Absolute Value Inequalities Calculator

What are absolute value inequalities?

An absolute value inequality asks not for the exact points where an expression's distance from zero equals some value, but for the entire range of x where that distance is less than, or greater than, some value.

Where the equation |ax + b| = c typically has two discrete solutions, the inequality |ax + b| < c typically has infinitely many solutions — every x in a continuous stretch of the number line. That shift, from a handful of specific points to a whole interval, is the main conceptual jump between solving absolute value equations and solving absolute value inequalities.

The four inequality forms this calculator handles are:

As with the equation case, a, b, and c are constants and x is the variable being solved for. The strategy for solving each of these builds directly on the two-case method used for absolute value equations, but the result is a range rather than a list of points.

It’s worth connecting this back to the equation |ax + b| = c for a moment, since the two topics are closely related. The equation’s solutions — when they exist — are the exact boundary points where distance from zero equals c precisely. The inequality asks about everything on one side of those boundary points or the other. In fact, one efficient way to sanity-check an inequality’s solution is to first solve the corresponding equation |ax + b| = c to find the boundary values, then test a single point from each region of the number line those boundaries create, to see which regions actually satisfy the inequality. This calculator does the full algebraic version of that process directly, but the boundary-then-test approach is a useful independent check by hand.

The two patterns: "less than" vs. "greater than"

The single most important idea in this topic is that “less than” and “greater than” absolute value inequalities produce fundamentally different-shaped solution sets, even though the algebra to get there looks similar.

“Less than” inequalities (|ax + b| < c or |ax + b| ≤ c) produce a bounded solution — a single continuous interval sitting between two numbers. This makes sense from the distance interpretation: “the distance from zero is less than c” describes a band close to zero, and once ax + b is translated back to x, that band becomes a single interval on the number line.

“Greater than” inequalities (|ax + b| > c or |ax + b| ≥ c) produce an unbounded solution — two separate rays heading off in opposite directions, away from a middle band that’s excluded. “The distance from zero is greater than c” describes everything far from zero, which after solving for x becomes two disconnected pieces rather than one interval.

Both rules assume c > 0; the boundary cases where c is zero or negative are covered in their own section further down, since they don’t follow this same split cleanly.

Step-by-step solving process

1. Isolate the absolute value expression on one side of the inequality, exactly as with equations.
2. Check the sign of c. If c is negative, the inequality resolves immediately without any case-splitting — see the special cases section below.
3. Determine which pattern applies. “Less than” (or ≤) becomes a compound inequality between −c and c. “Greater than” (or ≥) splits into two separate inequalities.
4. Solve for x in each part, remembering to flip the inequality direction if dividing or multiplying by a negative number — a rule that applies here exactly as it does in any linear inequality.
5. Write the solution as a compound inequality, in interval notation, or on a number line — whichever format the context calls for.

Walking through the process on |3x + 1| ≤ 10 makes each step concrete. The expression is already isolated (step 1), and c = 10 is positive, so no special case applies (step 2). Because the inequality is “less than or equal to,” the bounded pattern applies (step 3): −10 ≤ 3x + 1 ≤ 10. Solving for x means subtracting 1 from all three parts (−11 ≤ 3x ≤ 9) and then dividing by 3, which doesn’t require a sign flip since 3 is positive (−11/3 ≤ x ≤ 3, or roughly −3.67 ≤ x ≤ 3) — that’s step 4. The final answer can then be written as a compound inequality, as the interval [−11/3, 3], or sketched on a number line with closed circles at both ends, satisfying step 5.

Strict vs. non-strict inequalities

The difference between < and ≤ (or between > and ≥) is whether the boundary points themselves are included in the solution.

• Strict inequalities (< or >) exclude the boundary. The boundary points make the two sides exactly equal, which doesn’t satisfy a strict “less than” or “greater than” comparison.
• Non-strict inequalities (≤ or ≥) include the boundary. Equality is an acceptable outcome.

This distinction shows up visually as open circles (strict) versus filled circles (non-strict) when a solution set is drawn on a number line, and as parentheses versus square brackets in interval notation. It’s a small detail, but it’s the kind of detail that graders and automated answer-checkers check closely — getting the boundary value’s inclusion status backward is a very easy way to lose credit on an otherwise fully correct solution.

Special cases: c = 0 or c negative

The clean split into “bounded interval” or “two rays” above assumes c is a positive number. When c is zero or negative, the reasoning has to restart from the basic fact that |anything| is always ≥ 0.

When c is negative: a “less than” inequality (|expr| < c or |expr| ≤ c) has no solution, since a non-negative quantity can never be less than a negative number. A “greater than” inequality (|expr| > c or |expr| ≥ c) is true for every real number, since |expr| ≥ 0 is automatically greater than any negative c.

When c is exactly zero, each of the four forms behaves differently, and this is where it’s easy to make a subtle mistake:

That last row is the one most often gotten wrong, including in some other absolute value calculators: |ax + b| ≥ 0 is true for absolutely every real number, because an absolute value is always zero or positive — there is no x for which it fails. It can be tempting to treat ”≥” at c = 0 the same way as the strict ”> 0” case (which does exclude the single point where the expression is exactly zero), but the non-strict version includes that point rather than excluding it, since 0 ≥ 0 is a true statement. This calculator handles all four c = 0 cases distinctly and correctly.

The reasoning above generalizes beyond just c = 0: whenever an inequality’s non-strict version sits right at a boundary that the expression can actually reach, it’s worth pausing to check whether the boundary point itself belongs to the solution set, rather than defaulting to whatever pattern the strict version would suggest. Most of the time, non-strict and strict inequalities differ only by whether a single point is included — but at c = 0 specifically, that single point being included or excluded is the difference between “one excluded value” and “every real number without exception,” which is a much larger practical difference than it might first appear.

Graphical interpretation

As with the equation case, the graph of y = |ax + b| is a V-shape (or an upside-down V’s mirror image, still V-shaped, if a is negative), with its vertex at x = −b/a.

Solving an inequality is then a matter of asking where the V sits above or below a horizontal line y = c, rather than where it crosses that line exactly.

• |ax + b| < c asks where the V dips below the line y = c — which happens only near the bottom of the V, giving the single bounded interval around the vertex.
• |ax + b| > c asks where the V rises above the line y = c — which happens on both outer arms of the V, giving the two unbounded rays.

This graphical picture is a fast way to sanity-check an algebraic answer: a “less than” solution should always be a single connected interval, and a “greater than” solution should always be two separate unbounded pieces, never the other way around.

Writing the solution in interval notation

Interval notation is a compact way to describe a range of numbers using brackets and parentheses instead of inequality symbols.

For a bounded “less than” solution like −1 < x < 4, interval notation writes this as (−1, 4). If the inequality had been non-strict (−1 ≤ x ≤ 4), it becomes [−1, 4] instead.

For an unbounded “greater than” solution like x < −1 or x > 4, interval notation needs two intervals joined by the union symbol: (−∞, −1) ∪ (4, +∞). Infinity is never actually reached, so it always pairs with a parenthesis, never a bracket, regardless of whether the finite boundary is strict or not.

One more variant worth showing explicitly: a non-strict bounded solution, such as −2 ≤ x ≤ 6 (coming from an inequality like |x − 2| ≤ 4). Both boundary values are included here, so both get square brackets: [−2, 6]. Compare that to the strict version, −2 < x < 6, which would instead be written (−2, 6). The two look almost identical in interval notation — the only difference is round versus square — but they describe subtly different sets, since the strict version excludes exactly two points (the endpoints themselves) that the non-strict version includes.

Representing solutions on a number line

A number line picture makes the difference between bounded and unbounded solutions immediately visible, which is why this calculator draws one automatically alongside every result.

For a bounded solution, a single segment is shaded between the two boundary values, with a circle at each end — open (unfilled) for a strict inequality, closed (filled) for a non-strict one.

For an unbounded solution, two separate rays are shaded — one heading left from the lower boundary toward negative infinity, one heading right from the upper boundary toward positive infinity — again with open or closed circles at the two finite endpoints depending on strictness.

Worked examples

Example 1: |2x − 3| < 5

Rewrite as a compound inequality: −5 < 2x − 3 < 5. Add 3 to all three parts: −2 < 2x < 8. Divide by 2: −1 < x < 4. Interval notation: (−1, 4).

Example 2: |2x − 3| ≥ 5

The complement of example 1’s setup, but for the “greater than or equal” pattern: 2x − 3 ≤ −5 or 2x − 3 ≥ 5. Solving each: x ≤ −1 or x ≥ 4. Interval notation: (−∞, −1] ∪ [4, +∞).

Example 3: |x + 6| ≤ 0

Since the right-hand side is exactly 0 and the inequality is non-strict, the only solution is the single point where the expression equals zero: x = −6.

Example 4: |−4x + 8| > 12 (a negative coefficient)

The two-case split works the same way regardless of the sign of a: −4x + 8 < −12 or −4x + 8 > 12. Solving the first: −4x < −20, and dividing by −4 flips the inequality, giving x > 5. Solving the second: −4x > 4, and dividing by −4 again flips the inequality, giving x < −1. Putting the pieces in order: x < −1 or x > 5. It’s worth noticing that even though a was negative throughout, the final answer still comes out as a clean pair of rays extending away from the two boundary values — the sign of a affects the arithmetic along the way but not the overall bounded-vs-unbounded shape of the result, which is determined purely by the inequality type (< / ≤ vs. > / ≥).

Example 5: sanity-checking |x − 2| < 6 with a test point

Solving algebraically gives −6 < x − 2 < 6, so −4 < x < 8. As a quick independent check, pick a point clearly inside that range, like x = 0: |0 − 2| = 2, and 2 < 6 is true, confirming x = 0 belongs in the solution set. Then check a point outside the range, like x = 10: |10 − 2| = 8, and 8 < 6 is false, confirming x = 10 correctly falls outside the solution set. This kind of spot-check with one interior point and one exterior point is a fast way to catch an algebra mistake before finalizing an answer.

Common mistakes to avoid

Forgetting to flip the inequality sign when dividing by a negative. If solving −3x < 12 requires dividing both sides by −3, the direction flips: x > −4, not x < −4. This ordinary linear-inequality rule still applies inside each case of an absolute value inequality.

Mixing up which pattern produces a bounded interval and which produces two rays. A quick sanity check: “less than” keeps x close to the vertex (bounded), and “greater than” pushes x far from the vertex (unbounded, two rays). If a “less than” inequality is solved and the resulting interval notation has an infinity symbol in it, something went wrong.

Treating c = 0 the same way as c > 0. As covered in the special cases section, the clean “bounded vs. unbounded” split only holds when c is strictly positive. At c = 0, each of the four inequality types needs to be reasoned through individually rather than mechanically applying the general pattern.

Getting the boundary inclusion backward. A common transcription error is writing a strict inequality’s answer with square brackets, or a non-strict inequality’s answer with parentheses. Since this changes whether the exact boundary values themselves count as solutions, it’s worth double-checking against the original inequality symbol before finalizing an answer.

Trying to join the two rays of an unbounded solution with “and” instead of “or.” For a solution like x < −1 or x > 4, a value can’t simultaneously be less than −1 and greater than 4 — those two conditions describe different, non-overlapping parts of the number line, joined by “or,” not “and.” Writing “−1 > x > 4” as a single compound inequality (the way a bounded solution is written) is mathematically meaningless, since no real number satisfies both halves at once. Keeping the two pieces of an unbounded solution as clearly separate statements avoids this trap.

Real-world applications

Quality control and acceptable tolerance ranges. If a manufactured part’s length L must stay within 0.5 mm of a 50 mm target, the acceptable range is described by |L − 50| ≤ 0.5 — a bounded “less than or equal” inequality whose solution, 49.5 mm ≤ L ≤ 50.5 mm, defines the entire passing range rather than just its two edges.

Signal processing and error thresholds. Systems that monitor a signal for anomalies often flag any reading whose deviation from an expected baseline exceeds a threshold — an unbounded “greater than” inequality of exactly the type covered here, where the flagged region is everything outside a safe band rather than a single crossing point.

Speed and safety margins. A vehicle’s speed staying within a safe range of a target speed (say, within 5 mph of a 65 mph target on a stretch of road with variable conditions) is naturally expressed and solved as a bounded absolute value inequality, the same |speed − target| ≤ margin structure used throughout this guide.

Confidence intervals in statistics. While full statistical confidence intervals involve additional machinery, the core boundary-defining step — describing the set of values within a certain distance of an estimate — is structurally identical to solving |x − estimate| < margin, the bounded case covered in this article.

Engineering: structural load and stress limits. Materials engineers often specify that stress on a component must stay within a safe operating band around a design value, rather than exceed a single hard cutoff in only one direction — both excessive tension and excessive compression can cause failure. Writing this as |stress − design value| ≤ safe margin captures both failure directions in a single inequality, and the bounded solution it produces directly defines the safe operating range for that component.

Scheduling and time-window problems. A delivery window promising arrival “within 30 minutes of 2:00 PM” is naturally modeled as |arrival time − 2:00| ≤ 30 minutes, producing the bounded interval from 1:30 PM to 2:30 PM as the acceptable range — the same bounded “less than or equal” structure used throughout this guide, just with a time value in place of a plain number.

Absolute value inequalities vs. equations, revisited

It’s worth closing on the relationship between this topic and its companion, the Absolute Value Equation Calculator. An equation like |ax + b| = c answers a yes-or-no question at each specific x: does the distance equal exactly c there? An inequality answers a broader question across a whole range: is the distance less than, or greater than, c throughout that range? Because of this, inequality solutions almost always contain infinitely many values, while equation solutions are almost always a small finite list — typically zero, one, or two numbers. Recognizing which type of question a word problem is actually asking — an exact boundary, or an entire acceptable range — is usually the first and most important step in choosing which of the two tools to reach for.