Absolute Value Calculator
Solve absolute value equations and inequalities step by step, or graph y = a|x−h|+k with vertex, intercepts, and interval notation — everything you need for solving absolute value equations, inequalities, and modulus problems in one place.
Solving Absolute Value Equation — Result
Case 1 — Positive
Case 2 — Negative
Verification
Step-by-Step Working
Solving Absolute Value Inequality — Result
Number Line — Interval Notation Visualised
Compound Inequality
Interval Notation
Set-Builder Notation
Step-by-Step Working
a (steepness/direction)1
h (horizontal shift)0
k (vertical shift)0
The Key Rules for Absolute Value Equations
| Equation/Inequality | Solution Method | Solution Type |
|---|---|---|
| |ax+b| = c, c > 0 | ax+b = c OR ax+b = −c | Two solutions |
| |ax+b| = c, c = 0 | ax+b = 0 | One solution |
| |ax+b| = c, c < 0 | Impossible | No solution |
| |ax+b| < c, c > 0 | −c < ax+b < c (AND) | Bounded interval |
| |ax+b| < c, c ≤ 0 | Impossible | No solution |
| |ax+b| > c, c ≥ 0 | ax+b < −c OR ax+b > c | Union of rays |
| |ax+b| > c, c < 0 | Always true | All real numbers |
Interval Notation Guide
| Notation | Meaning | Number Line |
|---|---|---|
| (a, b) | a < x < b | Open circles at a and b |
| [a, b] | a ≤ x ≤ b | Closed circles at a and b |
| (a, b] | a < x ≤ b | Open at a, closed at b |
| (−∞, a) | x < a | Arrow left, open circle at a |
| [a, +∞) | x ≥ a | Closed circle at a, arrow right |
| (−∞,a)∪(b,+∞) | x < a or x > b | Two outward rays |
Absolute Value Properties
| Property | Formula |
|---|---|
| Non-negativity | |x| ≥ 0 for all x |
| Zero | |x| = 0 ↔ x = 0 |
| Even function | |−x| = |x| |
| Multiplicative | |xy| = |x||y| |
| Triangle inequality | |x+y| ≤ |x|+|y| |
| Distance | |x−a| = distance from x to a |
What Is Absolute Value? — Definition and Key Properties
This absolute value calculator handles everything related to absolute value: solving absolute value equations of the form |ax+b|=c, solving absolute value inequalities like |ax+b|<c or |ax+b|≥c with full interval notation, and graphing the absolute value function y=a|x−h|+k with the vertex, intercepts, and axis of symmetry all labeled.
Absolute value is defined as the distance a number is from zero on the number line. Because distance can never be negative, absolute value is always non-negative: |3|=3, |−3|=3, and |0|=0. Formally, |x| = x when x ≥ 0, and |x| = −x when x < 0.
Some students first encounter this concept under the name modulus — this is the same idea as absolute value, just a different name used more commonly in UK textbooks, engineering, and physics contexts. Whether you call it absolute value or modulus, the notation |x| and the underlying rules are identical.
The key properties of absolute value that make solving absolute value equations and solving absolute value inequalities possible are:
- Non-negativity: |x| ≥ 0 for every real number x
- Even function: |−x| = |x| — the graph of absolute value is symmetric about the y-axis
- Multiplicative property: |xy| = |x|·|y|
- Triangle inequality: |x+y| ≤ |x|+|y|
- Distance interpretation: |x−a| = the distance between x and a on the number line
The distance interpretation is the foundation for everything else on this page. |x−a| answers the question "how far is x from a?" — and this single idea explains why absolute value inequalities give intervals (less than = within a distance) versus unions (greater than = outside a distance).
Solving Absolute Value Equations — Step-by-Step
The fundamental approach to solving absolute value equations is recognising that |expression| = c creates two separate linear equations when c is positive, one equation when c equals zero, and no solution at all when c is negative. This absolute value calculator applies exactly this logic every time you solve an equation.
The Three Cases
- Case c > 0: Two solutions exist. Split into ax+b = c AND ax+b = −c, then solve each equation separately.
- Case c = 0: Exactly one solution exists, found by solving ax+b = 0.
- Case c < 0: No solution exists — absolute value can never equal a negative number.
Critical mistake to avoid: Always check whether c ≥ 0 before splitting into two cases. Solving |2x−1| = −3 by setting up 2x−1=3 and 2x−1=−3 is completely wrong — no solution exists because absolute value output is never negative. Skipping this check is the single most common error when solving absolute value equations.
Example 1 — Two Solutions: |2x − 3| = 7
- Check: c = 7 > 0, so two solutions exist.
- Case 1: 2x − 3 = 7 → 2x = 10 → x = 5
- Case 2: 2x − 3 = −7 → 2x = −4 → x = −2
- Verify: |2(5)−3| = |7| = 7 ✓ and |2(−2)−3| = |−7| = 7 ✓
Example 2 — One Solution: |x + 4| = 0
- Check: c = 0, so exactly one solution exists.
- x + 4 = 0 → x = −4
- Verify: |−4+4| = |0| = 0 ✓
Example 3 — No Solution: |3x − 1| = −2
- Check: c = −2 < 0.
- No solution — absolute value output cannot be negative, so this equation is impossible for any x.
Example 4 — Absolute Value Equals Absolute Value: |x − 5| = |2x + 1|
- Case 1: x − 5 = 2x + 1 → −6 = x → x = −6
- Case 2: x − 5 = −(2x + 1) → x − 5 = −2x − 1 → 3x = 4 → x = 4/3
- Solutions: x = −6 and x = 4/3
Solving Absolute Value Inequalities — The AND vs OR Rule
The single most important conceptual distinction when solving absolute value inequalities is this: |ax+b| < c uses AND logic (a bounded interval — x is WITHIN a distance of the center), while |ax+b| > c uses OR logic (a union of rays — x is OUTSIDE a distance of the center).
Memory aid: "Less Than → beTween (AND)" and "Greater Than → OR". The distance interpretation makes this intuitive: |x−3| < 5 means "x is within 5 units of 3," which gives the interval (−2, 8).
Example 1 — AND / Bounded Interval: |2x + 1| < 5
- c = 5 > 0, and the operator is < → bounded solution.
- Rewrite as a compound inequality: −5 < 2x + 1 < 5
- Subtract 1: −6 < 2x < 4
- Divide by 2: −3 < x < 2
- Interval notation: (−3, 2) Set-builder: {x | −3 < x < 2}
Example 2 — OR / Unbounded Union: |3x − 6| ≥ 9
- c = 9 > 0, operator is ≥ → unbounded union.
- Rewrite as OR: 3x − 6 ≤ −9 OR 3x − 6 ≥ 9
- Left: 3x ≤ −3 → x ≤ −1 Right: 3x ≥ 15 → x ≥ 5
- Interval notation: (−∞, −1] ∪ [5, +∞)
Example 3 — One-Point Solution: |x − 4| ≤ 0
- Since |x−4| is always ≥ 0, the only way |x−4| ≤ 0 is if |x−4| = 0.
- Solution: x = 4 — interval notation is the single point [4, 4].
Example 4 — All Real Numbers: |x + 2| > −1
- Since |x+2| is always ≥ 0, and 0 > −1 is always true, every real number satisfies this.
- Solution: all real numbers, interval notation (−∞, +∞).
Interval Notation and Number Lines for Absolute Value Solutions
Reading and writing interval notation correctly is a major source of confusion for students working through absolute value inequalities. Parentheses ( ) mean the endpoint is NOT included (open circle, strict inequality); brackets [ ] mean the endpoint IS included (closed/filled circle, non-strict inequality).
Every solution to an absolute value inequality can be written four equivalent ways: as a compound inequality, in interval notation, in set-builder notation, and as a number line diagram. Learning to convert fluently between all four representations of interval notation is essential.
| Compound | Interval Notation | Set-Builder |
|---|---|---|
| −3 < x < 2 | (−3, 2) | {x | −3<x<2} |
| x ≤ −1 or x ≥ 5 | (−∞,−1]∪[5,+∞) | {x|x≤−1 or x≥5} |
| x = 4 | [4, 4] | {4} |
| all reals | (−∞, +∞) | {x | x∈ℝ} |
The interval notation calculator built into this tool draws the number line automatically for every absolute value inequality you solve, showing exactly where the open or closed circles belong — eliminating the most common source of wrong interval notation.
Graphing Absolute Value Functions — y = a|x − h| + k
Every absolute value function graphs as a V-shape. The vertex sits at the point (h, k) — this is the sharp corner where the two straight-line arms of the graph meet. The graph opens upward if a > 0 and downward if a < 0.
- a controls steepness and direction — larger |a| means steeper arms; a < 0 flips the graph upside down.
- h shifts the vertex horizontally — note that |x−h| shifts RIGHT by h when h > 0.
- k shifts the vertex vertically — up if k > 0, down if k < 0.
To find x-intercepts, set y = 0 and solve |x−h| = −k/a. To find the y-intercept, substitute x = 0 into the equation.
Example: y = 2|x − 3| + 1
- Vertex: (3, 1) — opens upward since a = 2 > 0
- Minimum value: y = 1 at x = 3
- x-intercepts: none, since the minimum y-value (1) is already above the x-axis
- y-intercept: y = 2|0−3|+1 = 2(3)+1 = 7 → (0, 7)
Special Cases — No Solution, One Solution, All Real Numbers
These edge cases are the most exam-tested part of absolute value equations and absolute value inequalities.
|expression| = negative number → No solution. Absolute value output is never negative.
|expression| = 0 → Exactly one solution. Only zero has an absolute value of zero.
|expression| > negative number → All real numbers. Absolute value is always ≥ 0, which is always greater than any negative number.
|expression| < 0 → No solution. Nothing can be strictly less than zero.
|expression| ≥ 0 → All real numbers. Every absolute value satisfies this automatically.
Absolute Value in Real Life — Distance Interpretation
The number line distance model makes abstract absolute value rules concrete. |x−3| < 5 asks "how far is x from 3? Less than 5" → x is between −2 and 8. |x−7| > 2 asks "is x more than 2 units from 7?" → x < 5 or x > 9.
Engineering tolerance is a classic real-world application: a manufactured part must be within 0.05mm of a target 10mm measurement, written as |measurement − 10| ≤ 0.05. Error bounds in scientific measurement follow the same pattern: |measured − actual| < tolerance.
Common Mistakes With Absolute Value
- Forgetting to check if c < 0 before setting up cases — solving |2x+1| = −5 as two cases gives wrong answers; the correct answer is "no solution."
- Swapping the AND/OR rule — using AND for > and OR for < is backwards. Remember: Less Than → beTween (AND), Greater Than → OR.
- Forgetting to flip inequality signs when dividing a compound inequality by a negative number.
- Thinking |x+3| = |x|+3 — absolute value does NOT distribute over addition. This is a very common algebra error.
- Confusing the vertex shift direction — if the equation shows |x−h| and h > 0, the vertex shifts RIGHT, not left.
Worked Examples — All Three Tools
1. Solving Absolute Value Equations: |4x + 8| = 0
c = 0 → one solution. 4x+8=0 → x=−2.
2. Solving Absolute Value Equations: 3|x−2|+1 = 10
Isolate: |x−2| = 3. Two cases: x−2=3 → x=5; x−2=−3 → x=−1.
3. Solving Absolute Value Inequalities: |x+6| < 4
Compound: −4<x+6<4 → −10<x<−2. Interval notation: (−10,−2).
4. Solving Absolute Value Inequalities: |5x−10| ≥ 20
5x−10≤−20 or 5x−10≥20 → x≤−2 or x≥6. Interval notation: (−∞,−2]∪[6,+∞).
5. Graphing: y = |x+1| − 2
Vertex (−1,−2), opens upward, x-intercepts at x=1 and x=−3.
6. Graphing: y = −3|x−1|
Vertex (1,0), opens downward, steep arms with slope ±3.
7. Modulus Equation: |2x| = 10
a=2,b=0,c=10 → x=5 or x=−5. (This is identical to an absolute value problem — modulus is the same concept.)
8. Special Case: |x−9| > −5
Right side negative and operator is > → all real numbers.
9. Absolute Value = Absolute Value: |2x+3| = |x−1|
Case 1: 2x+3=x−1 → x=−4. Case 2: 2x+3=−(x−1) → 3x=−2 → x=−2/3.
10. Interval Notation Practice: |x| ≤ 6
−6≤x≤6 → interval notation [−6,6] → set-builder {x|−6≤x≤6}.
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