Math Help for Section 3.7, Page 156
Study
Tip
An absolute
value inequality of the form $|x|<a$ (or $|x|\le a$)
can be
solved with a double inequality, but an inequality of the
form $|x|>a$ (or $|x|\ge a$)
cannot. Instead, you
must solve two separate
inequalities, as demonstrated
in Example 7 on page 90.
Example
7: Tip
The solution set in set notation is $\{x|x\le
-{1\over3}\}\cup\{x|x\ge 3\}$.
Example
7: Check
The solution was found to be $x\le -{1\over3}$ or $x\ge 3$, so check
that
$x=-{1\over3}$ and $x=3$
satisfy the inequality and that $x=0$ does not.
| $x=-{1\over3}$: | |
| $\eqalign{|3x-4|\ge & 5 &{\small\color{red}\quad\quad \text{Write original inequality.}} \cr \left|3\left({\color{red}-{1\over3}}\right)-4\right| \overset{?}{\ge}& 5 &{\small\color{red}\quad\quad \text{Substitute }-{1\over3}\text{ for }x.} \cr |-1-4|\overset{?}{\ge}& 5 &{\small\color{red}\quad\quad \text{Multiply.}} \cr |-5|\overset{?}{\ge}& 5 &{\small\color{red}\quad\quad \text{Subtract.}} \cr 5\ge & 5 &{\small\color{red}\quad\quad \text{Solution checks. }\checkmark} } $ |
|
| $x=3$: | |
| $\eqalign{|3x-4|\ge & 5 &{\small\color{red}\quad\quad \text{Write original inequality.}} \cr |3({\color{red}3})-4| \overset{?}{\ge}& 5 &{\small\color{red}\quad\quad \text{Substitute }3\text{ for }x.} \cr |9-4|\overset{?}{\ge}& 5 &{\small\color{red}\quad\quad \text{Multiply.}} \cr |5|\overset{?}{\ge}& 5 &{\small\color{red}\quad\quad \text{Subtract.}} \cr 5\ge & 5 &{\small\color{red}\quad\quad \text{Solution checks. }\checkmark} } $ |
|
| $x=0$: | |
| $\eqalign{|3x-4|\ge & 5 &{\small\color{red}\quad\quad \text{Write original inequality.}} \cr |3({\color{red}0})-4| \overset{?}{\ge}& 5 &{\small\color{red}\quad\quad \text{Substitute }0\text{ for }x.} \cr |0-4|\overset{?}{\ge}& 5 &{\small\color{red}\quad\quad \text{Multiply.}} \cr |-4|\overset{?}{\ge}& 5 &{\small\color{red}\quad\quad \text{Subtract.}} \cr 4\not\ge & 5 &{\small\color{red}\quad\quad \text{Solution does not check. ✗ }} } $ |
|
Example
8: Tip
The solution set in set notation is $\{x|\,5.97\le x\le6.03\}$.
Example 8: Check
The solution was found to be $5.97\le x\le6.03$, so check
that
$x=6$
satisfies the inequality and that $x=5$ and $x=7$ do not.
| $x=6$: | |
| $\eqalign{\left|2-{x\over3}\right| \le & 0.01 &{\small\color{red}\quad\quad \text{Write original inequality.}} \cr \left|2-{{\color{red}6}\over3}\right| \overset{?}{\le}& 0.01 &{\small\color{red}\quad\quad \text{Substitute 6 for }x.} \cr |2-2|\overset{?}{\le}& 0.01 &{\small\color{red}\quad\quad \text{Divide.}} \cr |0|\overset{?}{\le}& 0.01 &{\small\color{red}\quad\quad \text{Subtract.}} \cr 0\le & 0.01 &{\small\color{red}\quad\quad \text{Solution checks. }\checkmark} } $ |
|
| $x=5$: | |
| $\eqalign{\left|2-{x\over3}\right| \le & 0.01 &{\small\color{red}\quad\quad \text{Write original inequality.}} \cr \left|2-{{\color{red}5}\over3}\right| \overset{?}{\le}& 0.01 &{\small\color{red}\quad\quad \text{Substitute 5 for }x.} \cr |2-1.\overline{6}|\overset{?}{\le}& 0.01 &{\small\color{red}\quad\quad \text{Write }{5\over3}\text{ as }1.\overline{6}.} \cr |0.\overline{3}|\overset{?}{\le}& 0.01 &{\small\color{red}\quad\quad \text{Subtract.}} \cr 0.\overline{3}\not\le & 0.01 &{\small\color{red}\quad\quad \text{Solution does not check. ✗}} } $ |
|
| $x=7$: | |
| $\eqalign{\left|2-{x\over3}\right| \le & 0.01 &{\small\color{red}\quad\quad \text{Write original inequality.}} \cr \left|2-{{\color{red}7}\over3}\right| \overset{?}{\le}& 0.01 &{\small\color{red}\quad\quad \text{Substitute 7 for }x.} \cr |2-2.\overline{3}|\overset{?}{\le}& 0.01 &{\small\color{red}\quad\quad \text{Write }{7\over3}\text{ as }2.\overline{3}.} \cr |-0.\overline{3}|\overset{?}{\le}& 0.01 &{\small\color{red}\quad\quad \text{Subtract.}} \cr 0.\overline{3}\not\le & 0.01 &{\small\color{red}\quad\quad \text{Solution does not check. ✗}} } $ |
|