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
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: $|x| \gt 12$
$x \gt 12$ 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: $|3x - 4| \le 7$
$$ \begin{align} 3x - 4 &\le 7 \\[5pt] 3x &\le 11 \\[5pt] x &\le \frac{11}{3} \end{align}$$
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: $|2x + 3| \ge 1$
$$ \begin{align} 2x + 3 &\ge 1 \\[5pt] 2x &\ge -2 \\[5pt] x &\ge -1 \end{align}$$
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: $-3|2x + 5| + 6 \ge -12$
$$ \begin{align} -3|2x + 5| + 6 &\ge -12 \\[5pt] -3|2x + 5| &\ge -18 \\[5pt] |2x + 5| &\le 6 \end{align}$$
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. |
$-2|5 - x| \lt -8$ Solutionrearrange: $|5 - x| \gt 4$ $$ \begin{align} \color{#E90F89}{\text{I.}} \: 5 - x &\gt 4 \\[5pt] -x &\gt -1 \\[5pt] x &\lt 1 \\[9pt] \color{#E90F89}{\text{II.}} \: x - 5 &\gt 4 \\[5pt] x &\gt 9 \end{align}$$
$x \in (-\infty, \; 1) \cup (9, \infty)$ |
| 9. |
$1 \lt \left| x - \frac{7}{3} \right| + \frac{11}{3}$ Solutionrearrange: $$ \begin{align} \frac{3 - 11}{3} &\lt \left|x - \frac{7}{3} \right| \\[5pt] -\frac{8}{3} &\lt \left|x - \frac{7}{3} \right| \end{align}$$ $$ \begin{align} \color{#E90F89}{\text{I.}} \: -\frac{8}{3} &\lt x - \frac{7}{3} \\[5pt] -\frac{1}{3} &\lt x \\[9pt] \color{#E90F89}{\text{II.}} \: -\frac{8}{3} &\lt \frac{7}{3} - x \\[5pt] -\frac{15}{3} &\lt -x \\[5pt] 5 &\gt x \end{align}$$
$x \in \left(-\frac{1}{3}, \; 5 \right)$ |
| 10. |
$4 + \left|3 - \frac{x}{2} \right| \ge 9$ Solutionrearrange 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. |
$|3x - 4| + 9 \gt 5$ Solutionrearrange 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. |
$3 + 4|3x + 7| \ge -89$ Solutionrearrange 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. |
$4|6 - 3x| + 8 \le 24$ Solutionrearrange 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. |
$7 - |2x - 5| \gt 4$ Solutionrearrange 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. |
$3 - 2|3x - 1| \ge -7$ Solutionrearrange 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. |
$4 - 6|-6 + 3x| \le -4$ Solutionrearrange 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. |
$6 - 3|1 - 4x| \lt -2$ Solutionrearrange 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. |
$\frac{|3 + x|}{7} \le 5$ Solutionrearrange 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. |
$\frac{|2 + 3x|}{3} \ge 4$ Solutionrearrange 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. |
$\left|x - \frac{1}{2} \right| - 5 \lt -1$ Solutionrearrange to: $$\left| x - \frac{1}{2} \right| \lt 4$$ $$ \begin{align} \color{#E90F89}{\text{I.}} \: x - \frac{1}{2} &\lt 4\\[5pt] x &\lt \frac{9}{2} \\[9pt] \color{#E90F89}{\text{II.}} \: \frac{1}{2} - x &\lt 4 \\[5pt] -x &\lt \frac{7}{2} \\[5pt] x &\gt -\frac{7}{2} \end{align}$$
$x \in \left(-\frac{7}{2}, \; \frac{9}{2} \right)$ |
Caveat
A caveat is a warning that some specific condition(s) or exception(s) might apply to a statement.
For example, Bovines produce most of the milk for human consumption in the U.S., with the caveat that milk is produced only by the females (cows).
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