In short

Absolute value inequalities come in two flavours. |x| < a asks "which numbers are less than a units from zero?" — the answer is an interval centred at zero. |x| > a asks "which numbers are more than a units from zero?" — the answer is two rays heading outward. The same patterns extend to |f(x)| < g(x) and |f(x)| > g(x), to the triangle inequality for bounding sums, and to graphical interpretations that turn every inequality into a picture.

A guitar string is in tune when its frequency is within 2 Hz of 440 Hz. Too high or too low and it sounds off. Write that constraint mathematically: |f - 440| < 2. That single inequality captures every acceptable frequency — from 438 Hz to 442 Hz — in one shot.

This is what absolute value inequalities do. Instead of asking "which number is exactly this far from a target?" they ask "which numbers are close enough?" or "which numbers are too far away?" The answer is always an interval (or a pair of rays), and the shape depends on which way the inequality points.

Solution of the inequality absolute value of x less than 5 shown as an interval on the number lineA horizontal number line from negative eight to eight. The open interval from negative five to positive five is shaded with a thick line. Open circles at negative five and positive five show that the endpoints are excluded. The label reads x belongs to the open interval negative five to five. −6 −4 −2 0 2 4 6 −5 5 x ∈ (−5, 5)
The inequality $|x| < 5$ picks out every number whose distance from zero is strictly less than $5$. The result is a single connected interval. The open circles at $\pm 5$ mean the boundary points are excluded — they satisfy $|x| = 5$, not $|x| < 5$.

Now flip the inequality. Ask for numbers more than five units from zero:

|x| > 5

This time the answer is two separate rays: x < -5 or x > 5. Everything outside the interval [-5, 5]. The number 7 works (distance 7 > 5). The number -10 works (distance 10 > 5). But 3 does not (distance 3 < 5).

Solution of the inequality absolute value of x greater than 5 shown as two raysA horizontal number line from negative eight to eight. Two thick rays extend outward from the open circles at negative five and positive five. The left ray goes to negative infinity, the right to positive infinity. The label shows x belongs to the union of open intervals negative infinity to negative five and five to infinity. −6 −4 −2 0 2 4 6 −5 5 x ∈ (−∞, −5) ∪ (5, ∞)
The inequality $|x| > 5$ selects everything that lies more than $5$ units from zero — two rays shooting outward in opposite directions. This is the complement of the interval from the previous figure.

These two pictures — the single interval for "less than" and the two rays for "greater than" — are the two fundamental shapes of every absolute value inequality. Everything else is a variation.

The two core patterns

The two absolute value inequality rules

For any real number a > 0:

|x| < a \quad \Longleftrightarrow \quad -a < x < a
|x| > a \quad \Longleftrightarrow \quad x < -a \;\;\text{ or }\;\; x > a

The same rules hold with \le and \ge in place of < and >, and the intervals become closed at the endpoints.

The "less than" pattern gives a conjunction — both conditions must hold simultaneously, so the answer is an interval. The "greater than" pattern gives a disjunction — either condition suffices, so the answer is two rays (a union of intervals).

Here is a useful mnemonic: "less than" means "close to the centre" (one connected piece), and "greater than" means "far from the centre" (two disconnected pieces). If you remember the shape, you will never confuse the two.

Now extend the pattern. Replace x with a linear expression f(x) = mx + n. The inequality |mx + n| < a still means -a < mx + n < a — you just solve a compound inequality instead of a bare one.

Take |3x - 6| < 9. This gives -9 < 3x - 6 < 9. Add 6 throughout: -3 < 3x < 15. Divide by 3: -1 < x < 5. The solution is x \in (-1, 5).

Take |2x + 1| \ge 7. This gives 2x + 1 \le -7 or 2x + 1 \ge 7. First branch: 2x \le -8, so x \le -4. Second branch: 2x \ge 6, so x \ge 3. The solution is x \in (-\infty, -4] \cup [3, \infty).

Inequalities of the form |f(x)| < g(x)

When the right-hand side is not a constant but a function of x, the pattern still applies — but with a twist. The inequality |f(x)| < g(x) is equivalent to

-g(x) < f(x) < g(x) \qquad \text{and} \qquad g(x) > 0

The second condition is essential. If g(x) \le 0, the inequality |f(x)| < g(x) has no solution in that region — because |f(x)| \ge 0 and a non-negative number cannot be strictly less than a non-positive one.

Consider |x - 2| < x. This becomes -x < x - 2 < x with the additional requirement x > 0.

The left part: -x < x - 2 gives 2 < 2x, so x > 1. The right part: x - 2 < x gives -2 < 0, which is always true. Combined with x > 0: the solution is x > 1, or x \in (1, \infty).

Check: at x = 3, |3 - 2| = 1 < 3. Correct. At x = 0.5, |0.5 - 2| = 1.5 > 0.5. Correctly excluded.

The triangle inequality for real numbers

You met the triangle inequality in Operations and Properties:

|a + b| \le |a| + |b|

This is not just a property — it is the most important inequality involving absolute values in all of mathematics. Its companion, the reverse triangle inequality, is

|a - b| \ge \big||a| - |b|\big|

Together, they let you bound absolute values of sums and differences without computing them exactly. Here is a concrete use. Suppose you know that |x - 3| < 0.01, and you want to bound |x^2 - 9|.

Factor: |x^2 - 9| = |x - 3| \cdot |x + 3|. You already know |x - 3| < 0.01. What about |x + 3|? By the triangle inequality,

|x + 3| = |(x - 3) + 6| \le |x - 3| + 6 < 0.01 + 6 = 6.01

So |x^2 - 9| < 0.01 \times 6.01 = 0.0601. Without computing x^2 at all, you bounded how far x^2 can wander from 9 given how close x is to 3.

This kind of reasoning — bounding a complicated expression using the triangle inequality — is the central technique of mathematical analysis. It is how limits, continuity, and convergence are proved. The idea starts here, with absolute value inequalities on the real line.

Graphical interpretation

Every absolute value inequality has a clean graphical reading. The inequality |f(x)| < a asks: where does the V-shaped graph of y = |f(x)| sit below the horizontal line y = a? The inequality |f(x)| > a asks: where does it sit above the line?

Graphical interpretation of absolute value inequalitiesTwo side-by-side panels. The left panel shows the V-shaped graph of y equals the absolute value of x minus 2 with a horizontal dashed line at y equals 3. The portion of the V below the line is highlighted, corresponding to the interval from negative 1 to 5 where the absolute value of x minus 2 is less than 3. The right panel shows the same graph with the portions above the line highlighted, corresponding to the two rays x less than negative 1 and x greater than 5 where the absolute value of x minus 2 is greater than 3. |x − 2| < 3 x 3 −1 2 5 |x − 2| > 3 x 3 −1 2 5
Left: $|x - 2| < 3$ selects the portion of the V below the line $y = 3$, giving the interval $(-1, 5)$. Right: $|x - 2| > 3$ selects the portions above the line, giving the two rays $(-\infty, -1) \cup (5, \infty)$. The graph makes the "one interval versus two rays" dichotomy visible at a glance.

This graphical reading works for nonlinear interiors too. If f(x) is quadratic, y = |f(x)| is a "reflected parabola" — the parts of the parabola that dip below the axis get flipped upward. The inequality |f(x)| < a then asks where this reflected graph sits below a horizontal line, which can produce more than one interval. The visual approach is often faster than algebraic case-splitting for complicated expressions.

An interactive view: how the solution set changes

Drag the slider below to change the value of a in the inequality |x - 2| < a. Watch how the solution interval expands as a grows and shrinks to a single point when a reaches zero.

Interactive absolute value inequality showing how the solution interval changesAn interactive graph showing the V-shaped curve y equals the absolute value of x minus 2 and a horizontal line y equals a controlled by a draggable point. As the reader drags the point up and down, the horizontal line moves and the solution interval on the x-axis changes accordingly. Readouts show the current value of a and the bounds of the interval. ↕ drag the red point
Drag the red point up or down to change the threshold $a$. As $a$ increases, the horizontal line rises and the solution interval $\big(2 - a,\; 2 + a\big)$ widens. As $a$ shrinks toward zero, the interval collapses to the single point $x = 2$. When $a$ is negative, there is no solution at all.

Two worked examples

Example 1: Solve |4x − 8| ≤ 12

This is a "less than or equal" type, so the solution is a closed interval.

Step 1. Apply the core pattern.

|4x - 8| \le 12 \quad \Longleftrightarrow \quad -12 \le 4x - 8 \le 12

Why: the absolute value of 4x - 8 is at most 12, so the expression itself lies between -12 and 12 inclusive.

Step 2. Add 8 to all three parts.

-12 + 8 \le 4x \le 12 + 8
-4 \le 4x \le 20

Why: the same-thing-to-all-parts rule for compound inequalities. Adding 8 shifts all three parts by the same amount, which preserves the inequality directions.

Step 3. Divide all three parts by 4 (positive, so no flip).

-1 \le x \le 5

Why: dividing by a positive number does not change the inequality direction. Each part is divided by the same positive value 4.

Step 4. Write the answer in interval notation.

x \in [-1, 5]

Result. The solution is the closed interval [-1, 5].

Graph of y equals absolute value of 4x minus 8 with horizontal line at y equals 12The V-shaped graph of y equals the absolute value of 4x minus 8 has its vertex at x equals 2, y equals 0. It rises steeply with slope 4 on each side. A horizontal dashed line at y equals 12 intersects the V at x equals negative 1 and x equals 5. The region where the V sits below or on the line is shaded between these two x-values, showing the solution interval. x y −1 1 2 3 5 12 y = 12 x = −1 x = 5
The V-shaped graph of $y = |4x - 8|$ sits below (or on) the dashed line $y = 12$ between $x = -1$ and $x = 5$. The thick segment on the $x$-axis marks the solution interval $[-1, 5]$. The filled dots at the intersection points reflect the $\le$ (the boundary is included).

The graph makes the answer visual: the V dips below the horizontal line across exactly one interval, and the boundary points are exactly the solutions of |4x - 8| = 12.

Example 2: Solve |x + 3| > 2x − 1

This is an |f(x)| > g(x) type where the right side depends on x. The approach is to consider two cases based on the sign of x + 3.

Step 1. Identify the critical point. The expression x + 3 changes sign at x = -3.

Why: case-splitting at the point where the interior of the absolute value is zero lets you remove the bars with certainty about the sign.

Step 2. Case 1: x \ge -3. Here |x + 3| = x + 3, so the inequality becomes:

x + 3 > 2x - 1 \implies 3 + 1 > 2x - x \implies 4 > x

Combined with x \ge -3: the solution from this case is -3 \le x < 4.

Why: both the case condition (x \ge -3) and the inequality solution (x < 4) must hold simultaneously. The intersection is [-3, 4).

Step 3. Case 2: x < -3. Here |x + 3| = -(x + 3) = -x - 3, so:

-x - 3 > 2x - 1 \implies -3 + 1 > 2x + x \implies -2 > 3x \implies x < -\tfrac{2}{3}

Combined with x < -3: the solution from this case is x < -3 (since x < -3 is already stricter than x < -2/3).

Why: both conditions are "x less than something." The stricter one is x < -3, so that is all that survives.

Step 4. Combine the two cases. From Case 1: [-3, 4). From Case 2: (-\infty, -3). Their union is (-\infty, 4).

Result. x \in (-\infty, 4), i.e., x < 4.

Graphs of y equals absolute value of x plus 3 and y equals 2x minus 1Two graphs plotted on the same axes. The V-shaped graph of y equals the absolute value of x plus 3 has its vertex at x equals negative 3 and rises with slope 1 on each side. The straight line y equals 2x minus 1 rises with slope 2. The two graphs intersect at x equals 4 where both equal 7. For all x less than 4, the V-shaped graph lies above the straight line, illustrating that the inequality holds for x less than 4. x y −3 −1 1 2 3 4 5 2 6 x = 4 |x + 3| 2x − 1
The V-shaped graph of $y = |x + 3|$ (dark curve) lies above the line $y = 2x - 1$ (red line) for all $x < 4$. The two meet at $x = 4$, where both equal $7$. For $x > 4$, the straight line overtakes the V, so the inequality fails. The solution $x < 4$ is exactly the region where the V sits above the line.

The picture tells the same story as the algebra: the V-shaped curve stays above the straight line until they meet at x = 4, and beyond that the line wins. The solution is everything to the left of their intersection.

Common confusions

A few traps that catch almost everyone the first time.

Going deeper

If you came here to learn how to solve absolute value inequalities and read their solutions as intervals, you have it — you can stop here. The rest of this section is for readers who want to see the triangle inequality in more generality and the connection to epsilon-delta proofs.

The triangle inequality in two dimensions

On the real line, the triangle inequality |a + b| \le |a| + |b| is a statement about distances on a one-dimensional number line. In the plane, the same inequality becomes

\|\mathbf{u} + \mathbf{v}\| \le \|\mathbf{u}\| + \|\mathbf{v}\|

where \|\cdot\| is the length of a vector. This is now literally a statement about triangles: if two vectors form two sides of a triangle, the third side (their sum) is shorter than or equal to the sum of the two sides. The equality case is when the two vectors point in exactly the same direction — the triangle collapses into a straight line.

The one-dimensional version you have been using is the special case where all vectors lie along a single line.

Absolute value and limits

The formal definition of a limit in calculus uses absolute value inequalities directly. To say that \lim_{x \to c} f(x) = L means: for every \varepsilon > 0, there exists a \delta > 0 such that

0 < |x - c| < \delta \implies |f(x) - L| < \varepsilon

Both conditions are absolute value inequalities. The first says "x is within \delta of c (but not equal to c)." The second says "f(x) is within \varepsilon of L." The triangle inequality is the tool used to prove most limits — the bounding technique from the |x^2 - 9| example earlier in this article is exactly the kind of move that appears in every epsilon-delta proof.

So absolute value inequalities are not just an algebra topic. They are the language in which the foundations of calculus are written.

Absolute value inequalities and interval arithmetic

When you solve |x - c| < r, you get the interval (c - r, c + r). This is the open ball of radius r centred at c on the real line. In one dimension, "balls" are just intervals — but the language of balls, radii, and centres generalises immediately to higher dimensions, where the same inequality \|\mathbf{x} - \mathbf{c}\| < r defines a disk (in 2D) or a sphere (in 3D).

The notation B(c, r) for this interval is standard in analysis and topology. Absolute value inequalities are the entry point to the entire theory of metric spaces.

Where this leads next

Absolute value inequalities connect outward in several directions.