6.3: Solving Absolute Value Inequalities and Writing Answers in Interval Notation

Last updated
Save as PDF

Page ID: 45193

Victoria Dominguez, Cristian Martinez, & Sanaa Saykali
Citrus College via ASCCC Open Educational Resources Initiative

$ \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}}$

The previous section taught how to solve absolute value equations. This section teaches how to solve absolute value inequalities. To do so, first consider the following two properties:

Definition: Properties of Absolute Value Inequalities

Property 1: For all positive number $b$, and all real numbers $p$ and $q$,

$|a| < b$ if and only if $−b < a < b$.

The solution set is of the form $(p,q)$, a single open interval.

$|a| ≤ b$ if and only if $−b ≤ a ≤ b$.

The solution set is of the form $[p,q]$, a single closed interval.

Before considering Property 2, it is important to define the union of two intervals. The union of any two intervals $A$ and $B$, is the set of elements in $A$, or $B$, or both. The union is represented with the symbol $∪$.

Property 2: For all positive number $b$, and all real numbers $p$ and $q$,

$|a| > b$ if and only if $a < −b$ or $a > −b$

The solution set is of the form $(−∞, p) ∪ (q, ∞)$, a disjoint interval.

$|a| ≥ b$ if and only if $a ≤ −b$ or $a ≥ b$.

The solution set is of the form $(−∞, p] ∪ [q, ∞)$, a disjoint interval.

Note that before the properties of inequalities are applied, isolate the absolute value expression on either side of the inequality.

Example 6.3.1

Solve the following inequalities and graph the solution set.

$|5x − 2| < 7$
$|8x − 6| < −1$
$2|x − 3| + 5 ≤ 9$

Solution

This is an absolute value expression less than a positive number of the form $|a| < b$. Apply Property 1(i) with $a = 5x − 2$ and $b = 7$.

$\begin{array} &&|5x − 2| < 7 &\text{Given} \\ &−7 < 5x − 2 < 7 &\text{Property 1 (i)} \end{array}$

To solve the inequality, isolate $x$. The previous step becomes,

$\begin{array} &&−5 < 5x < 9 &\text{Add \(2$ to all sides} \\ &−1 < x < \dfrac{9}{5} &\text{Divide all sides by $5$} \end{array}\)

The Solution set is the single open interval $\left(−1, \dfrac{9}{5} \right)$ and the graph is as shown in the figure below.

$clipboard_eee2ec390ebb3481038f3625204438819.png$

Recall that the absolute value of any number is the distance from $0$ to that number on the number line. This means, the absolute value of any number is always greater or equal to $0$.

This example gives $|8x − 6| < −1,$ which can not happen since a distance is never negative. So, the absolute value inequality has no solution and the solution set is the empty set, written $\phi$.

To solve $2|x − 3| + 5 ≤ 9$, isolate the absolute value.

$\begin{array} &&2|x − 3| + 5 ≤ 9 &\text{Given} \\ &2|x − 3| ≤ 4 &\text{Subtract \(5$ from both sides} \\ &|x − 3| ≤ 2 &\text{Divide both side by $2$} \end{array}\)

Now, $|x − 3| ≤ 2$ is of the form $|a| ≤ b$. Apply Property 1 (ii) with $a = x − 3$ and $b = 2$.

$\begin{array} &&|x − 3| ≤ 2 & \\&− 2 ≤ x − 3 ≤ 2 &\text{Property 1(ii)} \\ &1 ≤ x ≤ 5 & \end{array}$

The solution set is the single interval $[1, 5]$ and the graph is as shown in the figure below.

$clipboard_e335bc0a4c68c53a86ca3273caf045e67.png$

Example 6.3.2

Solve and graph the solution set.

$\left| \dfrac{6 − x}{10} \right| ≥ 3$
$2 < \left|\dfrac{3}{4} x − 3 \right| − 5$
$|2 − 4x| ≥ −7$

Solution

The absolute value inequality $\left| \dfrac{6 − x}{10} \right| ≥ 3$ is in the form of $|a| ≥ b$. Apply Property 2 (ii) with $a = \dfrac{6 − x}{10}$ and $b = 3$ to solve the inequality.

$\begin{array} & & &\left| \dfrac{6 − x}{10} \right| ≥ 3 &&\text{Given} \\ &\dfrac{6 − x}{10} ≤ −3 &\text{ or } &\dfrac{6 − x}{10} ≥ 3 &\text{Property 2 (ii)} \\ &6 − x ≤ −30 &\text{ or } &6 − x ≥ 30 &\text{Multiply by \(10$ both sides} \\ &−x ≤ −36 &\text{ or } &−x ≥ 24 &\text{Subtract $6$ from both sides} \\ &x ≥ 36 &\text{ or } &x ≤ −24 &\text{Multiply by $−1$} \end{array}\)

Note that since the inequalities were multiplied by a negative number, namely $−1$, the direction of the inequality changed.

The solution set is the union of the two intervals. Thus, $(−∞, −24] ∪ [36, ∞)$ is the solution set in interval notation. The graph of the solution is as shown in the figure below.

$clipboard_e994fe946d0c11e329dc3c0829b29b123.png$

Isolate the absolute value.

$\begin{array} &&2 < \left|\dfrac{3}{4} x − 3 \right| − 5 &\text{Given} \\ &7 < \left| \dfrac{3}{4} x − 3 \right| &\text{Add \(5$ to both sides} \end{array}\)

Note that the above inequality is read from right to left as "the absolute value of the expression $\dfrac{3}{4} x − 3$ is greater than $7$" or equivalently switch the order of the absolute value inequality to have $\dfrac{3}{4} x − 3 > 7$, which is a more familiar form to solve.

Now, $\dfrac{3}{4} x − 3 > 7$ is of the form $|a| > b$. Use Property 2 (ii) with $a = \dfrac{3x}{4} − 3$ and $b = 7$.

$\begin{array} && &\dfrac{3}{4} x − 3 > 7 &&\text{Given} \\ &\dfrac{3}{4} x − 3 < −7 &\text{ or } &\dfrac{3}{4}x − 3 > 7 &\text{Property 2 (ii)}\\ &\dfrac{3}{4} x < −4 &\text{ or } &\dfrac{3}{4} x > 10 &\text{Add \(3$ to all sides} \\ &x < −\dfrac{16}{3} &\text{ or } &x > \dfrac{40}{3} &\text{Multiply both sides by $\dfrac{4}{3}$.} \end{array}\)

The solution set is the union of the two intervals, $(− ∞, −\dfrac{16}{3}] ∪ [\dfrac{40}{3}, ∞)$. The graph of the solution is as shown in the figure below

$clipboard_efb943e2ff7407d1b4777246c86c52bee.png$

Since $|2 − 4x|$ is always greater than or equal to $0$ for any real numbers $x$ then, the absolute value inequality is true for all real numbers. Let $x$ be any real number, negative or positive, then the absolute value will either be $0$ or a positive number.

So, the solution set is all the real numbers on the number line, as shown in the figure below. The solution set in interval notation is $(−∞, ∞)$.

$clipboard_e0e607828723541fb83638051b94dbe29.png$

Exercise 6.3.1

Solve the following inequalities, write answers in interval notation, and graph the solution sets:

$|−6x + 1| < 20$
$\left| \dfrac{2}{3} x + 5 \right| > 5$
$\left| 5 − \dfrac{1}{4} x \right| < −71$
$2 \left| − x + \dfrac{4}{5} \right| ≤ \dfrac{5}{2}$
$−\dfrac{1}{7} < |x + 10| − 10$
$|−12 − 3x| < −0.6$
$\left|\dfrac{16 − 2x}{8} \right| ≥ 11$
$|2 − 6x| − 5 ≥ −9$
$\left| \dfrac{2}{3} x − \dfrac{1}{4} \right| ≤ \dfrac{1}{12}$
$|.02x + 5| < .02$
$\left| \dfrac{1}{2} − x \right| < 8$
$| − 6x + 9| − 5 < −6$