result is $a>0\text{ or }a<-1$

Its graph should have an open point at 2 and a shaded line going to the right of 2 and an open point at -1 and a shaded line going to the left of -1.

Explanation of Solution

Given:

  $-3\left|2-4u\right|+5<-13$

Calculation:

  $\begin{array}{l}-3\left|2-4u\right|+5<-13\\ -3\left|2-4u\right|+5-5<-13-5\text{ Subtract both sides by 5}\\ \frac{-3\left|2-4u\right|}{-3}<\frac{-18}{-3}\text{ Divide both sides by -3}\\ 2-4u>6\text{ Reverse inequality symbol}\\ 2-4u<-6\text{ or }2-4u>6\text{ Write a compound inequality}\\ 2-2-4u<-6-2\text{ or }2-2-4u<6-2\text{ Subtract both sides for both inequalities by 2}\\ \frac{-4u}{-4}<\frac{-8}{-4}\text{ or }\frac{-4u}{-4}<\frac{4}{-4}\text{ Divide both sides for both inequalities by -4}\\ a>0\text{ or }a<-1\end{array}$

Since the inequalities are joined by “or”, take the union of the inequalities as its solution set.

Its graph should have an open point at 2 and a shaded line going to the right of 2 and an open point at -1 and a shaded line going to the left of -1, as in:

  

Conclusion:

Therefore, its graph should have an open point at 2 and a shaded line going to the right of 2 and an open point at -1 and a shaded line going to the left of -1.

The required result is $a>0\text{ or }a<-1$