Chapter 3.3, Problem 2CFU

To determine

To state: the two cases that are considered while solving the given problem algebraically.

Answer to Problem 2CFU

The two cases arise that are considered are $\left|a\right|=-a$ when $a<0$ and $\left|a\right|=+a$ when $a\ge 0$ .

Explanation of Solution

Given information:

One variable absolute value inequality

Calculation:

Let’s assume $\left|x-2\right|-5<4$ an inequality for the given condition. Now the two cases arise, if $\left|x-2\right|<0$ , then $\left|x-2\right|=-\left(x-2\right)$ and if $\left|x-2\right|\ge 0$ , then $\left|x-2\right|=+\left(x-2\right)$ . Thus, the two solutions of x will be available for solving the absolute value inequality.

First case: $\left|x-2\right|<0$

  $\begin{array}{c}\left|x-2\right|-5<4\\ -\left(x-2\right)-5<4\\ -x+2-5<4\\ -x-3<4\\ -x<4+3\\ -x<7\\ x<-7\end{array}$

Second case: $\left|x-2\right|\ge 0$

  $\begin{array}{c}\left|x-2\right|-5<4\\ +\left(x-2\right)-5<4\\ x-2-5<4\\ x-7<4\\ x<4+7\\ x<11\end{array}$

Thus, for solving the absolute value inequality algebraically the two cases arise, i.e. $\left|a\right|=-a$ when $a<0$ and $\left|a\right|=+a$ when $a\ge 0$ .