Absolute Value Inequalities Calculator
Enter the five coefficients of your absolute value inequality in the form a|bx + c| + d (inequality) e and choose the inequality sign. The calculator solves for x, shows the solution set in interval notation, explains the AND/OR logic with a full step-by-step breakdown, and flags special cases such as no solution or all real numbers. All four inequality types are supported.
Formula
Worked example
1|1x - 3| + 0 < 5: isolate to get |x - 3| < 5, split into -5 < x - 3 < 5, add 3 to get -2 < x < 8. Solution: (-2, 8).
What is an absolute value inequality?
An absolute value inequality is an inequality that contains an absolute value expression, written with vertical bars such as |x - 3| < 5. The absolute value of a number is its distance from zero on the number line, so |x - 3| measures how far x is from 3. An absolute value inequality places a constraint on that distance: |x - 3| < 5 asks for all x values that are within 5 units of 3, which gives the interval from -2 to 8. This calculator handles the general form a|bx + c| + d (sign) e, covering every standard linear case by letting you set five coefficients and choose any of the four inequality signs.
How to solve absolute value inequalities step by step
The standard algorithm has four stages. First, isolate the absolute value on the left side by subtracting d and dividing by a; note that dividing by a negative number flips the inequality sign. Second, check the sign of the resulting right-hand side k: if the simplified form is |expr| < k with k negative there is no solution; if it is |expr| > k with k negative every real number is a solution. Third, split the valid cases into two branches. For less-than inequalities (< or <=) the split is -k < bx + c < k, giving an AND compound inequality; for greater-than inequalities (> or >=) the split is bx + c < -k OR bx + c > k, giving an OR union. Fourth, solve the resulting linear inequalities for x in each branch, being careful to flip signs if b is negative. Express the final answer as a compound inequality or in interval notation.
AND vs OR: choosing the right logic
The biggest mistake students make with absolute value inequalities is mixing up when to use AND and when to use OR. The rule is simple and follows directly from the distance interpretation. If the inequality is a "less than" type (|expr| < k or |expr| <= k), the expression must be simultaneously greater than -k and less than k, which is an AND condition: both must be true at once. The solution is a single bounded interval centered near the critical point. If the inequality is a "greater than" type (|expr| > k or |expr| >= k), the expression can satisfy either branch independently, which is an OR condition: at least one must be true. The solution is two separate unbounded rays extending outward to infinity in both directions. The AND answer is a closed region, the OR answer is an open region with a gap in the middle.
Special cases: no solution and all real numbers
Two edge cases require special handling. First, if after isolating the absolute value the right-hand side k is negative and the inequality is a less-than type, there is no solution: an absolute value is always zero or positive, so it can never be less than a negative number, and the solution set is empty. Second, if k is negative and the inequality is a greater-than type, every real number is a solution: any non-negative absolute value is automatically greater than a negative number, so no restriction on x exists. A third edge case arises when k equals zero: |expr| < 0 has no solution, |expr| <= 0 gives the single point where the expression equals zero, |expr| > 0 gives all reals except that single point, and |expr| >= 0 is always true. This calculator detects and reports all these cases automatically.
Interval notation and number line representation
Interval notation is the compact way to express a solution set. A round bracket ( or ) means the endpoint is not included (strict inequality < or >), and a square bracket [ or ] means the endpoint is included (non-strict inequality <= or >=). For AND solutions the result is one interval: (lo, hi) if strict, [lo, hi] if inclusive. For OR solutions the result is a union of two rays written with the union symbol U: (-inf, lo) U (hi, +inf) if strict, or (-inf, lo] U [hi, +inf) if inclusive. On a number line an AND solution shades the region between two points, while an OR solution shades the two outer regions with an unshaded gap between the boundary points.
Absolute value inequality solution rules
| Form | Condition | Split into | Solution type | Interval notation |
|---|---|---|---|---|
| |expr| < k | k > 0 | -k < expr < k (AND) | Bounded interval | (lo, hi) |
| |expr| <= k | k >= 0 | -k <= expr <= k (AND) | Bounded interval | [lo, hi] |
| |expr| > k | k >= 0 | expr < -k OR expr > k | Two rays (OR) | (-inf, lo) U (hi, +inf) |
| |expr| >= k | k >= 0 | expr <= -k OR expr >= k | Two rays (OR) | (-inf, lo] U [hi, +inf) |
| |expr| < k | k <= 0 | Impossible | No solution | empty set |
| |expr| > k | k < 0 | Always true | All real numbers | (-inf, +inf) |
| |expr| >= 0 | k = 0 | Always true | All real numbers | (-inf, +inf) |
These rules apply after isolating the absolute value: |bx + c| (sign) k, where k is the simplified right-hand side.
Frequently asked questions
What is the difference between AND and OR in absolute value inequalities?
The type of split depends on the direction of the inequality after isolating the absolute value. A less-than inequality (< or <=) produces an AND compound inequality: the expression inside the bars must be simultaneously between -k and +k. A greater-than inequality (> or >=) produces an OR union: the expression must be less than -k or greater than +k. Mixing these up is the most common source of errors.
What happens when the right-hand side is negative?
If after isolating the absolute value the right side is a negative number k, two special cases apply. For a less-than inequality, there is no solution because an absolute value is always at least zero and can never be less than a negative number. For a greater-than inequality, every real number is a solution because any absolute value is automatically greater than any negative number. This calculator detects both cases and reports them clearly.
How do I handle a negative coefficient outside the absolute value?
When you divide both sides by a negative coefficient a to isolate the absolute value, the inequality sign flips. For example, -2|x + 1| < 6 becomes |x + 1| > -3 after dividing by -2 (the sign < flips to >). This calculator handles the sign flip automatically when you enter a negative value for a.
How do I check my answer?
Pick a test value inside the claimed solution region and substitute it into the original inequality to verify it makes the inequality true. Then pick a test value outside the solution region and verify the inequality is false. For example, if the solution is -2 < x < 8, test x = 0 (should work) and x = 10 (should fail). This two-sided verification confirms both the solution and its boundary.
Does this calculator handle |bx + c| with negative b?
Yes. When b is negative, dividing by b during the final step flips the inequality signs in each branch. For example, |-2x + 4| < 6 splits into -6 < -2x + 4 < 6, which after subtracting 4 becomes -10 < -2x < 2, and dividing by -2 (with sign flip) gives -1 < x < 5. The steps panel shows each flip explicitly so you can follow along.