Absolute Value Calculator Guide: Evaluate Expressions, Solve Equations, and Solve Inequalities
Absolute value measures distance from zero on the real number line. Because distance cannot be negative, the absolute value of a real number is always zero or positive.
For example, 5 and −5 are both five units from zero, so |5| = 5 and |−5| = 5. This is more precise than the common shortcut that absolute value “makes a number positive,” because zero remains zero and positive values remain unchanged.
The formal definition is piecewise. For a real number x, |x| = x when x is nonnegative and |x| = −x when x is negative. The second branch turns a negative input into its positive distance from zero.
Absolute value also measures distance between two real numbers. The distance between a and b is |a − b|, which is equal to |b − a|. This makes absolute value useful in geometry, measurement error, numerical analysis, optimization, and algebra.
Absolute-value equations require special logic. If |x| = 7, then x can be 7 or −7 because both values lie seven units from zero. More generally, |A| = k with k > 0 produces two branches: A = k or A = −k.
If the right side is zero, only one branch remains because |A| = 0 implies A = 0. If the right side is negative, there is no real solution because an absolute value can never be negative.
Absolute-value inequalities also depend on whether the inequality describes points inside or outside a distance boundary. The inequality |x| < 4 means x is less than four units from zero, giving −4 < x < 4. By contrast, |x| > 4 describes points more than four units from zero, giving x < −4 or x > 4.
The same distance interpretation extends to shifted expressions. |x − 3| < 5 means the distance from x to 3 is less than 5, so x lies between −2 and 8.
This calculator therefore separates four tasks: evaluating an absolute-value expression, solving an absolute-value equation, solving an absolute-value inequality, and calculating absolute difference.
How to Calculate and Solve Absolute Value Problems
- Choose the problem type: Select evaluation, equation, inequality, or absolute difference.
- Enter the expression inside the absolute-value bars: The calculator evaluates or solves the complete expression between the bars rather than applying absolute value to only one term.
- Simplify inside the bars first: For numerical evaluation, apply ordinary order of operations inside the absolute value before taking the final magnitude.
- For equations, inspect the right-hand side: A positive right side usually creates two branches, zero creates one, and a negative right side produces no real solution.
- For inequalities, identify inside versus outside: Less-than inequalities describe points inside a distance interval; greater-than inequalities describe points outside it.
- Solve each branch separately: After removing the absolute-value notation correctly, solve the resulting ordinary equations or inequalities.
- Check solutions in the original problem: Substitution can identify algebra mistakes, particularly in more complex or nested expressions.
- Display interval notation when useful: Inequality solutions can be shown both as compound inequalities and interval notation.
Formula and variables
Absolute value is defined piecewise because it returns the nonnegative magnitude of a real number. For equations of the form |A| = k with k > 0, solve A = k and A = −k. For inequalities, |A| < k means −k < A < k, while |A| > k means A < −k or A > k when k is positive.
|x| = x if x ≥ 0, and |x| = −x if x < 0; Distance between a and b = |a − b|- x — Real-number input
- The value whose distance from zero is being measured.
- A — Absolute-value expression
- Any real-valued expression inside the absolute-value bars.
- k — Distance threshold
- A nonnegative value used in equations or inequalities such as |A| = k or |A| < k.
- a — First value
- One endpoint used in an absolute-difference calculation.
- b — Second value
- The other endpoint used in an absolute-difference calculation.
Scenario 1: Solve |2x − 3| = 7
A student needs to solve an absolute-value equation with a positive value on the right side.
- Equation
- |2x − 3| = 7
- Because the right side is positive, create two branches.
- Branch 1: 2x − 3 = 7.
- Add 3: 2x = 10.
- Divide by 2: x = 5.
- Branch 2: 2x − 3 = −7.
- Add 3: 2x = −4.
- Divide by 2: x = −2.
- Check: |2(5) − 3| = |7| = 7.
- Check: |2(−2) − 3| = |−7| = 7.
Result: The solutions are x = −2 and x = 5.
The two solutions occur because the expression 2x − 3 can be either 7 or −7 and still have absolute value 7.
Understanding your results
Absolute value
This is the nonnegative distance of the input from zero.
The result cannot be negative for a real-valued input.
Two-solution equation
When |A| equals a positive constant, A can equal the positive or negative version of that constant.
Both branches must be solved.
Single-solution equation
If |A| = 0, then A must equal zero.
There is no separate positive and negative zero branch.
No real solution
An equation such as |A| = −4 has no real solution.
Absolute value cannot produce a negative output.
Inequality interval
Less-than absolute-value inequalities generally describe an interval around a center.
Greater-than inequalities generally describe two outside rays.
Assumptions
- The calculator operates over real numbers unless complex-number support is explicitly added.
- Absolute value returns a nonnegative real number.
- Standard arithmetic precedence applies inside absolute-value bars.
- Equation branches are created only after the absolute-value expression has been isolated appropriately.
- Inequality transformations assume a nonnegative distance threshold where the standard distance interpretation applies.
- Interval endpoints are included for ≤ and ≥ and excluded for < and >.
Limitations
- Absolute value over real numbers is different from the complex modulus when complex-number inputs are allowed.
- The calculator should not remove absolute-value bars by simply deleting negative signs from individual terms.
- Equations with absolute values on both sides or multiple absolute-value expressions can require case analysis beyond the simple two-branch rule.
- Nested absolute values may need inside-out simplification before solving.
- Expressions involving variables in denominators, radicals, or other restricted domains require domain checks in addition to absolute-value logic.
- Graph-based interpretation can be useful but is not a substitute for algebraic verification.
- Squaring both sides of an absolute-value equation can introduce extraneous solutions when other constraints are present, so direct branch logic is usually clearer.
- Absolute difference measures magnitude of difference and discards direction. If direction matters, use signed difference instead.
Common mistakes
- Saying absolute value always makes a number positive.
- Forgetting that |0| = 0.
- Writing |−5| = −5.
- Assuming |a + b| = |a| + |b| in general.
- Assuming |a − b| = |a| − |b|.
- Solving only the positive branch of |A| = k.
- Creating two branches when k = 0.
- Trying to solve |A| = −k with a negative right side as though real solutions exist.
- Using AND instead of OR for a greater-than absolute-value inequality.
- Using OR instead of AND for a less-than absolute-value inequality.
- Forgetting to reverse an ordinary inequality when multiplying or dividing by a negative number during branch solving.
- Ignoring domain restrictions outside the absolute-value expression.
Practical use cases
Scenario 2: Evaluate an absolute value
Calculate |−18|.
The result is 18 because −18 is eighteen units from zero.
Scenario 3: Evaluate an expression inside absolute value
Calculate |5 − 12|.
First evaluate 5 − 12 = −7, then take the absolute value to obtain 7.
Scenario 4: Solve an equation with zero
Solve |3x + 6| = 0.
The inside expression must equal zero, so 3x + 6 = 0 and x = −2.
Scenario 5: No-solution equation
Solve |x − 4| = −3.
There is no real solution because absolute value cannot equal a negative number.
Scenario 6: Solve a less-than inequality
Solve |x| < 5.
The solution is −5 < x < 5.
Planning and decision guide
Absolute value is distance from zero
The number-line interpretation is the central concept.
Both 6 and −6 have absolute value 6 because they are six units from zero.
Scenario 7: Opposite numbers
|8| = 8.
|−8| = 8.
Absolute value is never negative
For every real x, |x| ≥ 0.
This immediately resolves many equation and inequality edge cases.
Scenario 8: Can |x| = −1?
No real x can satisfy the equation.
The left side is always nonnegative.
The piecewise definition explains the sign behavior
If x is already nonnegative, |x| = x.
If x is negative, |x| = −x, which produces a positive value.
Scenario 9: x = −12
Because x is negative, |x| = −x.
Therefore |−12| = −(−12) = 12.
Absolute value is idempotent
Applying absolute value twice changes nothing after the first application.
For every real x, ||x|| = |x|.
Scenario 10: Nested simple value
||−9|| = |9|.
The result is 9.
Negating an input does not change its absolute value
|−x| = |x|.
Opposite real numbers lie the same distance from zero.
Absolute value squared equals the square
|x|² = x² for real x.
Both sides remove the effect of sign.
The square root of x² is absolute value
For real x, √(x²) = |x|, not simply x.
The principal square root is nonnegative.
Scenario 11: x = −4
√((-4)²) = √16 = 4.
That equals |−4|, not −4.
Products distribute through absolute value
|ab| = |a||b|.
The magnitude of a product equals the product of the magnitudes.
Quotients also preserve magnitude multiplicatively
For b ≠ 0, |a/b| = |a|/|b|.
The denominator remains nonzero.
Absolute value does not distribute across addition
|a + b| generally does not equal |a| + |b|.
Instead, the triangle inequality states |a + b| ≤ |a| + |b|.
Scenario 12: Counterexample to distribution
Let a = 5 and b = −5.
|a + b| = |0| = 0, while |a| + |b| = 10.
The triangle inequality is fundamental
|a + b| ≤ |a| + |b|.
Geometrically, the direct distance cannot exceed the distance obtained by traveling in separate segments.
Absolute difference measures distance between values
The distance from a to b is |a − b|.
Because distance is symmetric, |a − b| = |b − a|.
Scenario 13: Distance between −3 and 8
|−3 − 8| = |−11| = 11.
The points are eleven units apart.
Absolute difference discards direction
8 − 3 = 5 and 3 − 8 = −5 show directional difference.
|8 − 3| and |3 − 8| both equal 5.
This makes absolute difference useful for error measurement
Absolute error is commonly written |measured − true|.
It measures magnitude of error without labeling one direction positive and the other negative.
Scenario 14: Measurement error
Measured value = 10.4 and reference value = 10.0.
Absolute error = |10.4 − 10.0| = 0.4.
An equation |x| = a asks for points a units from zero
For a > 0, there are two such real points.
They are x = a and x = −a.
Scenario 15: |x| = 12
x = 12 or x = −12.
Both points are twelve units from zero.
|x| = 0 has exactly one solution
Only zero is zero units from zero.
Therefore x = 0.
|x| = negative number has no real solution
Distance cannot be negative.
This should be detected before doing unnecessary algebra.
Shifted absolute value represents distance from a center
|x − c| measures the distance from x to c.
This is one of the most useful interpretations for equations and inequalities.
Scenario 16: |x − 4| = 3
x is three units from 4.
Therefore x = 1 or x = 7.
Scaled expressions can be solved after branch creation
|2x + 1| = 9 becomes 2x + 1 = 9 or 2x + 1 = −9.
Each resulting linear equation is solved normally.
The Algebra Calculator should handle broader equations
Absolute Value Calculator specializes in expressions involving | |.
The Algebra Calculator should handle general linear, quadratic, polynomial, and other supported equations.
Less-than inequalities describe an inside interval
For a > 0, |x| < a means −a < x < a.
Geometrically, x lies inside the radius-a interval centered at zero.
Scenario 17: |x| < 3
The solution is −3 < x < 3.
In interval notation: (−3, 3).
Less-than-or-equal includes the endpoints
|x| ≤ a means −a ≤ x ≤ a for a ≥ 0.
Interval notation uses square brackets at included endpoints.
Scenario 18: |x| ≤ 4
Solution: −4 ≤ x ≤ 4.
Interval notation: [−4, 4].
Greater-than inequalities describe outside regions
For a ≥ 0, |x| > a means x < −a or x > a.
The keyword is OR because the solution lies on either side of the central interval.
Scenario 19: |x| > 2
x < −2 or x > 2.
Interval notation: (−∞, −2) ∪ (2, ∞).
Greater-than-or-equal includes boundary points
|x| ≥ a means x ≤ −a or x ≥ a for positive a.
The endpoints satisfy the equality part.
Shifted inequalities describe distance from another point
|x − c| < r means x lies within distance r of c.
The interval is c − r < x < c + r.
Scenario 20: |x − 5| < 2
x lies within two units of 5.
Therefore 3 < x < 7.
Scenario 21: |x + 2| ≥ 6
The center is −2 because x + 2 = x − (−2).
Points at least six units from −2 satisfy x ≤ −8 or x ≥ 4.
Inequalities with a negative threshold can often be resolved immediately
Since |A| ≥ 0, an inequality such as |A| < −2 has no solution.
An inequality such as |A| > −2 is true for every real A.
Scenario 22: |x| ≤ −1
There is no real solution.
A nonnegative quantity cannot be less than or equal to −1.
Scenario 23: |x| ≥ −1
Every real number satisfies the inequality.
Absolute value is always at least zero, which is already greater than −1.
Zero thresholds have special inequality behavior
|x| < 0 has no real solution.
|x| ≤ 0 has x = 0 only.
Scenario 24: |x| > 0
Every real number except zero satisfies the inequality.
Interval notation: (−∞, 0) ∪ (0, ∞).
Nested absolute values are evaluated inside out
Resolve the innermost absolute-value expression first when performing numerical evaluation.
For symbolic solving, simplification may remove redundant outer absolute values.
Scenario 25: ||x| − 3|
The inner |x| is already nonnegative.
The outer absolute value measures how far |x| is from 3.
Nested equations can create more branches
An equation such as ||x| − 3| = 1 produces |x| − 3 = 1 or |x| − 3 = −1.
Those branches then create further absolute-value equations.
Scenario 26: Solve ||x| − 3| = 1
First branch: |x| = 4, giving x = ±4.
Second branch: |x| = 2, giving x = ±2.
Solutions: −4, −2, 2, 4.
Multiple absolute values may require case intervals
Expressions such as |x − 1| + |x + 2| divide the number line at points where the internal expressions change sign.
Each interval can be analyzed with the appropriate piecewise form.
Scenario 27: Breakpoints
For |x − 1| + |x + 2|, sign changes occur at x = 1 and x = −2.
The real line is therefore divided into x < −2, −2 ≤ x < 1, and x ≥ 1.
Absolute value defines a V-shaped graph
The parent function y = |x| has vertex at (0,0).
It consists of y = −x for x < 0 and y = x for x ≥ 0.
Horizontal shifts move the vertex
y = |x − h| has vertex at (h,0).
The inside sign is opposite the direction of horizontal translation.
Scenario 28: y = |x − 3|
The vertex moves from x = 0 to x = 3.
The graph represents distance from 3.
Vertical shifts move the entire graph
y = |x| + k has vertex at (0,k).
Adding k moves the graph vertically by k units.
Scaling changes steepness
y = a|x| stretches vertically when |a| > 1 and compresses when 0 < |a| < 1.
A negative a also reflects the graph across the x-axis.
Absolute-value equations can be interpreted graphically
Solving |x| = 3 means finding intersections of y = |x| and y = 3.
The two intersections occur at x = −3 and x = 3.
Graphing supports but does not replace exact solving
A graph can reveal the number and approximate location of solutions.
Algebra provides exact values when the equation is solvable symbolically.
Absolute value is continuous but not differentiable at its vertex
The graph y = |x| has no jump or break at zero.
Its left-hand slope is −1 and right-hand slope is +1, so the ordinary derivative does not exist at x = 0.
Piecewise form explains the corner
The function changes algebraic rule at the sign boundary.
That switch creates the characteristic V-shaped corner.
Absolute value appears in interval notation
|x − c| < r describes an open interval centered at c with radius r.
|x − c| ≤ r describes the corresponding closed interval.
Scenario 29: Interval centered at 10 with radius 3
|x − 10| ≤ 3.
Equivalent interval: [7,13].
Absolute value also appears in tolerances
A manufacturing rule such as |measured − target| ≤ tolerance means the measurement must remain within a specified distance of the target.
This interpretation is exactly the same mathematics as an absolute-value inequality.
Scenario 30: Manufacturing tolerance
|x − 50| ≤ 0.2 means the measurement must lie between 49.8 and 50.2.
The target is 50 and the allowed distance is 0.2.
Use the Scientific Calculator for numeric absolute value inside advanced expressions
Scientific Calculator can evaluate values such as |sin(−2)| numerically.
Absolute Value Calculator should specialize in distance interpretation, equations, inequalities, and step-by-step transformations.
The strongest result explains the transformation
Do not return only x = −2, 5.
Show why two branches exist, solve each branch, and verify the solutions against the original absolute-value condition.
Frequently asked questions
What is absolute value?
Absolute value is the nonnegative distance of a real number from zero.
What is the absolute value of −5?
5.
What is the absolute value of 5?
5.
What is the absolute value of 0?
0.
Can absolute value be negative?
No. The absolute value of a real number is always greater than or equal to zero.
Why is |−7| = 7?
Because −7 is seven units from zero on the number line.
What is the formal definition of absolute value?
|x| = x when x ≥ 0 and |x| = −x when x < 0.
Is absolute value the same as making a number positive?
That shortcut works for negative inputs but is incomplete. Positive numbers remain unchanged and zero remains zero.
What is absolute difference?
The absolute difference between a and b is |a − b|, which measures the distance between the two values.
Is |a − b| the same as |b − a|?
Yes. Distance is symmetric.
How do I solve |x| = 5?
x = 5 or x = −5.
How do I solve |x| = 0?
x = 0.
How do I solve |x| = −5?
There is no real solution because absolute value cannot be negative.
How do I solve |2x − 1| = 9?
Solve 2x − 1 = 9 and 2x − 1 = −9, giving x = 5 and x = −4.
Why do absolute-value equations have two answers?
A positive distance from zero usually corresponds to two opposite values.
Do all absolute-value equations have two solutions?
No. A zero right side can produce one solution, a negative right side produces none, and more complex equations can produce other solution counts.
How do I solve |x| < 5?
−5 < x < 5.
How do I solve |x| ≤ 5?
−5 ≤ x ≤ 5.
How do I solve |x| > 5?
x < −5 or x > 5.
How do I solve |x| ≥ 5?
x ≤ −5 or x ≥ 5.
Why does less than use AND?
Because x must lie between the two distance boundaries.
Why does greater than use OR?
Because x can lie beyond the boundary on either side.
How do I solve |x − 3| < 4?
Interpret it as distance from 3 less than 4, giving −1 < x < 7.
What does |x − a| mean?
It measures the distance from x to a.
What is ||−5||?
5. Applying absolute value again does not change an already nonnegative value.
Is |−x| equal to |x|?
Yes for every real x.
Is |a + b| equal to |a| + |b|?
Not in general. The correct general relationship is the triangle inequality |a + b| ≤ |a| + |b|.
Is |ab| equal to |a||b|?
Yes.
Is |a/b| equal to |a|/|b|?
Yes when b is nonzero.
Why is sqrt(x²) equal to |x|?
Because the principal square root is nonnegative. If x is negative, sqrt(x²) returns −x, which equals |x|.
What does the graph of y = |x| look like?
It is a V-shaped graph with vertex at the origin.
Why does y = |x| have a corner?
Its piecewise slope is −1 on the left and +1 on the right, so the slope changes abruptly at zero.
What is the vertex of y = |x − 4|?
(4,0).
Can absolute value represent measurement tolerance?
Yes. |measured − target| ≤ tolerance means the measurement must remain within that distance of the target.
Can absolute value be used for error?
Yes. Absolute error is commonly the magnitude |measured value − reference value|.
Can the Scientific Calculator evaluate absolute value?
Yes for numerical expressions. Use this calculator when you need absolute-value equations, inequalities, or conceptual steps.
Can this calculator solve general equations?
It specializes in absolute-value problems. Use the Algebra Calculator for broader equation solving.
How accurate is an absolute value calculator?
The mathematics is exact for symbolic linear cases and valid numeric inputs. Numerical approximation limitations can arise when the inner expression itself uses floating-point calculations.
Sources and review
- Absolute Value Equations — OpenStax. Accessed 2026-09-01.
- Linear Inequalities and Absolute Value Inequalities — OpenStax. Accessed 2026-09-01.
- Absolute Value Functions — OpenStax. Accessed 2026-09-01.
- Absolute Value — Encyclopedia of Mathematics. Accessed 2026-09-01.
Reviewed 2026-09-01 by Dr Akawak Ejigu, DBA.