15.7 Absolute Value Equations and Inequalities

In this section, we review some of the basic concepts involving the absolute value of a real number $x$ . There are a few different ways to define absolute value and in this section we choose the following definition. (Absolute value will be revisited in much greater depth in Section 1.3 where we present what one can think of as the “precise” definition.)

For example, $vertical bar 5 vertical bar equals 5$ and $| - 5 | = 5$ , since each is $5$ units from $0$ on the number line:

Coordinate-plane figure. — **Figure 15.100**

Graphically why $| - 5 | = 5$ and $vertical bar 5 vertical bar equals 5$

Computationally, the absolute value `makes negative numbers positive', though we need to be a little cautious with this description. While $| - 7 | = 7$ , $| 5 - 7 | \neq 5 + 7$ . The absolute value acts as a grouping symbol, so $| 5 - 7 | = | - 2 | = 2$ , which makes sense since $5$ and $7$ are two units away from each other on the number line:

Graphically why $| 5 - 7 | = 2$

We list some of the operational properties of absolute value below.

Theorem

Properties of Absolute Value: Let $a$ and $b$ be real numbers and let $n$ be an integer.¹

Product Rule: $| a b | = | a | | b |$
Power Rule: $| a^{n} | = {| a |}^{n}$ whenever $a^{n}$ is defined
Quotient Rule: $| \frac{a}{b} | = \frac{| a |}{| b |}$ , provided $b not equal to 0$

The proof of Theorem A.4 is difficult, but not impossible, using the distance definition of absolute value or even the `it makes negatives positive' notion. It is, however, much easier if one uses the “precise” definition given in Section 1.3 so we will revisit the proof then. For now, let's focus on how to solve basic equations and inequalities involving the absolute value.

Absolute Value Equations

Thinking of absolute value in terms of distance gives us a geometric way to interpret equations. For example, to solve $vertical bar x vertical bar equals 3$ , we are looking for all real numbers $x$ whose distance from $0$ is $3$ units. If we move three units to the right of $0$ , we end up at $x equals 3$ . If we move three units to the left, we end up at $x = - 3$ . Thus the solutions to $vertical bar x vertical bar equals 3$ are $x = \pm 3$ .

The solutions to $vertical bar x vertical bar equals 3$ are $x = \pm 3$ .

Thinking this way gives us the following.

Theorem

Absolute Value Equations: Suppose $x$ , $y$ and $c$ are real numbers.

$vertical bar x vertical bar equals 0$ if and only if $x equals 0$ .
For $c greater than 0$ , $| x | = c$ if and only if $x = c$ or $x = - c$ .
For $c less than 0$ , $| x | = c$ has no solution.
$| x | = | y |$ if and only if $x = y$ or $x = - y$ .
(That is, if two numbers have the same absolute values, they are either the same number or exact opposites of each other.)

Theorem A.5 is our main tool in solving equations involving the absolute value, since it allows us a way to rewrite such equations as compound linear equations.

Strategy for Solving Equations Involving Absolute Value

In order to solve an equation involving the absolute value of a quantity $| X |$ :

Isolate the absolute value on one side of the equation so it has the form $| X | = c$ .
Apply Theorem A.5.

The techniques we use to `isolate the absolute value' are precisely those we used in Section A.4 to isolate the variable when solving linear equations. Time for some practice.

Example 1

Solve each of the following equations.

$| 3 x - 1 | = 6$
$\frac{3 - | y + 5 |}{2} = 1$
$3 | 2 t + 1 | - \sqrt{5} = 0$
$4 - | 5 w + 3 | = 5$
$| 3 - x \sqrt[3]{12} | = | 4 x + 1 |$
$| t - 1 | - 3 | t + 1 | = 0$

The equation $| 3 x - 1 | = 6$ is of already in the form $| X | = c$ , so we know that either $3 x - 1 = 6$ or $3 x - 1 = - 6$ . Solving the former gives us at $x = \frac{7}{3}$ and solving the latter yields $x = - \frac{5}{3}$ . We may check both of these solutions by substituting them into the original equation and showing that the arithmetic works out.
We begin solving $\frac{3 - | y + 5 |}{2} = 1$ by isolating the absolute value to put it in the form $| X | = c$ .
$\begin{matrix} \frac{3 - | y + 5 |}{2} & = & 1 \\ 3 - | y + 5 | & = & 2 & Multiply by 2 \\ - | y + 5 | & = & - 1 & Subtract 3 \\ | y + 5 | & = & 1 & Divide by - 1 \end{matrix}$
At this point, we have $y + 5 = 1$ or $y + 5 = - 1$ , so our solutions are $y = - 4$ or $y = - 6$ . We leave it to the reader to check both answers in the original equation.
As in the previous example, we first isolate the absolute value. Don't let the $the square root of 5$ throw you off - it's just another real number, so we treat it as such:
$\begin{matrix} 3 | 2 t + 1 | - \sqrt{5} & = & 0 \\ 3 | 2 t + 1 | & = & \sqrt{5} & Add \sqrt{5} \\ | 2 t + 1 | & = & \frac{\sqrt{5}}{3} & Divide by 3 \end{matrix}$
From here, we have that $2 t + 1 = \frac{\sqrt{5}}{3}$ or $2 t + 1 = - \frac{\sqrt{5}}{3}$ . The first equation gives $t = \frac{\sqrt{5} - 3}{6}$ while the second gives $t = \frac{- \sqrt{5} - 3}{6}$ thus we list our answers as $t = \frac{- 3 \pm \sqrt{5}}{6}$ . The reader should enjoy the challenge of substituting both answers into the original equation and following through the arithmetic to see that both answers work.
Upon isolating the absolute value in the equation $4 - | 5 w + 3 | = 5$ , we get $| 5 w + 3 | = - 1$ . At this point, we know there cannot be any real solution. By definition, the absolute value is a distance, and as such is never negative. We write `no solution' and carry on.
Our next equation already has the absolute value expressions (plural) isolated, so we work from the principle that if $| x | = | y |$ , then $x = y$ or $x = - y$ . Thus from $| 3 - x \sqrt[3]{12} | = | 4 x + 1 |$ we get two equations to solve:
$3 - x \sqrt[3]{12} = 4 x + 1, and 3 - x \sqrt[3]{12} = - (4 x + 1)$
Notice that the right side of the second equation is $- (4 x + 1)$ and not simply $- 4 x + 1$ . Remember, the expression $4 ⁢ x plus 1$ represents a single real number so in order to negate it we need to negate the entire expression $- (4 x + 1)$ . Moving along, when solving $3 - x \sqrt[3]{12} = 4 x + 1$ , we obtain $x = \frac{2}{4 + \sqrt[3]{12}}$ and the solution to $3 - x \sqrt[3]{12} = - (4 x + 1)$ is $x = \frac{4}{\sqrt[3]{12} - 4}$ . As usual, the reader is invited to check these answers by substituting them into the original equation.
We start by isolating one of the absolute value expressions: $| t - 1 | - 3 | t + 1 | = 0$ gives $| t - 1 | = 3 | t + 1 |$ . While this resembles the form $| x | = | y |$ , the coefficient $3$ in $3 ⁢ vertical bar t plus 1 vertical bar$ prevents it from being an exact match. Not to worry - since $3$ is positive, $3 equals vertical bar 3 vertical bar$ so
$3 | t + 1 | = | 3 | | t + 1 | = | 3 (t + 1) | = | 3 t + 3 | .$
Hence, our equation becomes $| t - 1 | = | 3 t + 3 |$ which results in the two equations: $t - 1 = 3 t + 3$ and $t - 1 = - (3 t + 3)$ . The first equation gives $t = - 2$ and the second gives $t = - \frac{1}{2}$ . The reader is encouraged to check both answers in the original equation.

Absolute Value Inequalities

We now turn our attention to solving some basic inequalities involving the absolute value. Suppose we wished to solve $vertical bar x vertical bar less than 3$ . Geometrically, we are looking for all of the real numbers whose distance from $0$ is less than $3$ units. We get $minus 3 less than x less than 3$ , or in interval notation, $open parenthesis minus 3, 3 close parenthesis$ . Suppose we are asked to solve $vertical bar x vertical bar greater than 3$ instead. Now we want the distance between $x$ and $0$ to be greater than $3$ units. Moving in the positive direction, this means $x greater than 3$ . In the negative direction, this puts $x less than minus 3$ . Our solutions would then satisfy $x less than minus 3$ or $x greater than 3$ . In interval notation, we express this as $(- \infty, - 3) \cup (3, \infty)$ .

The solution to $vertical bar x vertical bar less than 3$ is $open parenthesis minus 3, 3 close parenthesis$

The solution to $vertical bar x vertical bar greater than 3$ is $(- \infty, - 3) \cup (3, \infty)$

Generalizing this notion, we get the following:

Theorem

Inequalities Involving Absolute Value: Let $c$ be a real number.

If $c greater than 0$ , $| x | < c$ is equivalent to $- c < x < c$ .
If $c less than or equal to 0$ , $| x | < c$ has no solution.
If $c greater than 0$ , $| x | > c$ is equivalent to $x < - c$ or $x > c$ .
If $c less than or equal to 0$ , $| x | > c$ is true for all real numbers.

If the inequality we're faced with involves ` $\leq$ ' or ` $\geq$ ,' we can combine the results of Theorem A.6 with Theorem A.5 as needed.

Strategy for Solving Inequalities Involving Absolute Value

In order to solve an inequality involving the absolute value of a quantity $| X |$ :

Isolate the absolute value on one side of the inequality.
Apply Theorem A.6.

Example 2

Solve the following inequalities.

$| x - \sqrt[4]{5} | > 1$
$\frac{4 - 2 | 2 x + 1 |}{4} \geq - \sqrt{3}$
$| 2 x - 1 | \leq 3 | 4 - 8 x | - 10$
$| 2 x - 1 | \leq 3 | 4 - 8 x | + 10$
$2 less than vertical bar x minus 1 vertical bar less than or equal to 5$
$| 10 x - 5 | + | 10 - 5 x | \leq 0$

From Theorem A.6, $| x - \sqrt[4]{5} | > 1$ is equivalent to $x - \sqrt[4]{5} < - 1$ or $x - \sqrt[4]{5} > 1$ . Solving this compound inequality, we get $x < - 1 + \sqrt[4]{5}$ or $x > 1 + \sqrt[4]{5}$ . Our answer, in interval notation, is: $(- \infty, - 1 + \sqrt[4]{5}) \cup (1 + \sqrt[4]{5}, \infty)$ . As with linear inequalities, we can only partially check our answer by selecting values of $x$ both inside and outside of the solution intervals to see which values of $x$ satisfy the original inequality and which do not.
Our first step in solving $\frac{4 - 2 | 2 x + 1 |}{4} \geq - \sqrt{3}$ is to isolate the absolute value.
$\begin{matrix} \frac{4 - 2 | 2 x + 1 |}{4} & \geq & - \sqrt{3} \\ 4 - 2 | 2 x + 1 | & \geq & - 4 \sqrt{3} & Multiply by 4 \\ - 2 | 2 x + 1 | & \geq & - 4 - 4 \sqrt{3} & Subtract 4 \\ | 2 x + 1 | & \leq & \frac{- 4 - 4 \sqrt{3}}{- 2} & Divide by - 2, reverse the inequality \\ | 2 x + 1 | & \leq & 2 + 2 \sqrt{3} & Reduce \end{matrix}$
Since we're dealing with ` $\leq$ ' instead of just ` $<$ ,' we can combine Theorems A.6 and A.5 to rewrite this last inequality as:² $- (2 + 2 \sqrt{3}) \leq 2 x + 1 \leq 2 + 2 \sqrt{3}$ . Subtracting the ` $1$ ' across both inequalities gives $- 3 - 2 \sqrt{3} \leq 2 x \leq 1 + 2 \sqrt{3}$ , which reduces to $\frac{- 3 - 2 \sqrt{3}}{2} \leq x \leq \frac{1 + 2 \sqrt{3}}{2}$ . In interval notation this reads as $[\frac{- 3 - 2 \sqrt{3}}{2}, \frac{1 + 2 \sqrt{3}}{2}]$ .
There are two absolute values in $| 2 x - 1 | \leq 3 | 4 - 8 x | - 10$ , so we cannot directly apply Theorem A.6 here. Notice, however, that $| 4 - 8 x | = | (- 4) (2 x - 1) |$ . Using this, we get:
$\begin{matrix} | 2 x - 1 | & \leq & 3 | 4 - 8 x | - 10 \\ | 2 x - 1 | & \leq & 3 | (- 4) (2 x - 1) | - 10 & Factor \\ | 2 x - 1 | & \leq & 3 | - 4 | | 2 x - 1 | - 10 & Product Rule \\ | 2 x - 1 | & \leq & 12 | 2 x - 1 | - 10 \\ - 11 | 2 x - 1 | & \leq & - 10 & Subtract 12 | 2 x - 1 | \\ | 2 x - 1 | & \geq & \frac{10}{11} & Divide by - 11 and reduce \end{matrix}$
Now we are allowed to invoke Theorems A.5 and A.6 and write the equivalent compound inequality: $2 x - 1 \leq - \frac{10}{11}$ or $2 x - 1 \geq \frac{10}{11}$ . We get $x \leq \frac{1}{22}$ or $x \geq \frac{21}{22}$ , which when written with interval notation becomes $(- \infty, \frac{1}{22}] \cup [\frac{21}{22}, \infty)$ .
The inequality $| 2 x - 1 | \leq 3 | 4 - 8 x | + 10$ differs from the previous example in exactly one respect: on the right side of the inequality, we have ` $plus 10$ ' instead of ` $minus 10$ .' The steps to isolate the absolute value here are identical to those in the previous example, but instead of obtaining $| 2 x - 1 | \geq \frac{10}{11}$ as before, we obtain $| 2 x - 1 | \geq - \frac{10}{11}$ . This latter inequality is always true. (Absolute value is, by definition, a distance and hence always $0$ or greater.) Thus our solution to this inequality is all real numbers.
To solve $2 less than vertical bar x minus 1 vertical bar less than or equal to 5$ , we rewrite it as the compound inequality: $2 less than vertical bar x minus 1 vertical bar$ and $vertical bar x minus 1 vertical bar less than or equal to 5$ . The first inequality, $2 less than vertical bar x minus 1 vertical bar$ , can be re-written as $vertical bar x minus 1 vertical bar greater than 2$ so it is equivalent to $x - 1 < - 2$ or $x minus 1 greater than 2$ . Thus the solution to $2 less than vertical bar x minus 1 vertical bar$ is $x less than minus 1$ or $x greater than 3$ , which in interval notation is $(- \infty, - 1) \cup (3, \infty)$ . For $vertical bar x minus 1 vertical bar less than or equal to 5$ , we combine the results of Theorems A.5 and A.6 to get $- 5 \leq x - 1 \leq 5$ so that $minus 4 less than or equal to x less than or equal to 6$ , or $open bracket minus 4, 6 close bracket$ .
Our solution to $2 less than vertical bar x minus 1 vertical bar less than or equal to 5$ is comprised of values of $x$ which satisfy both parts of the inequality, so we intersect $(- \infty, - 1) \cup (3, \infty)$ with $open bracket minus 4, 6 close bracket$ to get our final answer $[- 4, - 1) \cup (3, 6]$ .
Our first hope when encountering $| 10 x - 5 | + | 10 - 5 x | \leq 0$ is that we can somehow combine the two absolute value quantities as we'd done in earlier examples. We leave it to the reader to show, however, that no matter what we try to factor out of the absolute value quantities, what remains inside the absolute values will always be different.
At this point, we take a step back and look at the equation in a more general way: we are adding two absolute values together and wanting the result to be less than or equal to $0$ . The absolute value of anything is always $0$ or greater, so there are no solutions to: $| 10 x - 5 | + | 10 - 5 x | < 0$ .
Is it possible that $| 10 x - 5 | + | 10 - 5 x | = 0$ ? Only if there is an $x$ where $| 10 x - 5 | = 0$ and $| 10 - 5 x | = 0$ at the same time.³ The first equation holds only when $x = \frac{1}{2}$ , while the second holds only when $x equals 2$ . Alas, we have no solution.⁴

The astute reader will have noticed by now that the authors have done nothing in the way of explaining why anyone would ever need to know this stuff. Go back and read the New Preface and the introduction to the Appendix. These sections are designed to review skills and concepts that you've already learned. Thus the deeper applications are in the main body of the text as opposed to here in the Appendix.

We close this section with an example of how the properties in Theorem A.4 are used in Calculus. Here, ` $ε$ ' is the Greek letter `epsilon' and it represents a positive real number. Those of you who will be taking Calculus in the future should become very familiar with this type of algebraic manipulation.

$\begin{matrix} | \frac{8 - 4 x}{3} | & < & ε \\ \frac{| 8 - 4 x |}{| 3 |} & < & ε & Quotient Rule \\ \frac{| - 4 (x - 2) |}{3} & < & ε & Factor \\ \frac{| - 4 | | x - 2 |}{3} & < & ε & Product Rule \\ \frac{4 | x - 2 |}{3} & < & ε \\ \frac{3}{4} \cdot \frac{4 | x - 2 |}{3} & < & \frac{3}{4} \cdot ε & Multiply by \frac{3}{4} \\ | x - 2 | & < & \frac{3}{4} ε \end{matrix}$

Exercises

In Exercises -, solve the equation.

$vertical bar x vertical bar equals 6$
Answer
$x = - 6$ or $x equals 6$
$| 3 t - 1 | = 10$
Answer
$t = - 3$ or $t = \frac{11}{3}$
$| 4 - w | = 7$
Answer
$w = - 3$ or $w equals 11$
$4 - | y | = 3$
Answer
$y = - 1$ or $y equals 1$
$2 | 5 m + 1 | - 3 = 0$
Answer
$m = - \frac{1}{2}$ or $m = \frac{1}{10}$
$| 7 x - 1 | + 2 = 0$
Answer
No solution
$\frac{5 - | x |}{2} = 1$
Answer
$x = - 3$ or $x equals 3$
$\frac{2}{3} | 5 - 2 w | - \frac{1}{2} = 5$
Answer
$w = - \frac{13}{8}$ or $w = \frac{53}{8}$
$| 3 t - \sqrt{2} | + 4 = 6$
Answer
$t = \frac{\sqrt{2} \pm 2}{3}$
$\frac{| 2 v + 1 | - 3}{4} = \frac{1}{2} - | 2 v + 1 |$
Answer
$v = - 1$ or $v equals 0$
$| 2 x + 1 | = \frac{| 2 x + 1 | - 3}{2}$
Answer
No solution
$\frac{| 3 - 2 y | + 4}{2} = 2 - | 3 - 2 y |$
Answer
$y = \frac{3}{2}$
$| 3 t - 2 | = | 2 t + 7 |$
Answer
$t = - 1$ or $t equals 9$
$| 3 x + 1 | = | 4 x |$
Answer
$x = - \frac{1}{7}$ or $x equals 1$
$| 1 - \sqrt{2} y | = | y + 1 |$
Answer
$y equals 0$ or $y = \frac{2}{\sqrt{2} - 1}$
$| 4 - x | - | x + 2 | = 0$
Answer
$x equals 1$
$| 2 - 5 z | = 5 | z + 1 |$
Answer
$z = - \frac{3}{10}$
$\sqrt{3} | w - 1 | = 2 | w + 1 |$
Answer
$w = \frac{\sqrt{3} \pm 2}{\sqrt{3} \mp 2}$
See footnote⁵

In Exercises -, solve the inequality. Write your answer using interval notation.

$vertical bar 3 ⁢ x minus 5 vertical bar less than or equal to 4$
Answer
$[\frac{1}{3}, 3]$
$vertical bar 7 ⁢ t plus 2 vertical bar greater than 10$
Answer
$(- \infty, - \frac{12}{7}) \cup (\frac{8}{7}, \infty)$
$| 2 w + 1 | - 5 < 0$
Answer
$open parenthesis minus 3, 2 close parenthesis$
$| 2 - y | - 4 \geq - 3$
Answer
$open parenthesis minus ∞, 1 close bracket union open bracket 3, ∞ close parenthesis$
$| 3 z + 5 | + 2 < 1$
Answer
No solution
$2 | 7 - v | + 4 > 1$
Answer
$open parenthesis minus ∞, ∞ close parenthesis$
$3 - | x + \sqrt{5} | < - 3$
Answer
$(- \infty, - 6 - \sqrt{5}) \cup (6 - \sqrt{5}, \infty)$
$vertical bar 5 ⁢ t vertical bar less than or equal to vertical bar t vertical bar plus 3$
Answer
$[- \frac{3}{4}, \frac{3}{4}]$
$| w - 3 | < | 3 - w |$
Answer
No solution
$2 less than or equal to vertical bar 4 minus y vertical bar less than 7$
Answer
$open parenthesis minus 3, 2 close bracket union open bracket 6, 11 close parenthesis$
$1 less than vertical bar 2 ⁢ w minus 9 vertical bar less than or equal to 3$
Answer
$open bracket 3, 4 close parenthesis union open parenthesis 5, 6 close bracket$
$3 > 2 | \sqrt{3} - x | > 1$
Answer
$(\frac{2 \sqrt{3} - 3}{2}, \frac{2 \sqrt{3} - 1}{2}) \cup (\frac{2 \sqrt{3} + 1}{2}, \frac{2 \sqrt{3} + 3}{2})$
With help from your classmates, solve:
1. $| 5 - | 2 x - 3 | | = 4$
2. $| 5 - | 2 x - 3 | | < 4$
Answer
1. $x = - 3$ , or $x equals 1$ , or $x equals 2$ , or $x equals 6$
2. $open parenthesis minus 3, 1 close parenthesis union open parenthesis 2, 6 close parenthesis$

See page if you don't remember what an integer is. ↩
Note the use of parentheses: $- (2 + 2 \sqrt{3})$ as opposed to $- 2 + 2 \sqrt{3}$ . ↩
Do you see why? ↩
Not for lack of trying, however! ↩
That is, $w = \frac{\sqrt{3} + 2}{\sqrt{3} - 2}$ or $w = \frac{\sqrt{3} - 2}{\sqrt{3} + 2}$ ↩

Adapted from Precalculus, Preliminary 4th Edition (integrated calculus), by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.