The Absolute Value Inequality Calculator provided by Hesapstan solves structured inequalities of the form p|ax+b|+q [<, ≤, >, ≥] r step by step. It does not support multiple absolute value groups, nonlinear inner expressions such as |x²−1|, graphing, or interval unions beyond the supported two-piece patterns.
The inequality is first isolated as |ax+b| op c
The calculator works with the form p|ax+b|+q [op] r. The first step is to isolate the absolute value expression: |ax+b| [op'] c, where c=(r−q)/p. If p is negative, the comparison operator is flipped during isolation.
Absolute value measures distance from zero. Writing |x| < 5 means x is less than 5 units away from zero — that is, x must stay between −5 and 5. This distance interpretation explains why less-than absolute value inequalities produce an AND interval. Similarly, |x| > 5 means x is more than 5 units from zero, which pushes x either below −5 or above 5.
Expressions such as |x+1| + |x−2| < 5, |x²−1| < 3, or graph-based problems are outside this calculator's scope. The runtime solves only the structured p|ax+b|+q [op] r form.
Less-than inequalities produce an AND interval
When |u| < c or |u| ≤ c and c is positive, u must stay between two bounds: −c < u < c or −c ≤ u ≤ c. That is why the solution is usually one continuous interval.
|x| < 5 means −5 < x < 5, not two separate points. In interval notation this is (-5, 5).
Greater-than inequalities produce an OR union
When |u| > c or |u| ≥ c and c is positive, u must be far enough from zero on either side. The solution splits into two parts: u < −c or u > c; with inclusive operators, u ≤ −c or u ≥ c.
For |x| ≥ 3, the solution is x ≤ −3 or x ≥ 3, written as (−∞, −3] ∪ [3, +∞).
Dividing by a negative coefficient flips the operator
After the absolute value rule is applied, the calculator solves linear inequalities involving ax+b. If a is negative, dividing by a changes the direction of the inequality. This is one of the most common mistakes in manual solutions.
- Start: |-x+1| < 5
- Apply AND rule: −5 < −x+1 < 5
- Subtract 1 from all parts: −6 < −x < 4
- Divide by −1 (flip the operator): 6 > x > −4, which is −4 < x < 6
- Interval notation: (−4, 6)
- Set-builder notation: { x ∈ ℝ | −4 < x < 6 }
Dividing or multiplying both sides of an inequality by a negative number reverses the direction of the inequality. Here −6 < −x < 4 becomes 6 > x > −4 after dividing by −1, which in standard form is −4 < x < 6.
Zero and negative thresholds create special cases
An absolute value cannot be negative. Therefore, if the isolated threshold c is negative, less-than cases may have no solution while greater-than cases may become all real numbers. When c equals zero, strict and inclusive inequalities behave differently.
|u| < 0 is impossible; |u| ≤ 0 gives one point; |u| > 0 gives all real numbers except the zero of u; |u| ≥ 0 gives all real numbers.
The final answer is shown in interval and set-builder notation
The runtime shows the result in interval notation, such as (-4, 6) or (−∞, -2] ∪ [3, +∞), and also in set-builder notation such as { x ∈ ℝ | ... }.
Parentheses mean an endpoint is excluded. Square brackets mean an endpoint is included. Strict inequalities use open endpoints; non-strict inequalities include the appropriate endpoint.
The result is notation, not a graph
This calculator presents the solution set in interval notation such as (−4, 6) or (−∞, −3] ∪ [3, +∞), and in set-builder notation such as { x ∈ ℝ | ... }. It does not draw a number line or graph.
Interval notation is the symbolic equivalent of a number-line diagram. The interval (−4, 6) corresponds to the open region between −4 and 6 on the number line. Square brackets indicate that an endpoint is included.
Frequently Asked Questions
What type of absolute value inequality does this calculator solve?
It solves one-group structured inequalities of the form p|ax+b|+q [<, ≤, >, ≥] r, with a and p both nonzero.
Why do < and ≤ become an AND statement?
Because |u| < c means u is between −c and c. Both bounds must be true at the same time.
Why do > and ≥ become an OR statement?
Because |u| > c means u is either less than −c or greater than c. Either side is enough.
Why does the operator flip?
When solving a linear inequality, dividing by a negative number reverses the comparison direction. This often happens when the coefficient of x is negative.
What do parentheses and square brackets mean in interval notation?
Parentheses exclude an endpoint; square brackets include it. The choice depends on whether the inequality is strict or inclusive.
What if the isolated c value is negative?
The result depends on the operator. Less-than cases may have no solution, while greater-than cases may become all real numbers.
Can it solve |x+1| + |x−2| < 5?
No. That has two absolute value groups. This calculator supports only the p|ax+b|+q [op] r structure.
Does this calculator draw a graph?
No. It returns algebraic steps, interval notation, and set-builder notation, but it does not draw a graph or number line.