Search

Text Color

Margin Size

Font Type

Enable Dyslexic Font

1.4: Absolute value inequalities

Last updated

May 2, 2022
Save as PDF
- 1.3: Inequalities and intervals
- 1.5: Exercises

Thomas Tradler and Holly Carley
CUNY New York City College of Technology via New York City College of Technology at CUNY Academic Works

$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$

$\newcommand{\id}{\mathrm{id}}$ $\newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $\newcommand{\range}{\mathrm{range}\,}$

$\newcommand{\RealPart}{\mathrm{Re}}$ $\newcommand{\ImaginaryPart}{\mathrm{Im}}$

$\newcommand{\Argument}{\mathrm{Arg}}$ $\newcommand{\norm}[1]{\| #1 \|}$

$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$\newcommand{\Span}{\mathrm{span}}$

$\newcommand{\id}{\mathrm{id}}$

$\newcommand{\Span}{\mathrm{span}}$

$\newcommand{\kernel}{\mathrm{null}\,}$

$\newcommand{\range}{\mathrm{range}\,}$

$\newcommand{\RealPart}{\mathrm{Re}}$

$\newcommand{\ImaginaryPart}{\mathrm{Im}}$

$\newcommand{\Argument}{\mathrm{Arg}}$

$\newcommand{\norm}[1]{\| #1 \|}$

$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$\newcommand{\Span}{\mathrm{span}}$ $\newcommand{\AA}{\unicode[.8,0]{x212B}}$

$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vectorC}[1]{\textbf{#1}}$

$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$

$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$

$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$

$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$

$\newcommand{\avec}{\mathbf a}$

$\newcommand{\bvec}{\mathbf b}$

$\newcommand{\cvec}{\mathbf c}$

$\newcommand{\dvec}{\mathbf d}$

$\newcommand{\dtil}{\widetilde{\mathbf d}}$

$\newcommand{\evec}{\mathbf e}$

$\newcommand{\fvec}{\mathbf f}$

$\newcommand{\nvec}{\mathbf n}$

$\newcommand{\pvec}{\mathbf p}$

$\newcommand{\qvec}{\mathbf q}$

$\newcommand{\svec}{\mathbf s}$

$\newcommand{\tvec}{\mathbf t}$

$\newcommand{\uvec}{\mathbf u}$

$\newcommand{\vvec}{\mathbf v}$

$\newcommand{\wvec}{\mathbf w}$

$\newcommand{\xvec}{\mathbf x}$

$\newcommand{\yvec}{\mathbf y}$

$\newcommand{\zvec}{\mathbf z}$

$\newcommand{\rvec}{\mathbf r}$

$\newcommand{\mvec}{\mathbf m}$

$\newcommand{\zerovec}{\mathbf 0}$

$\newcommand{\onevec}{\mathbf 1}$

$\newcommand{\real}{\mathbb R}$

$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$

$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$

$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$

$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$

$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$

$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$

$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$

$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$

$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$

$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$

$\newcommand{\bcal}{\cal B}$

$\newcommand{\ccal}{\cal C}$

$\newcommand{\scal}{\cal S}$

$\newcommand{\wcal}{\cal W}$

$\newcommand{\ecal}{\cal E}$

$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$

$\newcommand{\gray}[1]{\color{gray}{#1}}$

$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$

$\newcommand{\rank}{\operatorname{rank}}$

$\newcommand{\row}{\text{Row}}$

$\newcommand{\col}{\text{Col}}$

$\renewcommand{\row}{\text{Row}}$

$\newcommand{\nul}{\text{Nul}}$

$\newcommand{\var}{\text{Var}}$

$\newcommand{\corr}{\text{corr}}$

$\newcommand{\len}[1]{\left|#1\right|}$

$\newcommand{\bbar}{\overline{\bvec}}$

$\newcommand{\bhat}{\widehat{\bvec}}$

$\newcommand{\bperp}{\bvec^\perp}$

$\newcommand{\xhat}{\widehat{\xvec}}$

$\newcommand{\vhat}{\widehat{\vvec}}$

$\newcommand{\uhat}{\widehat{\uvec}}$

$\newcommand{\what}{\widehat{\wvec}}$

$\newcommand{\Sighat}{\widehat{\Sigma}}$

$\newcommand{\lt}{<}$

$\newcommand{\gt}{>}$

$\newcommand{\amp}{&}$

$\definecolor{fillinmathshade}{gray}{0.9}$

Using the notation from the previous section, we now solve inequalities involving the absolute value. These inequalities may be solved in three steps:

Step 1: Solve the corresponding equality. The solution of the equality divides the real number line into several subintervals.
Step 2: Using step 1, check the inequality for a number in each of the subintervals. This check determines the intervals of the solution set.
Step 3: Check the endpoints of the intervals.

Here are some examples for the above solution method.

Example $\PageIndex{1}$

Solve for $x$ :

$|x+7|<2$
$|3x-5|\geq 11$
$|12-5x|\leq 1$

Solution

We follow the three steps described above. In step 1, we solve the corresponding equality, $|x+7|=2$ . $x+7=2$ & $x+7=-2$ $\begin{array}{l|l} x+7=2 & x+7=-2 \\ \Longrightarrow x=-5 & \Longrightarrow x=-9 \end{array} \nonumber$ The solutions $x=-5$ and $x=-9$ divide the number line into three subintervals:

Now, in step 2, we check the inequality for one number in each of these subintervals.

$\begin{array}{c|c|c} \text { Check: } \quad x=-10 & \text { Check: } \quad x=-7 & \text { Check: } \quad x=0 \\ |(-10)+7| \stackrel{?}{<} 2 & |(-7)+7| \stackrel{?}{<} 2 & |0+7| \stackrel{?}{<} 2 \\ |-3| \stackrel{?}{<} 2 & |0| \stackrel{?}{<} 2 & |7| \stackrel{?}{<} 2 \\ 3 \stackrel{?}{<} 2 & 0 \stackrel{?}{<} 2 & 7 \stackrel{?}{<} 2 \\ \text { false } & \text { true } & \text { false } \end{array} \nonumber$

Since $x=-7$ in the subinterval given by $-9<x<-5$ solves the inequality $|x+7|<2$ , it follows that all numbers in the subinterval given by $-9<x<-5$ solve the inequality. Similarly, since $x=-10$ and $x=0$ do not solve the inequality, no number in these subintervals will solve the inequality. For step 3, we note that the numbers $x=-9$ and $x=-5$ are not included as solutions since the inequality is strict (that is we have $<$ instead of $\leq$ ).The solution set is therefore the interval $S=(-9,-5)$ . The solution on the number line is:

$clipboard_ea1d81614b0d8e2564fd8ec56ba182dbf.png$

We follow the steps as before. First, in step 1, we solve . $\begin{array}{l|l} 3 x-5=11 & 3 x-5=-11 \\ \Longrightarrow 3 x=16 & \Longrightarrow 3 x=-6 \\ \Longrightarrow x=\dfrac{16}{3} & \Longrightarrow x=-2 \end{array} \nonumber$

The two solutions $x=-2$ and $x=\dfrac{16}{3}=5\dfrac {1}{3}$ divide the number line into the subintervals displayed below. $x<-2 \hspace{1in} -2<x<5\frac 1 3 \hspace{1in} 5\frac 1 3<x \nonumber$

$clipboard_e0d7878e0eca72b0ee120cafa8166dd75.png$

For step 2, we check a number in each subinterval. This gives:

$\begin{array}{c|c|c|c} \text { Check: } x=-3 & \text { Check: } \quad x=1 & \text { Check: } \quad x=6 \\ |3 \cdot(-3)-5| \stackrel{?}{\geq} 11 & |3 \cdot 1-5| \stackrel{?}{\geq} 11 & |3 \cdot 6-5| \stackrel{?}{\geq} 11 \\ |-9-5| \stackrel{?}{\geq} 11 & |3-5| \stackrel{?}{\geq} 11 & |18-5| \stackrel{?}{\geq} 11 \\ |-14| \stackrel{?}{\geq} 11 & |-2| \stackrel{?}{\geq} 11 & |13| \stackrel{?}{\geq} 11 \\ 14 \stackrel{?}{\geq} 11 & 2 \stackrel{?}{\geq} 11 & 13 \stackrel{?}{\geq} 11 \\ \text { true } & \text { false } & \text { true } \end{array} \nonumber$

For step 3, note that we include $-2$ and $5\dfrac {1}{3}$ in the solution set since the inequality is “greater than or equal to” (that is $\geq$ , as opposed to $>$ ). Furthermore, the numbers $-\infty$ and $\infty$ are not included, since $\pm\infty$ are not real numbers.

The solution set is therefore the union of the two intervals: $S=\Big(-\infty,-2\Big]\cup \Big[5\dfrac {1}{3}, \infty\Big) \nonumber$

To solve , we first solve the equality . $\begin{array}{l|l} 12-5 x=1 & 12-5 x=-1 \\ \Longrightarrow-5 x=-11 & \Longrightarrow-5 x=-13 \\ \Longrightarrow x=\frac{-11}{-5}=2.2 & \Longrightarrow x=\frac{-13}{-5}=2.6 \end{array} \nonumber$

This divides the number line into three subintervals, and we check the original inequality $|12-5x|\leq 1$ for a number in each of these subintervals.

$\begin{array}{c|c|c|c} \text {Interval: } \quad x<2.2 & \text {Interval: } \quad 2.2<x<2.6 & \text {Interval: } \quad 2.6<x\\ \text {Check: } \quad x=1 & \text {Check: } \quad x=2.4 & \text {Check: } \quad x=3 \\ |12-5 \cdot 1| \stackrel{?}{\leq} 1 & |12-5 \cdot 1| \stackrel{?}{\leq} 1 & |12-5 \cdot 3| \stackrel{?}{\leq} 1 \\ |12-5| \stackrel{?}{\leq} 1 & |12-12| \stackrel{?}{\leq} 1 & |12-15| \stackrel{?}{\leq} 1 \\ |7| \stackrel{?}{\leq} 1 & |0| \stackrel{?}{\leq} 1 & |-3| \stackrel{?}{\leq} 1 \\ 7 \stackrel{?}{\leq} 1 & 0 \stackrel{?}{\leq} 1 & 3 \stackrel{?}{\leq} 1 \\ \text { false } & \text { true } & \text { false } \end{array} \nonumber$

The solution set is the interval $S=[2.2,2.6]$ , where we included $x=2.2$ and $x=2.6$ since the original inequality “less than or equal to” ( $\leq$ ) includes the equality.

Note

Alternatively, whenever you have an absolute value inequality you can turn it into two inequalities.

Here are a couple of examples.

Example $\PageIndex{2}$

Solve for $x$ : $|12-5x|\leq 1$

Solution

Note that $|12-5x|\leq 1$ implies that

$-1\leq 12-5x\leq1 \nonumber$

so that

$-13\leq -5x\leq -11 \nonumber$

and by dividing by $-5$ (remembering to switch the direction of the inequalities when multiplying or dividing by a negative number) we see that

$\frac{13}{5}\geq x\geq \frac{11}{5} \nonumber$

or in interval notation, we have the solution set

$S=\left[\frac{11}{5},\frac{13}{5}\right] \nonumber$

Example $\PageIndex{3}$

If $|x+6|>2$ then either $x+6>2$ or $x+6<-2$ so that either $x>-4$ or $x<-8$ so that in interval notation the solution is $S=(-\infty,-8)\cup(-4,\infty)$ .

Solution

There is a geometric interpretation of the absolute value on the number line as the distance between two numbers:

$clipboard_e8a95ea5bc21fc953df9681bc64a8644c.png$

distance between $a$ and $b$ is $|b-a|$ which is also equal to $|a-b|$

This interpretation can also be used to solve absolute value equations and inequalities.

Example $\PageIndex{4}$

Solve for $x$ :

$|x-6|=4$
$|x-6|\leq 4$
$|x-6|\geq 4$

Solution

Consider the distance between $x$ and $6$ to be $4$ on a number line:

$clipboard_e17c977283e3ff391111c7da91e948997.png$

There are two solutions, $x=2$ or $x=10$ . That is, the distance between $2$ and $6$ is $4$ and the distance between $10$ and $6$ is $4$ .

Numbers inside the braces above have distance $4$ or less. The solution is given on the number line as:

$clipboard_e54e786ef1671d92c3b9291ecb0fc38f1.png$

In interval notation, the solution set is the interval $S=[2,10]$ . One can also write that the solution set consists of all $x$ such that $2\leq x\leq 10$ .

Numbers outside the braces above have distance $4$ or more. The solution is given on the number line as:

$clipboard_ea020fa310cf3fcba8ba44ff547e18ed9.png$

In interval notation, the solution set is the interval $(-\infty,2]$ and $[10,\infty)$ , or in short it is the union of the two intervals:

$S= (\infty,2]\cup [10,\infty) \nonumber$

One can also write that the solution set consists of all $x$ such that $x\leq 2$ or $x\geq 10$ .

Example 1.4.1\PageIndex{1}

Note

Example 1.4.2\PageIndex{2}

Example 1.4.3\PageIndex{3}

Example 1.4.4\PageIndex{4}

Support Center

How can we help?

Example $\PageIndex{1}$

Example $\PageIndex{2}$

Example $\PageIndex{3}$

Example $\PageIndex{4}$