The Absolute Value Inequality Calculator solves |ax + b| < c and |ax + b| > c, expressing solutions as inequalities, interval notation, and number line descriptions. Handles both conjunction (AND) and disjunction (OR) types with full step-by-step reasoning for algebra and precalculus students.
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The calculator for absolute value inequalities solves both types of absolute value inequality — the conjunction type (|ax + b| < c, meaning "within c units of zero") and the disjunction type (|ax + b| > c, meaning "more than c units from zero") — providing solutions in inequality notation, interval notation, and number line description with complete step-by-step reasoning.
The direction of the inequality sign determines the solution structure:
The geometric interpretation is intuitive: "less than" means the expression stays close to zero (inside a distance), while "greater than" means it moves far from zero (outside a distance). The absolute value equation calculator solves the boundary case where the expression equals c exactly.
For |ax + b| < c with c > 0:
For the OR type (greater than), the same algebra applies to each separate inequality independently. Use this online calculator to verify manual solutions or explore how changing parameters affects the solution interval.
Several special cases alter the expected solution structure:
The inequality calculator handles general (non-absolute-value) linear and quadratic inequalities, and the algebra equation solvers category provides the full toolkit for inequality and equation problem solving.
Absolute value inequalities model tolerance specifications directly. A machined part must satisfy |diameter − 25.00| ≤ 0.05 mm — meaning the diameter must fall within [24.95, 25.05] mm. Any part outside this range is rejected. In signal processing, |signal − baseline| > threshold defines anomaly detection. In statistics, confidence intervals and hypothesis testing use the same mathematical structure: an observation is "within tolerance" if its standardized deviation falls inside an acceptance region defined by an absolute value inequality.
Case 1: Less-than type $$|ax + b| < c$$
This means: $$-c < ax + b < c$$
Subtract b: $$-c - b < ax < c - b$$
Divide by a (flip if a < 0): $$\frac{-c - b}{a} < x < \frac{c - b}{a} \quad (a > 0)$$
Equivalently: the center is $$x_0 = -\frac{b}{a}$$ and the half-width is $$w = \frac{c}{|a|}$$, giving the interval $$(x_0 - w, \; x_0 + w)$$.
Case 2: Greater-than type $$|ax + b| > c$$
This means: $$ax + b > c \quad \text{OR} \quad ax + b < -c$$
Solving each: $$x > \frac{c - b}{a} \quad \text{OR} \quad x < \frac{-c - b}{a} \quad (a > 0)$$
The solution is: $$(-\infty, \; x_0 - w) \cup (x_0 + w, \; +\infty)$$
In both cases, the bounds are the same — only the interpretation (between vs. outside) differs.
The center (−b/a) is the value of x where the expression ax+b equals zero — the midpoint of the solution interval. The half-width (c/|a|) is the distance from the center to each bound. For the less-than type, the solution is the interval between the lower and upper bounds: all x values within half-width of the center. For the greater-than type, the solution is everything outside that interval: x must be more than half-width away from the center. The bounds are always symmetric around the center point.
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|2x+3| < 7 → -7 < 2x+3 < 7 → -10 < 2x < 4 → -5 < x < 2. Solution: (-5, 2).
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|x-4| > 3 → x-4>3 or x-4<-3 → x>7 or x<1. Solution: (-∞,1) ∪ (7,∞).
The less-than type gives a bounded interval (a segment) as the solution — values close to the center. The greater-than type gives two unbounded rays — values far from the center. They are complementary: if one is the solution to < c, the other (plus the boundary) is the solution to ≥ c.
The absolute value is always ≥ 0. If c ≤ 0, then |ax+b| < c has no solution (nothing is less than a non-positive bound), and |ax+b| > c is satisfied by all x (or all except one point if c = 0). The interesting cases require c > 0.
For |ax+b| < c: the answer is (lower_bound, upper_bound). For |ax+b| > c: the answer is (-∞, lower_bound) ∪ (upper_bound, ∞). Use brackets [ ] instead of parentheses for ≤ and ≥ to include the endpoints.
The center is −b/a, which is the value of x where the expression inside the absolute value equals zero. It is the midpoint of the solution interval for the less-than type and the midpoint of the excluded interval for the greater-than type.
A tolerance specification like 'within ±0.05 of 10.0' translates to |x - 10.0| < 0.05. The center is 10.0, the half-width is 0.05, giving the acceptable range (9.95, 10.05). Items outside this range fail quality control.
For |ax+b| < c with c < 0: yes, the solution set is empty. For |ax+b| > c with c < 0: the solution is all real numbers. For c = 0: |ax+b| < 0 is empty, |ax+b| > 0 is all x except x = -b/a.
The bounds are the same. For ≤ and ≥, include the boundary points (use closed brackets in interval notation). For |ax+b| ≤ c: [-5, 2]. For |ax+b| ≥ c: (-∞, 1] ∪ [7, ∞).
The sign of a affects the order of solutions but not the final interval. The center −b/a and half-width c/|a| formulas handle both positive and negative a correctly. The solution interval is always (center − half_width, center + half_width).
|ax+b| < c means the signed expression ax+b is within distance c of zero. Equivalently, |x − (−b/a)| < c/|a| means x is within distance c/|a| of the center point −b/a. Every absolute value inequality is a distance question.
Inequalities like |ax+b| < |cx+d| require case analysis (splitting based on the sign of each expression) and are not handled by this calculator. You would need to consider the intervals where each expression changes sign and solve piece by piece.
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