Absolute-value
Inequalities

The absolute-value function

The absolute value of a quantity is represented by surrounding it with vertical bars. Some examples are:

\begin{align} |3| &= 3 \\[5pt] |-3| &= 3 \\[5pt] |(-1)| &= 1 \end{align}

\begin{align} |7x| &= 7|x| \\[5pt] |-7x| &= 7|x| \\[5pt] |-0.2| &= 0.2 \end{align}

$f(x) = |x|$ is a function because it assigns to each x one and only one y value. Here is a graph of the function $y = |x|$ or $f(x) = |x|.$

Notice that $f(-1) = f(1) = 1.$ If you're not too familiar with the language of functions, just notice that the absolute value bars "cancel" any negative signs inside.

Solving absolute-value equations

Consider the absolute value equation

$$|x| = 8$$

There are two solutions, x = 8 and x = -8, because the absolute value function will convert either to positive 8.   So $x = ±8.$

Let's take a look at another one:

$$|x| = -5$$

This equation has no solution because the absolute value function can "return" only positive values, so the result of -5 is impossible.

We use the same method to solve more complicated absolute-value equalities. For example,

$$|x - 4| = 11$$

To solve such an equation, we notice that both $(x - 4)$ and $-(x - 4)$ must solve it, so we have $x - 4 = 11$ and $-x + 4 = 11,$ giving $x = 15$ and $x = -7.$

Absolute-value inequalities

We solve absolute-value inequalities just like equalities, except for the usual caveat that if we multiply or divide by a negative number, we must change the direction of the inequality symbol, < ↔ >,   ≤ ↔ ≥.

The best way to learn about absolute-value inequalities is by example, so here are a few examples, with some practice problems (with solutions) at the end.

Example 1

Solve:   $|x| \gt 12$

Solution: We begin by dividing this inequality into two:

$x \gt 12$   and   $-x \gt 12.$

The first inequality is straightforward. For the second, we divide (or multiply) by -1, remembering to switch the inequality sign, to get $x \lt -12.$

We can sketch a graph of this solution like this:

We can sketch a graph of this solution like this:

The set of possible solutions is the union (collection) of two sets. We write it as

$$x \in (-\infty, -12) \cup (12, \infty).$$

Example 2

Solve:   $|3x - 4| \le 7$

Solution: We can separate this inequality into two separate ones, $3x - 4 \le 7$   and   $-(3x - 4) \le 7.$ Now to solve each of these:

\begin{align} 3x - 4 &\le 7 \\[5pt] 3x &\le 11 \\[5pt] x &\le \frac{11}{3} \end{align}

The second is

\begin{align} -(3x - 4) &\le 7 \\[5pt] -3x + 4 &\le 7 \\[5pt] -3x &\le 3 \\[5pt] x &\ge -1 \end{align}

The number line for this solution like this:

Notice that the solutions to both equations are in the region where the individual solutions overlap. The set notation for this set of solutions is

$$x \in \left[-1, \; \frac{11}{3} \right]$$

Remember that square brackets mean that the endpoints are included in the solution set.

Example 3

Solve:   $|2x + 3| \ge 1$

Solution: First separate this inequality into two separate ones, $2x + 3 \ge 1$   and   $-(2x + 3) \ge 1.$ Now to solve each of these. The first is:

\begin{align} 2x + 3 &\ge 1 \\[5pt] 2x &\ge -2 \\[5pt] x &\ge -1 \end{align}

The second is:

\begin{align} -(2x + 3) &\ge 1 \\[5pt] 2x + 3 &\le -1 \\[5pt] 2x &\le -4 \\[5pt] x &\le -2 \end{align}

The number line for this solution like this:

The set notation for this set of solutions is

$$x \in (-\infty, \; -2] \cup [-1, \; \infty)$$

Remember that square brackets mean that the endpoints are included in the solution set. We use round brackets on ∞ limits because ∞ isn't really a number that could be included.

Example 4

Solve:   $-3|2x + 5| + 6 \ge -12$

Solution: First we need to do a little algebra on this one to isolate the absolute value expression:

\begin{align} -3|2x + 5| + 6 &\ge -12 \\[5pt] -3|2x + 5| &\ge -18 \\[5pt] |2x + 5| &\le 6 \end{align}

Now we can separate that inequality into two. You can think of this as "resolving" the absolute value. The first is

\begin{align} 2x + 5 &\le 6 \\[5pt] 2x &\le 1 \\[5pt] x &\le \frac{1}{2} \end{align}

The second is:

\begin{align} -(2x + 5) &\le 6 \\[5pt] 2x + 5 &\ge -6 \\[5pt] 2x &\ge -11 \\[5pt] x &\ge -\frac{11}{2} \end{align}

The number line for this solution like this:

The set notation for this set of solutions is

$$x \in \left[-\frac{11}{2}, \; \frac{1}{2}\right]$$

Practice problems

Solve the inequality for x, show the set of solutions on a number line and write the solutions in set notation.

 1 $|x| \lt 5$ Solution \begin{align} \color{#E90F89}{\text{I.}} \: x &\lt 5 \\[9pt] \color{#E90F89}{\text{II.}} \: -x &\lt 5 \\[5pt] x &\gt -5 \end{align} $x \in (-5, \; 5)$ 2 $|x + 2| \le 4$ Solution \begin{align} \color{#E90F89}{\text{I.}} \: x + 2 &\le 4 \\[5pt] x &\le 2 \\[9pt] \color{#E90F89}{\text{II.}} \: -(x + 2) &\le 4 \\[5pt] x + 2 &\ge -4 \\[5pt] x &\ge -6 \end{align} $x \in [-6, \; 2]$ 3 $|3x + 7| \le 19$ Solution \begin{align} \color{#E90F89}{\text{I.}} \: 3x + 7 &\le 19 \\[5pt] 3x &\le 12 \\[5pt] x &\le 4 \\[9pt] \color{#E90F89}{\text{II.}} \: -(3x + 7) &\le 19 \\[5pt] 3x + 7 &\ge -19 \\[5pt] 3x &\ge -26 \\[5pt] x &\ge \frac{-26}{3} \approx -9 \end{align} $x \in [-9, \; 4]$ 4 $|5x - 2| \gt 19$ Solution \begin{align} \color{#E90F89}{\text{I.}} \: 5x - 2 &\gt 19 \\[5pt] 5x &\gt 21 \\[5pt] x &\gt \frac{21}{5} \\[9pt] \color{#E90F89}{\text{II.}} \: 2 - 5x &\gt 19 \\[5pt] -5x &\gt 17 \\[5pt] x &\lt \frac{-17}{5} \end{align} Note that $2 - 5x = -(5x - 2).$ $x \in (-\infty, \; -\frac{17}{5}) \cup (\frac{21}{5}, \; \infty)$ 5 $\left| \frac{3x - 2}{9} \right| \ge 1$ Solution \begin{align} \color{#E90F89}{\text{I.}} \: \frac{3x - 2}{9} &\ge 1 \\[5pt] 3x - 2 &\ge 9 \\[5pt] 3x &\ge 11 \\[5pt] x &\ge \frac{11}{3} \\[9pt] \color{#E90F89}{\text{II.}} \: \frac{-(3x - 2)}{9} &\gt 1 \\[5pt] -(3x - 2) &\ge 9 \\[5pt] 3x - 2 &\le -9 \\[5pt] 3x &\le -7 \\[5pt] x &\lt \frac{-7}{3} \end{align} $x \in (-\infty, \; -\frac{7}{3}] \cup [\frac{11}{3}, \; \infty)$ 6 $\left| 3 - \frac{3}{4} x \right| \ge 9$ Solution \begin{align} \color{#E90F89}{\text{I.}} \: 3 - \frac{3}{4}x &\ge 9 \\[5pt] -\frac{3}{4}x &\ge 6 \\[5pt] x &\le \frac{6 \cdot 4}{-3} \\[5pt] x &\le -8 \\[9pt] \color{#E90F89}{\text{II.}} \: \frac{3}{4}x - 3 &\ge 9 \\[5pt] \frac{4}{4}x &\ge 12 \\[5pt] x &\ge \frac{12 \cdot 4}{3} \\[5pt] x &\ge 16 \end{align} $x \in (-\infty, \; -8] \cup [16, \infty)$ 7 $5|2x + 3| -3 \ge 8$ First isolate the absolute value expression: \begin{align} 5|2x + 3| -3 &\ge 8 \\[5pt] 5|2x + 3| &\ge 11 \\[5pt] |2x + 3| &\ge \frac{11}{5} \end{align} Solution \begin{align} \color{#E90F89}{\text{I.}} \: 2x + 3 &\ge \frac{11}{5}\\[5pt] 2x &\ge \frac{11-15}{5} \\[5pt] x &\ge \frac{-2}{5} \\[9pt] \color{#E90F89}{\text{II.}} \: -(2x + 3) &\ge \frac{11}{5}\\[5pt] 2x + 3 &\le -\frac{11}{5} \\[5pt] 2x &\le \frac{-11-15}{5} \\[5pt] 2x &\le \frac{-26}{5} \\[5pt] x &\le \frac{-13}{5} \end{align} $x \in (-\infty, \; -\frac{13}{5}] \cup [-\frac{2}{5}, \infty)$ 8 $-2|5 - x| \lt -8$ Solution rearrange:   $|5 - x| \gt 4$ \begin{align} \color{#E90F89}{\text{I.}} \: 5 - x &\gt 4 \\[5pt] -x &\gt -1 \\[5pt] x &\lt 1 \\[9pt] \color{#E90F89}{\text{II.}} \: x - 5 &\gt 4 \\[5pt] x &\gt 9 \end{align} $x \in (-\infty, \; 1) \cup (9, \infty)$ 9 $1 \lt \left| x - \frac{7}{3} \right| + \frac{11}{3}$ Solution rearrange: \begin{align} \frac{3 - 11}{3} &\lt \left|x - \frac{7}{3} \right| \\[5pt] -\frac{8}{3} &\lt \left|x - \frac{7}{3} \right| \end{align} \begin{align} \color{#E90F89}{\text{I.}} \: -\frac{8}{3} &\lt x - \frac{7}{3} \\[5pt] -\frac{1}{3} &\lt x \\[9pt] \color{#E90F89}{\text{II.}} \: -\frac{8}{3} &\lt \frac{7}{3} - x \\[5pt] -\frac{15}{3} &\lt -x \\[5pt] 5 &\gt x \end{align} $x \in \left(-\frac{1}{3}, \; 5 \right)$ 10 $4 + \left|3 - \frac{x}{2} \right| \ge 9$ Solution rearrange to: $$\left|3 - \frac{x}{2} \right| \ge 5$$ \begin{align} \color{#E90F89}{\text{I.}} \: 3 - \frac{x}{2} &\ge 5 \\[5pt] -\frac{x}{2} &\ge 2 \\[5pt] x &\le -4 \\[9pt] \color{#E90F89}{\text{II.}} \: \frac{x}{2} - 3 \ge 5 \\[5pt] \frac{x}{2} \ge 8 \\[5pt] x &\ge 16 \end{align} $x \in (-\infty, \; -4] \cup [16, \; \infty)$

 11 $|3x - 4| + 9 \gt 5$ Solution rearrange to: $$|3x - 4| \gt -4$$ \begin{align} \color{#E90F89}{\text{I.}} \: 3x - 4 &\gt -4 \\[5pt] 3x &\gt 0 \\[5pt] x &\gt 0 \\[9pt] \color{#E90F89}{\text{II.}} \: 4 - 3x &\gt -4 \\[5pt] -3x &\gt -8 \\[5pt] x &\lt \frac{8}{3} \end{align} $x \in \left(0, \; \frac{8}{3} \right)$ 12 $3 + 4|3x + 7| \ge -89$ Solution rearrange to: $$|3x + 7| \ge \frac{-89 - 3}{4} = 23$$ \begin{align} \color{#E90F89}{\text{I.}} \: 3x + 7 &\ge 23 \\[5pt] 3x &\ge 16 \\[5pt] x &\ge \frac{16}{3} \\[9pt] \color{#E90F89}{\text{II.}} \: -(3x + 7) &\ge 23 \\[5pt] 3x + 7 &\le -23 \\[5pt] 3x &\le -30 \\[5pt] x &\le -10 \end{align} $x \in (-\infty, \; 10] \cup \left[\frac{16}{3}, \; \infty \right]$ 13 $4|6 - 3x| + 8 \le 24$ Solution rearrange to: $$|6 - 3x| \le 4$$ \begin{align} \color{#E90F89}{\text{I.}} \: 6 - 3x &\le 4 \\[5pt] -3x &\le -2 \\[5pt] x &\ge \frac{2}{3} \\[9pt] \color{#E90F89}{\text{II.}} \: 3x - 6 &\le 4 \\[5pt] 3x &\le 10 \\[5pt] x &\le \frac{10}{3} \end{align} $x \in \left[\frac{2}{3}, \; \frac{10}{3} \right]$ 14 $7 - |2x - 5| \gt 4$ Solution rearrange to: $$|2x - 5| \lt 3$$ \begin{align} \color{#E90F89}{\text{I.}} \: 2x - 5 &\lt 3 \\[5pt] 2x &\lt 8 \\[5pt] x &\lt 4 \\[9pt] \color{#E90F89}{\text{II.}} \: 5 - 2x &\lt 3 \\[5pt] -2x &\lt -2 \\[5pt] x &\gt 1 \end{align} $x \in \left[\frac{2}{3}, \; \frac{10}{3} \right]$ 15 $3 - 2|3x - 1| \ge -7$ Solution rearrange to: $$|3x - 1| \le 5$$ \begin{align} \color{#E90F89}{\text{I.}} \: 3x - 1 &\le 5 \\[5pt] 3x &\le 6 \\[5pt] x &\le 2 \\[9pt] \color{#E90F89}{\text{II.}} \: 1 - 3x &\le 5 \\[5pt] -3x &\le 4 \\[5pt] x &\ge -\frac{4}{3} \end{align} $x \in \left[-\frac{4}{3}, \; 2 \right]$ 16 $4 - 6|-6 + 3x| \le -4$ Solution rearrange to: $$|3x - 6| \ge \frac{4}{3}$$ \begin{align} \color{#E90F89}{\text{I.}} \: 3x - 6 &\ge \frac{4}{3} \\[5pt] 3x &\ge \frac{22}{3} \\[5pt] x &\ge \frac{22}{9} \\[9pt] \color{#E90F89}{\text{II.}} \: 6 - 3x &\ge \frac{4}{3} \\[5pt] -3x &\ge -\frac{14}{3} \\[5pt] x &\le \frac{14}{9} \end{align} $x \in \left(-\infty, \; \frac{14}{9}\right] \cup \left[ \frac{22}{9}, \; \infty \right)$ 17 $6 - 3|1 - 4x| \lt -2$ Solution rearrange to: $$|1 - 4x| \gt \frac{8}{3}$$ \begin{align} \color{#E90F89}{\text{I.}} \: 1 - 4x &\gt \frac{8}{3} \\[5pt] -4x &\gt \frac{5}{3} \\[5pt] x &\lt -\frac{5}{12} \\[9pt] \color{#E90F89}{\text{II.}} \: 4x - 1 &\gt \frac{8}{3} \\[5pt] 4x &\gt \frac{11}{3} \\[5pt] x &\gt \frac{11}{12} \end{align} $x \in \left(-\infty, \; -\frac{5}{12}\right) \cup \left( \frac{11}{12}, \; \infty \right)$ 18 $\frac{|3 + x|}{7} \le 5$ Solution rearrange to: $$|3 + x| \le 5$$ \begin{align} \color{#E90F89}{\text{I.}} \: 3 + x &\le 5 \\[5pt] x &\le 2 \\[9pt] \color{#E90F89}{\text{II.}} \: -(3 + x) &\gt 5 \\[5pt] 3 + x &\ge -5 \\[5pt] x &\ge -8 \end{align} $x \in [-8, \; 2]$ 19 $\frac{|2 + 3x|}{3} \ge 4$ Solution rearrange to: $$|2 + 3x| \ge 12$$ \begin{align} \color{#E90F89}{\text{I.}} \: 2 + 3x &\ge 12 \\[5pt] 3x &\ge 10 \\[5pt] x &\ge \frac{10}{3} \\[9pt] \color{#E90F89}{\text{II.}} \: -(2 + 3x) &\ge 12 \\[5pt] 2 + 3x &\le -12 \\[5pt] 3x &\le -14 \\[5pt] x &\le -\frac{14}{3} \end{align} $x \in \left(-\infty, \; -\frac{14}{3}\right] \cup \left[ \frac{10}{3}, \; \infty \right)$ 20 $\left|x - \frac{1}{2} \right| - 5 \lt -1$ Solution rearrange to: $$\left| x - \frac{1}{2} \right| \lt 4$$ \begin{align} \color{#E90F89}{\text{I.}} \: x - \frac{1}{2} &\lt 4\\[5pt] x &\lt \frac{9}{2} \\[9pt] \color{#E90F89}{\text{II.}} \: \frac{1}{2} - x &\lt 4 \\[5pt] -x &\lt \frac{7}{2} \\[5pt] x &\gt -\frac{7}{2} \end{align} $x \in \left(-\frac{7}{2}, \; \frac{9}{2} \right)$

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