Absolute Value Calculator
This absolute value calculator solves equations like |2x − 3| = 7 and inequalities like |x + 1| ≤ 4, shows both cases step by step, and graphs the answer on a number line. Enter a plain number to get its absolute value.
Solve an absolute value problem
Type the problem with vertical bars for absolute value. Supported signs: =, <, >, <=, >= (or ≤, ≥).
Examples: |2x - 3| = 7, 2|x - 1| + 3 = 11, |x + 1| <= 4, -15
Answer:
Steps:
How to use the absolute value calculator
- Type your problem using vertical bars, for example |2x - 3| = 7. Coefficients outside the bars are fine, as in 2|x - 1| + 3 = 11.
- Choose the relation you need: = for an equation, or <, <=, >, >= for an inequality.
- Press Solve with Steps. The calculator isolates the absolute value, splits it into cases, solves each case, and checks equation answers.
- Read the answer in inequality form, in interval notation, and on the number line. Filled dots mean an endpoint is included; open dots mean it is not.
To evaluate a single number, just type it (for example -15 or |3 - 8|). The solver handles one absolute value of a linear expression in x; for quadratics inside the bars, or two absolute values in one equation, use the photo solver linked below.
How it works: the absolute value rules
Absolute value measures distance from zero: |x| = x when x ≥ 0 and |x| = −x when x < 0. Because distance is never negative, every problem reduces to one of three patterns once the absolute value is isolated on one side (written below as |A| with a number c > 0 on the other side):
| Form | Rewrite as | Shape of the answer |
|---|---|---|
| |A| = c | A = c or A = −c | Two points |
| |A| < c | −c < A < c | One interval between two values (AND) |
| |A| > c | A < −c or A > c | Two rays pointing outward (OR) |
Special cases: if c is negative, |A| = c and |A| < c have no solution, while |A| > c is true for every real number. If c = 0, |A| = 0 only when A = 0. A handy memory aid: "less thAND" (one connected piece) and "greatOR" (two pieces).
Worked examples
Worked example 1: |2x − 3| = 7
The expression inside is 7 units from zero, so split into two equations: 2x − 3 = 7 or 2x − 3 = −7. The first gives 2x = 10, so x = 5. The second gives 2x = −4, so x = −2. Check: |2(5) − 3| = |7| = 7 and |2(−2) − 3| = |−7| = 7. Answer: x = 5 or x = −2.
Worked example 2: 2|x − 1| + 3 = 11
Isolate the absolute value first. Subtract 3: 2|x − 1| = 8. Divide by 2: |x − 1| = 4. Now split: x − 1 = 4 gives x = 5, and x − 1 = −4 gives x = −3. Answer: x = 5 or x = −3.
Worked example 3: |x + 1| ≤ 4
"Less than or equal" means x + 1 stays within 4 units of zero: −4 ≤ x + 1 ≤ 4. Subtract 1 from all three parts: −5 ≤ x ≤ 3. Both endpoints are included. Answer: [−5, 3].
Worked example 4: |3x − 2| > 7
"Greater than" gives two separate inequalities: 3x − 2 < −7 or 3x − 2 > 7. The first becomes 3x < −5, so x < −5/3. The second becomes 3x > 9, so x > 3. Answer: (−∞, −5/3) ∪ (3, ∞).
Worked example 5: |5 − x| < 3
Write −3 < 5 − x < 3. Subtract 5: −8 < −x < −2. Divide every part by −1 and reverse both signs: 8 > x > 2, which reads 2 < x < 8 from smallest to largest. Answer: (2, 8).
Common mistakes to avoid
- Splitting before isolating. In 2|x − 1| + 3 = 11, do not write 2(x − 1) + 3 = ±11. Get |x − 1| alone first, then split.
- Dropping the negative case. |x − 1| = 4 has two answers (5 and −3). Solving only x − 1 = 4 loses half the solution.
- Ignoring a negative right side. |x + 4| = −2 has no solution; do not solve x + 4 = ±2 out of habit.
- Writing a "greater than" answer as one interval. A statement such as −7 > 3x − 2 > 7 is impossible. Greater-than problems give two pieces joined by "or".
- Forgetting to flip the sign. Dividing or multiplying an inequality by a negative number, as in Example 5, reverses the inequality signs.
Frequently asked questions
What is absolute value?
The absolute value of a number is its distance from zero on the number line, so it is never negative. |7| = 7 and |−7| = 7 because both numbers are 7 units from 0. Formally, |x| = x when x ≥ 0 and |x| = −x when x < 0.
Why do absolute value equations usually have two solutions?
Two different numbers are the same distance from zero: c and −c. So |A| = c with c > 0 splits into A = c or A = −c, and each case normally gives its own value of x. For example, |2x − 3| = 7 gives x = 5 or x = −2.
Can an absolute value equation have no solution?
Yes. After you isolate the absolute value, if it equals a negative number (for example |x + 4| = −2) there is no real solution, because a distance cannot be negative. By the same logic, |x + 4| < −2 has no solution and |x + 4| > −2 is true for every real number.
How do you solve |x| < a compared with |x| > a?
For a > 0, |x| < a means x is within a units of zero, so −a < x < a, one interval. |x| > a means x is more than a units from zero, so x < −a or x > a, two separate rays joined by a union. Use ≤ or ≥ and square brackets when the endpoints are included.
How do I write the answer in interval notation?
A "less than" answer such as −5 ≤ x ≤ 3 becomes [−5, 3]. A "greater than" answer such as x < −5/3 or x > 3 becomes (−∞, −5/3) ∪ (3, ∞). Square brackets include an endpoint, parentheses exclude it, and infinity always gets a parenthesis.
Is |a + b| the same as |a| + |b|?
No. Absolute value does not distribute over addition. For example, |3 + (−5)| = |−2| = 2, but |3| + |−5| = 8. The triangle inequality says |a + b| ≤ |a| + |b|, with equality only when a and b have the same sign (or one is zero).
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Snap a photo and get step-by-step help →How to solve absolute-value equations
Formula
|x| = a means x = a or x = −a, for a ≥ 0
Absolute value represents distance from zero, so a positive target usually creates two cases.
Worked example
|x − 3| = 5 gives x − 3 = 5 or x − 3 = −5, so x = 8 or x = −2.
Common mistake
An absolute value cannot equal a negative number over the real numbers.
