Chapter P.2, Problem 56E

Solution

Describe the real set that satisfies the given inequality.

  $\left(-1,5\right)$

Given:

  $\left|x-2\right|<3$

Concept Used:

• For any x :

  $\left|x\right|=\left\{\begin{array}{l}x\text{ if }x\text{ is positive}\\ -x\text{ if }x\text{ is negative}\end{array}\right.$

• Properties of inequality:

  $b\le c\Rightarrow b\pm a\le c\pm a$

  $b\le c\Rightarrow ab\le ac,\text{ if }a>0$

  $b\le c\Rightarrow ab\ge ac,\text{ if }a<0$

Calculation:

In order to describe the real set that satisfies the given inequality

  $\left|x-2\right|<3$ use for any x :

  $\left|x\right|=\left\{\begin{array}{l}x\text{ if }x\text{ is positive}\\ -x\text{ if }x\text{ is negative}\end{array}\right.$

Here, $x-2$ is positive if

  $\begin{array}{c}x-2>0\\ \Rightarrow x>2\end{array}$

And, $x-2$ is negative if

  $\begin{array}{c}x-2<0\\ \Rightarrow x<2\end{array}$

i.e. using

  $\left|x\right|=\left\{\begin{array}{l}x\text{ if }x\text{ is positive}\\ -x\text{ if }x\text{ is negative}\end{array}\right.$

  $\left|x-2\right|$ can be written as:

  $\left|x-2\right|=\left\{\begin{array}{l}x-2\text{ if }x>2\\ -\left(x-2\right)\text{ if }x<2\end{array}\right.$

Thus, the given inequality can be written as:

  $\begin{array}{c}\text{if }x>2\\ x-2<3\\ x-2+2<3\\ x<5,\\ \text{if }x<2\\ -\left(x-2\right)<3\\ -x+2<3\\ -x+2-2<3-2\\ -x<1\\ x>-1\end{array}$

Thus, the solution of given inequality is:

  $x>-1\text{ and }x<5$

Thus, the solution of given inequality is every number strictly less than 5 and strictly greater than $-1$ , this can be represented in interval form, as

  $\left(-1,5\right)$