Absolute Value

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Quick Reference


Textbook	OpenStax College Algebra 2e
Chapter	Chapter 2: Equations and Inequalities
Section	2.7 Linear Inequalities and Absolute Value Inequalities
Companion	OpenStax Precalculus 2e, Section 1.6 Absolute Value Functions (piecewise form)
Pages	Page numbers pending faculty verification.

Try This First

Before any formula, place these four numbers on a number line: $3$, $-3$, $7$, and $-7$.

Answer with the picture, not a rule:

How many steps is $-3$ from the point $0$?
How many steps is $3$ from the point $0$?
Which is farther from $0$: the point at $-7$ or the point at $3$?

Check your picture

The point $-3$ sits three steps to the left of $0$, so it is $3$ steps away.
The point $3$ sits three steps to the right of $0$, so it is also $3$ steps away.
The point $-7$ is seven steps from $0$; the point $3$ is three steps from $0$. So $-7$ is farther from $0$ even though $-7$ is the smaller number.

The count of steps from $0$ is the absolute value. Notice that the count is never negative, because a distance is never negative.

There is no need to rush this picture. Spending a minute seeing why $-3$ and $3$ return the same distance is doing the real work of the lesson.

Absolute Value Is Distance

Picture a process that takes any number as input and returns one output: how far that input sits from the point $0$ on the number line. That output is the absolute value of the input.

Distance does not care about direction. Walking three steps left and walking three steps right both move you three steps, so the input $-3$ and the input $3$ return the same output, $3$. A count of steps is never negative.

This is why the absolute value of a number is always nonnegative. The only way the distance from $0$ can be zero is if the input already sits at $0$.

Two representations of the same idea, side by side:

Number-line view. The absolute value of a point is its distance from the marker at $0$.

        |x| = 3            |x| = 3
   <------ 3 ------>  <------ 3 ------>
 ----+----+----+----+----+----+----+----
    -3        |        0        |   +3
              far                far

Symbolic view. Write the operation with the bars: $\lvert x \rvert$ reads as “the distance from $x$ to $0$.”

Translate between the two views before moving on. The number-line distance $3$ corresponds to the symbol $\lvert 3 \rvert = 3$ and also to $\lvert -3 \rvert = 3$. Two different inputs, one shared distance, one shared output.

Prerequisite Hub

Builds on:

Prerequisite	Why it is needed
Linear Equations (`linear-equations`)	Solving an absolute value equation reduces to solving two ordinary linear equations
Solving Inequalities (`solving-inequalities`)	An absolute value inequality is rewritten as a compound linear inequality

Unlocks:

Next skill	What it adds
Absolute Value Functions (`m141-absolute-value-functions`)	Turns the distance operation into a function with a V-shaped graph, a vertex, and transformations

graph LR
    A["Linear<br/>Equations"]
    B["Solving<br/>Inequalities"]
    C["Absolute<br/>Value"]
    D["Absolute Value<br/>Functions"]

    A --> C
    B --> C
    C --> D

    style C fill:#d1fae5,stroke:#a565f0,stroke-width:3px

The Official Definition

Absolute value. The absolute value of a number is its distance from zero on the number line; it is always nonnegative.

Source: OpenStax College Algebra 2e, Section 2.7 Linear Inequalities and Absolute Value Inequalities (and Precalculus 2e 1.6 piecewise form).

The same definition written as a piecewise rule, which is the form used when the input is an expression rather than a single number:

\[ \lvert x \rvert = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases} \]

Read the second case carefully. When $x$ is negative, $-x$ is positive. For the input $x = -3$, the rule returns $-x = -(-3) = 3$. The minus sign does not make the output negative; it flips a negative input back to its positive distance. This step trips up most readers, so it earns its own callout below.

Three consequences follow directly from the definition:

$\lvert x \rvert \ge 0$ for every input $x$.
$\lvert x \rvert = 0$ exactly when $x = 0$.
$\lvert x \rvert = \lvert -x \rvert$, since an input and its opposite are the same distance from $0$.

Distance Between Two Points

The distance idea extends past $0$. The expression $\lvert x - a \rvert$ measures the distance from the point $x$ to the point $a$ on the number line.

This one reading drives every equation and inequality that follows:

$\lvert x - a \rvert = r$ means the distance from $x$ to $a$ equals $r$. Two points sit exactly $r$ units from $a$: the point $a + r$ and the point $a - r$. That is why such an equation splits into two cases.
$\lvert x - a \rvert < r$ means the distance from $x$ to $a$ is less than $r$. Every point within $r$ units of $a$ forms the open interval $(a - r,\ a + r)$.
$\lvert x - a \rvert > r$ means the distance from $x$ to $a$ is more than $r$. Those points lie either far to the left ($x < a - r$) or far to the right ($x > a + r$).

Solving Absolute Value Equations

An equation $\lvert X \rvert = r$ with $r \ge 0$ asks: which inputs sit exactly $r$ units from $0$? There are two, namely $X = r$ and $X = -r$. So every such equation splits into two ordinary equations.

Worked Example 1: A clean split

Solve $\lvert x - 4 \rvert = 6$.

Predict first. The expression measures the distance from $x$ to $4$. A distance of $6$ should land you $6$ units on either side of $4$, so expect answers near $4 - 6 = -2$ and $4 + 6 = 10$.

Now solve. Split into the two cases:

\[ x - 4 = 6 \qquad \text{or} \qquad x - 4 = -6 \]

\[ x = 10 \qquad \text{or} \qquad x = -2 \]

Check against the prediction. The solutions $x = 10$ and $x = -2$ match the predicted points exactly. Verify in the original equation: $\lvert 10 - 4 \rvert = \lvert 6 \rvert = 6$, and $\lvert -2 - 4 \rvert = \lvert -6 \rvert = 6$. Both hold.

Worked Example 2: Isolate before splitting

Solve $3\lvert 2x + 1 \rvert - 5 = 7$.

Predict first. The bars are tangled with a coefficient and a constant. Nothing can split until the absolute value stands alone, so expect a first step that has nothing to do with cases.

Isolate the absolute value.

\[ 3\lvert 2x + 1 \rvert = 12 \]

\[ \lvert 2x + 1 \rvert = 4 \]

Now split.

\[ 2x + 1 = 4 \qquad \text{or} \qquad 2x + 1 = -4 \]

\[ 2x = 3 \qquad \text{or} \qquad 2x = -5 \]

\[ x = \frac{3}{2} \qquad \text{or} \qquad x = -\frac{5}{2} \]

Check. Test $x = \frac{3}{2}$: the inside is $2 \cdot \frac{3}{2} + 1 = 4$, so $3\lvert 4 \rvert - 5 = 12 - 5 = 7$. Test $x = -\frac{5}{2}$: the inside is $2 \cdot \left(-\frac{5}{2}\right) + 1 = -4$, so $3\lvert -4 \rvert - 5 = 12 - 5 = 7$. Both produce $7$.

Worked Example 3: When there is no solution

Solve $\lvert x + 2 \rvert = -5$.

Predict first. An absolute value reports a distance, and a distance is never negative. Asking for a distance of $-5$ asks for something that cannot happen, so expect no solution.

Confirm with the definition. Since $\lvert x + 2 \rvert \ge 0$ for every input $x$, it can never equal $-5$. There is no solution; the solution set is empty.

One way to see this is through the definition ($\lvert X \rvert \ge 0$). Another way is the number-line picture: no point can sit a negative number of steps from $-2$. Both routes reach the same conclusion. Which felt more convincing to you, and how would you convince a classmate?

Solving Absolute Value Inequalities

The inside of the bars again measures a distance, and the inequality sign decides whether you want points close to a center or points far from it.

Worked Example 4: A “less than” inequality (a sandwich)

Solve $\lvert x - 3 \rvert < 5$.

Predict first. This asks for every point whose distance from $3$ is under $5$. Those points crowd around $3$, reaching out almost $5$ units each way, so expect a single interval centered near $3$.

Rewrite as a compound inequality. “Distance less than $5$” means the inside lies strictly between $-5$ and $5$:

\[ -5 < x - 3 < 5 \]

Add $3$ to all three parts:

\[ -2 < x < 8 \]

Check against the prediction. The interval $(-2,\ 8)$ is centered at $3$ and stretches $5$ units each way, matching the prediction. Spot-check the center: $\lvert 3 - 3 \rvert = 0 < 5$, true. Spot-check the boundary at $x = 8$: $\lvert 8 - 3 \rvert = 5$, which is not less than $5$, so $8$ is correctly excluded.

Worked Example 5: A “greater than” inequality (two rays)

Solve $\lvert 2x + 1 \rvert \ge 7$.

Predict first. This asks for points whose distance is at least $7$, so expect the answer to break into two far-apart pieces, not one interval.

Rewrite as two inequalities. “Distance at least $7$” means the inside is at least $7$ or at most $-7$:

\[ 2x + 1 \ge 7 \qquad \text{or} \qquad 2x + 1 \le -7 \]

Solve each:

\[ 2x \ge 6 \qquad \text{or} \qquad 2x \le -8 \]

\[ x \ge 3 \qquad \text{or} \qquad x \le -4 \]

Check. The solution is $x \le -4$ or $x \ge 3$, written $(-\infty,\ -4] \cup [3,\ \infty)$, two rays as predicted. Spot-check $x = 3$: $\lvert 2 \cdot 3 + 1 \rvert = \lvert 7 \rvert = 7 \ge 7$, true.

The two patterns side by side

Inequality form	Plain reading	Rewrite	Shape of the answer
$\lvert X \rvert < r$	distance under $r$	$-r < X < r$	one interval (a sandwich)
$\lvert X \rvert > r$	distance over $r$	$X < -r$ or $X > r$	two rays (a split)

A “less than” sign keeps the points together; a “greater than” sign pushes them apart. Picture which one you are solving before writing a single line. One way to remember the difference: a small distance hugs the center, while a large distance escapes to both sides.

Common Misconceptions

Common misconception

absolute value flips the sign of any number to make it positive. The operation reports a distance, so a positive input keeps its value rather than flipping. For the input $5$, the output is $\lvert 5 \rvert = 5$, not $-5$. The piecewise rule uses the first case ($x \ge 0$ returns $x$). Sign-flipping only applies to the negative case. A clean test of the idea: $\lvert 5 \rvert + \lvert -5 \rvert = 5 + 5 = 10$, never $0$.

Common misconception

the absolute value of a sum equals the sum of the absolute values. The bars do not distribute across addition. Test it on a small example: $\lvert 3 + (-5) \rvert = \lvert -2 \rvert = 2$, while $\lvert 3 \rvert + \lvert -5 \rvert = 3 + 5 = 8$. The two results, $2$ and $8$, are not equal. The distance of a combined input is not the same as adding two separate distances. Treat the inside of the bars as a single quantity and simplify it fully before applying the absolute value.

Common misconception

$-x$ is automatically a negative number. The symbol $-x$ means the opposite of $x$, and the opposite of a negative number is positive. In the piecewise definition, the case $x < 0$ returns $-x$, which is positive precisely because $x$ was negative. For $x = -3$, the rule returns $-x = -(-3) = 3$, a positive distance. Read $-x$ as “flip whatever $x$ is,” not as “a negative value.”

Practice Problems

Level 1 Reading the Definition

Evaluate $\lvert -9 \rvert$, $\lvert 0 \rvert$, and $\lvert 4 \rvert$.

Show Answer

$\lvert -9 \rvert = 9$ (the point $-9$ is nine steps from $0$).

$\lvert 0 \rvert = 0$ (the point $0$ is zero steps from itself).

$\lvert 4 \rvert = 4$ (the point $4$ is four steps from $0$).

Level 2 A Clean Equation Split

Solve $\lvert x + 1 \rvert = 8$.

Thought Process

The inside measures distance from the point $-1$. A distance of $8$ lands $8$ units on either side, so predict answers near $-1 + 8 = 7$ and $-1 - 8 = -9$. Split into two cases.

Show Answer

Split:

\[ x + 1 = 8 \qquad \text{or} \qquad x + 1 = -8 \]

\[ x = 7 \qquad \text{or} \qquad x = -9 \]

Check: $\lvert 7 + 1 \rvert = \lvert 8 \rvert = 8$ and $\lvert -9 + 1 \rvert = \lvert -8 \rvert = 8$. The solution set is $\{-9,\ 7\}$.

Level 3 Isolate, Then Split

Solve $2\lvert x - 5 \rvert + 1 = 11$.

Thought Process

Nothing splits while the bars are tangled with a coefficient and a constant. Isolate the absolute value first, then split.

Show Answer

Isolate:

\[ 2\lvert x - 5 \rvert = 10 \qquad \Longrightarrow \qquad \lvert x - 5 \rvert = 5 \]

Split:

\[ x - 5 = 5 \qquad \text{or} \qquad x - 5 = -5 \]

\[ x = 10 \qquad \text{or} \qquad x = 0 \]

Check: $2\lvert 10 - 5 \rvert + 1 = 2 \cdot 5 + 1 = 11$, and $2\lvert 0 - 5 \rvert + 1 = 2 \cdot 5 + 1 = 11$. The solution set is $\{0,\ 10\}$.

Level 4 A "Less Than" Inequality

Solve $\lvert 3x - 2 \rvert \le 7$ and write the answer in interval notation.

Thought Process

A “less than or equal to” sign keeps the points together, so predict one closed interval. Rewrite as a sandwich, then solve all three parts at once.

Show Answer

Rewrite as a compound inequality:

\[ -7 \le 3x - 2 \le 7 \]

Add $2$ to all three parts:

\[ -5 \le 3x \le 9 \]

Divide by $3$:

\[ -\frac{5}{3} \le x \le 3 \]

The solution is the closed interval $\left[ -\dfrac{5}{3},\ 3 \right]$.

Spot-check the endpoint $x = 3$: $\lvert 3 \cdot 3 - 2 \rvert = \lvert 7 \rvert = 7 \le 7$, true.

Level 5 A "Greater Than" Inequality, Reversed

Solve $5 - \lvert x + 4 \rvert < 1$ and write the answer in interval notation. Then explain in one sentence why the answer is two pieces rather than one.

Thought Process

The absolute value is buried inside a subtraction, so isolate it first. Watch the inequality direction when multiplying by a negative number. Once isolated, decide whether the result is a “less than” sandwich or a “greater than” split.

Show Answer

Isolate the absolute value. Subtract $5$:

\[ -\lvert x + 4 \rvert < -4 \]

Multiply both sides by $-1$, which reverses the inequality sign:

\[ \lvert x + 4 \rvert > 4 \]

This is a “greater than” form, so it splits into two pieces:

\[ x + 4 > 4 \qquad \text{or} \qquad x + 4 < -4 \]

\[ x > 0 \qquad \text{or} \qquad x < -8 \]

In interval notation: $(-\infty,\ -8) \cup (0,\ \infty)$.

The answer is two pieces because a “greater than” condition asks for points whose distance from the center $-4$ exceeds $4$, and those points fall on both far sides of $-4$ rather than in one band around it.

Spot-check $x = 0$: $5 - \lvert 0 + 4 \rvert = 5 - 4 = 1$, which is not less than $1$, so $0$ is correctly excluded (the interval is open at $0$).

Mastery Checklist

Novice (Level 1-2):

I can evaluate the absolute value of any number using the distance idea
I can state why an absolute value is always nonnegative
I can split a clean equation $\lvert X \rvert = r$ into its two cases

Competent (Level 3-4):

I can isolate the absolute value before splitting an equation
I can recognize when an absolute value equation has no solution
I can rewrite a “less than” inequality as a single compound inequality

Proficient (Level 5):

I can rewrite a “greater than” inequality as two separate inequalities
I can isolate a buried absolute value and track the inequality direction through a sign flip
I can explain why the answer shape (one interval versus two rays) follows from the inequality sign

Connections

Looking back:

Linear Equations supply the two ordinary equations each absolute value equation splits into.
Solving Inequalities supply the compound and reversed-sign moves used on absolute value inequalities.

Looking ahead:

Absolute Value Functions turn this distance operation into a function with a V-shaped graph, a vertex, and transformations.

Real-world connections:

Tolerance in manufacturing: a part of target length $L$ with allowance $t$ is acceptable when $\lvert x - L \rvert \le t$.
Error bounds: a measured value within $t$ of a true value $T$ satisfies $\lvert \text{measured} - T \rvert \le t$.

Resources

Primary text: OpenStax College Algebra 2e, Section 2.7 Linear Inequalities and Absolute Value Inequalities. https://openstax.org/books/college-algebra-2e/pages/2-7-linear-inequalities-and-absolute-value-inequalities
Companion text: OpenStax Precalculus 2e, Section 1.6 Absolute Value Functions (piecewise form and graph). https://openstax.org/books/precalculus-2e/pages/1-6-absolute-value-functions

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Last updated: 2026-06-16