Absolute Value Equations and Inequalities

Absolute value is one of those ideas that looks trivial — "just make it positive" — and then quietly trips up thousands of students the moment an equation or inequality gets involved. The secret is to stop thinking of absolute value as a button that strips minus signs, and start thinking of it as distance. Distance is never negative, distance can be measured in two directions, and almost every rule you need falls out of that single picture. Master the distance interpretation and you will never again wonder why $|x| = 5$ has two answers or why one inequality gives an "and" while another gives an "or."

Learning Objectives

• Define absolute value both algebraically and as distance on the number line.
• Solve absolute-value equations using the two-case method.
• Solve absolute-value inequalities of the "less than" (and) and "greater than" (or) types.
• Graph absolute-value functions and identify the vertex and slope.
• Recognize equations and inequalities that have no solution or all real numbers as solutions.
• Diagnose and correct the most common sign and case errors.

Quick Answer

The absolute value $|x|$ is the distance from $x$ to $0$ on the number line, so it is always $\ge 0$. To solve $|A| = b$ (with $b \ge 0$), split into two cases: $A = b$ or $A = -b$. For $|A| < b$, the expression is trapped within distance $b$ of zero, giving the compound inequality $-b < A < b$ (an "and"). For $|A| > b$, the expression is farther than $b$ from zero, giving $A < -b$ or $A > b$. If the right side is negative, an equation or "less than" statement has no solution, while a "greater than" statement is true for all real numbers. The graph of $y = |x|$ is a V with its vertex at the origin.

Where It Came From

The need behind absolute value is older than the symbol for it: people always needed to talk about how big a quantity is regardless of direction. A merchant who is $50 short and a merchant who is $50 ahead are both "$50 away from breaking even" — the magnitude matters independently of the sign. Physics has the same need constantly: speed is the magnitude of velocity, distance is the magnitude of displacement.

For most of mathematical history this idea lived informally. Negative numbers themselves were distrusted well into the 1700s, so a dedicated notion of "size ignoring sign" only became pressing once negatives were fully accepted and once mathematicians began measuring errors and distances between numbers. The decisive push came in the 19th century from analysis — the rigorous study of limits and convergence. To say a sequence gets arbitrarily close to a limit, you need a precise way to say "the gap between these two numbers is small," and that gap is $|a - b|$, a distance that must not care which number is larger.

The notation and name were formalized by Karl Weierstrass, who introduced the vertical-bar symbol $|x|$ (from the German Betrag, meaning amount or magnitude) around 1841 as he built the epsilon-delta foundations of calculus. So absolute value was not invented to punish algebra students — it was invented to make the phrase "distance between two numbers" into something you could compute and reason about. Every rule in this page is ultimately a statement about that distance.

Absolute Value as Distance

The formal definition is a piecewise rule:

$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$

That second line looks alarming — how can $-x$ be positive? Remember that if $x$ is negative, then $-x$ flips it to positive. For example, if $x = -7$, then $|x| = -(-7) = 7$. The definition is just machinery for "return the size."

The far more useful mental model is distance from zero. $|x|$ is how far $x$ sits from $0$ on the number line, and distance is always non-negative. This instantly explains the two-answer phenomenon: the numbers exactly $5$ units from zero are $5$ and $-5$, so $|x| = 5$ has two solutions.

Even better, $|a - b|$ is the distance between $a$ and $b$. Check it: the distance between $3$ and $10$ should be $7$, and indeed $|3 - 10| = |-7| = 7$. This is symmetric, exactly as distance should be: $|a - b| = |b - a|$.

Worked example. Interpret $|x - 4| = 3$ in words and solve it by picture. It says "$x$ is a distance of $3$ from $4$." Starting at $4$ and stepping $3$ units each way gives $4 + 3 = 7$ and $4 - 3 = 1$. So $x = 7$ or $x = 1$ — no algebra required once you trust the distance picture.

Solving Absolute-Value Equations: The Two-Case Method

When the picture is not obvious, use the two-case method. To solve $|A| = b$ where $A$ is any expression and $b \ge 0$:

$A = b \quad \text{or} \quad A = -b$

Solve each case separately; both results are (usually) valid.

Worked example 1. Solve $|2x - 3| = 11$.

Case 1: $2x - 3 = 11 \Rightarrow 2x = 14 \Rightarrow x = 7$.

Case 2: $2x - 3 = -11 \Rightarrow 2x = -8 \Rightarrow x = -4$.

Check: $|2(7) - 3| = |11| = 11$ ✓ and $|2(-4) - 3| = |-11| = 11$ ✓. Solutions: $x = 7$ and $x = -4$.

Worked example 2 (isolate first). Solve $3|x + 2| - 5 = 7$.

You must isolate the absolute value before splitting into cases. Add $5$: $3|x + 2| = 12$. Divide by $3$: $|x + 2| = 4$.

Now split. Case 1: $x + 2 = 4 \Rightarrow x = 2$. Case 2: $x + 2 = -4 \Rightarrow x = -6$. Solutions: $x = 2$ and $x = -6$.

Worked example 3 (no solution). Solve $|x - 1| = -6$.

Absolute value is a distance, and distance can never be negative. There is no solution. You should recognize this immediately without splitting into cases.

Solving Absolute-Value Inequalities

This is where the distance picture pays off the most. There are two flavors, and the direction of the inequality decides which.

"Less than" — the AND case. $|A| < b$ (with $b > 0$) means "$A$ is within distance $b$ of zero," so $A$ is trapped between $-b$ and $b$:

$|A| < b \iff -b < A < b$

Worked example 4. Solve $|x - 5| \le 3$.

This says $x$ is within $3$ of $5$, so $-3 \le x - 5 \le 3$. Add $5$ across all three parts: $2 \le x \le 8$. The solution is the interval $[2, 8]$ — a single connected chunk, which matches the "trapped" picture.

"Greater than" — the OR case. $|A| > b$ (with $b > 0$) means "$A$ is farther than $b$ from zero," so $A$ lies out past $b$ on either side:

$|A| > b \iff A < -b \ \text{ or } \ A > b$

Worked example 5. Solve $|2x + 1| \ge 7$.

Case 1: $2x + 1 \ge 7 \Rightarrow 2x \ge 6 \Rightarrow x \ge 3$.

Case 2: $2x + 1 \le -7 \Rightarrow 2x \le -8 \Rightarrow x \le -4$.

The solution is $x \le -4$ or $x \ge 3$ — two separate rays, written $(-\infty, -4] \cup [3, \infty)$.

Edge cases worth memorizing. If $b$ is negative: $|A| < b$ has no solution (a distance can't be less than a negative), while $|A| > b$ is true for all real numbers (every distance exceeds a negative). Do not blindly apply the two-case template to these — think about the distance meaning first.

Graphing Absolute-Value Functions

The graph of $y = |x|$ is a V with its vertex (lowest point) at the origin. For $x \ge 0$ it is the line $y = x$; for $x < 0$ it is the line $y = -x$. The two half-lines meet at a sharp corner, which is why $|x|$ is continuous everywhere but not differentiable at $x = 0$.

The general form $y = a|x - h| + k$ has its vertex at $(h, k)$. The value $a$ stretches the V (steeper if $|a| > 1$) and flips it downward if $a$ is negative. Notice the sign trap: $y = |x - 3|$ shifts right by $3$, not left, because the vertex sits where the inside equals zero, i.e. $x = 3$.

Worked example 6. Sketch $y = |x - 3| + 2$ and connect it to inequalities. The vertex is at $(3, 2)$, opening upward with slope $\pm 1$. Now read a solution off the graph: solving $|x - 3| + 2 < 5$ means finding where the V dips below the horizontal line $y = 5$. Setting $|x - 3| + 2 = 5$ gives $|x - 3| = 3$, so $x = 0$ or $x = 6$; the V is below the line between them, giving $0 < x < 6$. The graph and the algebra agree.

Real-World Applications

• Manufacturing tolerances. A bolt specified as $10 \pm 0.2$ mm is acceptable when its length $L$ satisfies $|L - 10| \le 0.2$. Quality control is literally an absolute-value inequality.
• Error and accuracy. In science and engineering, the error of an estimate is $|\text{measured} - \text{true}|$. Saying an estimate is accurate to within a bound is an absolute-value statement.
• Distance and navigation. How far apart two mile markers are, or how far a temperature strays from a target thermostat setting, are absolute values of differences.
• Finance. Tracking how far a portfolio's daily return deviates from its average — the size of the swing regardless of gain or loss — uses $|r - \bar{r}|$, which is the backbone of mean absolute deviation.
• Programming and signal processing. Amplitude of a signal, deadbands in control systems ("ignore inputs within $\pm 0.05$"), and clamping logic are all coded as absolute-value conditions.

Common Mistakes

Mistake 1: Splitting into cases before isolating the absolute value. Students see $3|x + 2| - 5 = 7$ and immediately write $3(x+2) - 5 = 7$. Why it's wrong: the two-case rule only applies to $|A| = b$, with the absolute value alone on one side. Correction: undo everything outside the bars first (here, add $5$, divide by $3$) to reach $|x + 2| = 4$, then split.

Mistake 2: Forgetting the negative case entirely. Solving $|x - 1| = 6$ as just $x - 1 = 6$ gives only $x = 7$ and silently loses $x = -5$. Why it's wrong: two points on the number line are that far away, not one. Correction: always write both $A = b$ and $A = -b$.

Mistake 3: Mixing up "and" with "or." Writing $|x| > 4$ as $-4 < x < 4$. Why it's wrong: that interval is the set of numbers close to zero, the exact opposite of "far from zero." Also, $-4 < x < 4$ secretly requires $x$ to be both greater than $-4$ and less than $4$, which is impossible to satisfy as an "or." Correction: "greater than" splits into two outward rays joined by or; "less than" gives one interval joined by and.

Comparison and Connections

The single most useful distinction is less-than versus greater-than inequalities, because they produce structurally different answers.

Statement	Distance meaning	Rewrites as	Solution shape
$\lvert A\rvert = b$ ($b \ge 0$)	exactly $b$ from zero	$A = b$ or $A = -b$	two points
$\lvert A\rvert < b$ ($b > 0$)	within $b$ of zero	$-b < A < b$ (and)	one interval
$\lvert A\rvert > b$ ($b > 0$)	farther than $b$	$A < -b$ or $A > b$	two rays
$\lvert A\rvert = b$ ($b < 0$)	impossible	—	no solution
$\lvert A\rvert > b$ ($b < 0$)	always true	—	all reals

Absolute value on the real line generalizes to magnitude in higher settings: the modulus $|z|$ of a complex number and the length (norm) of a vector are the same distance idea in more dimensions. The piecewise definition also makes absolute value the first natural example of a function with a corner, foreshadowing where derivatives fail to exist.

Practice Questions

Recall

State the two-case rewrite of $|A| = b$ and the rewrite of $|A| < b$. Answer: $|A| = b$ becomes $A = b$ or $A = -b$; $|A| < b$ becomes $-b < A < b$.

Understanding

Explain in one sentence why $|x| > -2$ is true for every real number. Answer: $|x|$ is a distance, so it is always $\ge 0$, and every non-negative number is greater than $-2$; hence all reals satisfy it.

Application

Solve $|3x - 6| = 9$. Answer: Case 1: $3x - 6 = 9 \Rightarrow x = 5$. Case 2: $3x - 6 = -9 \Rightarrow 3x = -3 \Rightarrow x = -1$. Solutions: $x = 5$ and $x = -1$.

Analysis

Solve $2|x + 1| - 3 < 5$ and describe the solution as an interval. Answer: Add $3$: $2|x + 1| < 8$; divide by $2$: $|x + 1| < 4$. This gives $-4 < x + 1 < 4$, so $-5 < x < 3$, the interval $(-5, 3)$.

FAQ

Why does an absolute-value equation usually have two answers? Because two different points on the number line are the same distance from a center. $|x| = 5$ is asking which numbers are $5$ units from zero, and both $5$ and $-5$ qualify.

When does an absolute-value equation have only one answer? When the two cases collapse to the same value. $|x - 4| = 0$ has just $x = 4$, since the only point at distance zero from $4$ is $4$ itself.

How do I know whether to use "and" or "or"? Look at the inequality direction after isolating the absolute value. "Less thAND" (less than → and) keeps you close, giving one interval; "greatOR than" (greater than → or) sends you far, giving two rays. Those mnemonics are cheesy but they stick.

What if the number on the other side is negative? Think distance. $|A| = -3$ and $|A| < -3$ are impossible (no solution). $|A| > -3$ is always true (every distance beats a negative). Never run the two-case template on these — decide by meaning.

Why can't I just drop the bars and solve normally? Dropping the bars assumes the inside is already positive, which throws away half the solutions and can flip inequality logic. The bars encode a genuine either/or branching that you must preserve with cases.

Quick Revision

• $|x|$ = distance from $x$ to $0$; always $\ge 0$. $|a - b|$ = distance between $a$ and $b$.
• Equation $|A| = b$ ($b \ge 0$): $A = b$ or $A = -b$. Isolate the bars first.
• $|A| < b$: $-b < A < b$ (and, one interval).
• $|A| > b$: $A < -b$ or $A > b$ (two rays).
• Right side negative: equations and "$<$" have no solution; "$>$" is all reals.
• Graph of $y = a|x - h| + k$ is a V with vertex $(h, k)$; shift is opposite the sign inside.

Related Topics

Prerequisites

• Algebra overview

Related Topics

• Linear equations and inequalities
• Piecewise functions and graphing transformations

Next Topics

• Quadratic equations and inequalities
• Complex numbers and modulus