Absolute Value Inequality Solver
Enter an absolute-value inequality of the form |ax + b| OP c. The solver handles the three regimes: c < 0 (always-true or always-false depending on direction), c = 0 (single-point and complement cases) and c > 0 (the standard compound or two-ray solution).
What this solver does
An absolute-value inequality asks which values of x keep a distance below or above a threshold. Because |ax + b| measures how far ax + b sits from zero, |ax + b| ≤ 5 is the question “which x keep that expression within five units of zero?” — and the answer is a set of x, not a single number.
This solver takes the coefficients and the direction, splits the absolute value into its cases, and reports the solution in standard interval notation.
It solves inequalities of the exact shape |ax + b| OP c, where OP is one of <, ≤, > or ≥. That covers the form used in almost every algebra course and in the ε-δ definitions that follow in analysis, where |x − a| < δ is the standard way to write “x is within δ of a”.
The useful part is the case analysis, not the arithmetic. A “less than” inequality collapses to one bounded interval; a “greater than” opens outward into two rays. Getting that backwards is the classic error, so the solver names its case before giving the answer.
How to use this calculator
- Isolate the absolute value first The solver expects the bars alone on the left. If your inequality reads 3|2x − 1| + 4 ≤ 19, subtract 4 and divide by 3 by hand to reach |2x − 1| ≤ 5 before entering anything. It does not perform that rearrangement for you.
- Enter a and b from inside the bars a is the coefficient of x and b the constant, both read with their signs. For |2x − 1| enter a = 2 and b = −1.
- Enter c, the right-hand side c may be negative or zero — those are meaningful cases the solver handles explicitly rather than rejecting.
- Pick the direction from the dropdown The four options are <, ≤, > and ≥. Strict and non-strict directions produce the same endpoints but different brackets, and the solver shows the distinction in its output.
The formula, and where it comes from
|ax + b| < c ⟺ −c < ax + b < c |ax + b| > c ⟺ ax + b > c or ax + b < −c
Both equivalences follow from absolute value as distance from zero. If that distance is less than c, then ax + b lies strictly between −c and c — one compound statement, so one interval. If it exceeds c, then ax + b lies beyond c in one direction or beyond −c in the other — two disjoint statements, so a union.
Solving the bounded case for x gives endpoints at (−c − b)/a and (c − b)/a. When a is negative, dividing by it reverses the inequality and those two endpoints arrive in the opposite order, so the solver sorts them before reporting the interval rather than assuming the first is the smaller.
What each input means
- a Coefficient of x inside the bars — form field “a (x coefficient inside)”
- Scales the expression before the absolute value is taken. If a is zero the expression no longer depends on x, and the inequality reduces to a fixed statement |b| OP c that is either true for every real x or for none.
- b Constant inside the bars — form field “b (constant inside)”
- Shifts the expression. The point where ax + b equals zero, namely x = −b/a, is the centre of the solution interval in the bounded case and the excluded point in the c = 0 case.
- c Right-hand side — form field “c (right-hand side)”
- The threshold distance. Positive values give the standard interval or two-ray answers; zero and negative values are degenerate cases with their own rules, described below.
- OP Inequality direction — form field “Inequality”
- Chooses between the bounded cases (< and ≤) and the outward cases (> and ≥). Strict directions give open endpoints, written with round brackets; non-strict give closed endpoints, written with square brackets.
Worked examples
Every number below is produced by the same calculation engine the tool above runs. Nothing here is typed by hand, so the walkthrough cannot drift from what you get when you enter the same values yourself.
A bounded interval
Solve |2x − 1| ≤ 5. The distance from 2x − 1 to zero must be at most five, so the answer should be a single closed interval centred on the point where 2x − 1 vanishes.
Inputs a (x coefficient inside) = 2, b (constant inside) = -1, c (right-hand side) = 5, Inequality = le
- Inequality |2x − 1| ≤ 5
- Compound inequality -5 ≤ 2x − 1 ≤ 5
- Solution [-2, 3]
Result [-2, 3]
The square brackets record that both endpoints are included, because the direction was ≤ rather than <; the strict direction gives the same two numbers with round brackets instead.
The midpoint of the interval is x = 0.5, exactly where 2x − 1 vanishes. That is a reliable check: the bounded solution is always centred on −b/a with half-width c divided by |a|.
Two rays, and a negative threshold
Solve |3x| < −1. Nothing here is malformed — a and c are ordinary numbers — but the threshold is negative, which changes the nature of the answer entirely.
Inputs a (x coefficient inside) = 3, b (constant inside) = 0, c (right-hand side) = -1, Inequality = lt
- Inequality |3x + 0| < -1
- Negative right-hand side An absolute value is never negative, so |…| cannot be less than a negative number.
Result no solution
An absolute value is never negative, so no x makes |3x| smaller than −1 and the set is empty. The solver reports this as a reasoned case rather than an error, because “no solution” is the correct answer, not a failed calculation.
Reverse the direction and the same negative threshold makes the inequality true for every real x. A negative c therefore always gives one of these two extremes, never a finite interval.
Reading the result
Interval notation in the answer
Round brackets mark an excluded endpoint and square brackets an included one, so (1, 4] means every x greater than 1 and up to and including 4. Infinite ends always take a round bracket, because infinity is a direction rather than a value that can be attained.
Unions of two rays
For > and ≥ the answer contains ∪, joining two separate pieces. Values between them fail the inequality, which is why a single chain like 4 < x < 1 is meaningless and union notation exists.
The degenerate answers
Three phrases can replace an interval. “No solution” means the set is empty; “all real x” means every real satisfies it; and c = 0 with a > direction gives all reals except one point, written as two open rays.
When you would use this
Tolerance and specification limits
A part specified as 20 mm ± 0.05 mm is exactly |x − 20| ≤ 0.05. Entering a = 1, b = −20, c = 0.05 returns the acceptance interval; reversing the direction returns the reject region.
Limits and continuity in analysis
The ε-δ definition of a limit is built from |x − a| < δ and |f(x) − L| < ε. Converting those distance statements into explicit intervals is the mechanical half of every early analysis proof.
Assumptions and limitations
What this calculator assumes
- The inequality is already in the form |ax + b| OP c, with the absolute value isolated on the left.
- a, b and c are real numbers entered exactly as given; the solver does not parse an expression typed as text.
Where it stops being the right tool
- Only one absolute value, and only a linear expression inside it. Forms such as |x² − 4| < 5 or |x − 1| + |x + 2| ≤ 6 need a different method, because they split into more than two cases.
- The absolute value must already be isolated. Inequalities with coefficients or constants outside the bars must be rearranged by hand first.
Common mistakes
Turning a “greater than” into a single interval
Why it happens. The compound form −c < ax + b < c is memorised from the bounded case and then applied to the > direction as well, producing a chain like 4 < x < 1 that has no solutions and cannot be right.
How to avoid it. Check the direction before writing anything down. Less-than gives one interval, greater-than gives a union of two rays. If your chain reads backwards, you have applied the wrong case.
Forgetting to reverse the inequality when a is negative
Why it happens. Dividing both sides by a negative number flips the direction, and the step is easy to skip when the division is done mentally rather than written out.
How to avoid it. Let the solver do the division and compare with your own endpoints. If your interval is the correct width but the wrong way round, the missing flip is the reason.
Treating a negative right-hand side as an error
Why it happens. A negative c looks like a typing mistake, so it gets corrected to a positive value before the case is considered.
How to avoid it. Enter it as written. A negative threshold is a legitimate case with a definite answer — empty for < and ≤, all reals for > and ≥ — and knowing that saves the work of solving.
Frequently asked questions
How is |ax + b| < c rewritten?
As the compound inequality −c < ax + b < c, which is then solved for x in one pass. The result is a single bounded interval whose centre is x = −b/a and whose half-width is c divided by |a|.
Why does |ax + b| > c give two intervals instead of one?
Because the expression can be far from zero in either direction. The condition splits into ax + b > c or ax + b < −c, two statements that cannot both hold, so the solution is the union of two rays with a gap between them.
What happens when c is negative?
An absolute value is never negative, so the answer is one of two extremes rather than an interval: a < or ≤ inequality has no solution, and a > or ≥ inequality is satisfied by every real number.
What if c is exactly zero?
All four directions are handled separately. |ax + b| < 0 has no solution; ≤ 0 holds only at the single point x = −b/a; > 0 holds everywhere except that point; and ≥ 0 holds for every real x.