Algebra • Inequalities & Interval Analysis

Absolute Value Inequality Grapher

Solve and graph absolute value inequalities step-by-step. Features dual interactive visualizations: a 1D real number line with open/closed endpoints and an orthonormal 2D Cartesian graph with shaded solution zones.

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Last Updated: September 2026
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Verified Accurate: Real Analysis & Metric Topology
Inequalities • Absolute Value Grapher Bounded Interval
Curriculum Presets: Click to load & graph
Form: |ax + b| [Sign] c Configure Parameters
Interval Notation
[-1, 5]
Bounded Closed Interval
Set-Builder Notation
{ x | -1 ≤ x ≤ 5 }
Formal Solution Set
Boundary Points
x = -1, x = 5
Closed Endpoints (Solid Disc)
Vertex of |ax + b|
(2, 0)
Root Point (-b/a, 0)

1D Real Number Line Plot

Solid (Included) vs Open (Excluded) Endpoints

2D Cartesian Graph: y₁ = |ax + b| vs y₂ = c

1:1 Orthonormal Aspect Ratio • Shaded Solution Zone
y = |ax + b|
y = c
Solution Region

Step-by-Step Algebraic Solution

Rigorous Derivation
Direct Answer & Overview
Verified Educational Guide

Absolute Value Inequality Direct Overview

An absolute value inequality |ax + b| ≤ c defines the set of all real numbers whose linear transformation lies within distance c of the origin. When c > 0, 'less than' inequalities (|u| ≤ c) split into bounded conjunction intervals -c ≤ u ≤ c, while 'greater than' inequalities (|u| ≥ c) split into unbounded disjunction unions u ≤ -c OR u ≥ c.

Primary Mathematical Formula Universal Conjunction and Disjunction Splitting Principles
Standard Equation
ƒ(x)
Q.E.D.
∣ax+b∣≤c  ⟺  −c≤ax+b≤c,∣ax+b∣≥c  ⟺  ax+b≤−c  or  ax+b≥c|ax + b| \le c \iff -c \le ax + b \le c, \qquad |ax + b| \ge c \iff ax + b \le -c \;\text{or}\; ax + b \ge c
Requires c > 0 for standard intervals. If c < 0, |u| < c has no solution (∅), and |u| > c holds for all real numbers (ℝ).
Exact Formula
Input Parameters
Required
1
Linear Expression (|ax + b|): Scalar coefficient a and constant term b defining the inner argument.
2
Inequality Operator: Strict or non-strict comparison (≤, <, ≥, >).
3
Boundary Constant (c): The real scalar boundary threshold on the right-hand side.
Expected Outputs
Calculated
Interval Notation: The exact continuous or disjoint union interval representation (e.g. [-1, 5] or (-∞, -3) ∪ (2, ∞)).
Set-Builder Notation: Formal mathematical set definition {x | ...}.
Critical Roots / Boundary Points: The intersection x-coordinates solving ax + b = ±c.
1D Real Number Line Plot: Interactive visual with shaded intervals and solid/open endpoint indicators.
2D Cartesian Plot: V-shaped graph y = |ax + b| intersecting horizontal line y = c.
Worked Numerical Example
Instant Verification
Solve and graph |2x - 4| ≤ 6.
→ Step 1: Set up double inequality for 'less than or equal' (conjunction): -6 ≤ 2x - 4 ≤ 6. Step 2: Add 4 to all parts: -2 ≤ 2x ≤ 10. Step 3: Divide by 2: -1 ≤ x ≤ 5. Step 4: Write in interval notation: [-1, 5]. Step 5: Graph with solid discs at -1 and 5 and a shaded segment between them.
[-1, 5] (Critical points x = -1, x = 5)

Absolute Value Metric Definition & Distance Concept

In real analysis, the absolute value (or modulus) of a real number $x$, denoted $|x|$, is defined piecewise as:

|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}

Geometrically, $|x - k|$ represents the Euclidean distance between the point $x$ and the reference point $k$ along the 1D real number line. When we assert an inequality such as $|x - k| \le c$, we are stating that the physical distance between $x$ and the anchor $k$ cannot exceed the radius $c$. For linear system inequalities involving multiple variables, our Feasible Region of Linear Inequalities visualizer extends these concepts into higher dimensions.

Conjunction vs. Disjunction: The 'Less-Than' vs 'Greater-Than' Rules

The algebraic structure of the solution depends entirely on whether the inequality operator points toward or away from the absolute value term (assuming $c > 0$):

Less Than: Conjunction (AND)
|u| < c \iff -c < u < c

A conjunction bounds the expression between two opposite symmetric limits. The solution forms a single continuous bounded interval:

x \in (x_{\min}, x_{\max}) \quad \text{or} \quad [x_{\min}, x_{\max}]
Greater Than: Disjunction (OR)
|u| > c \iff u < -c \;\text{or}\; u > c

A disjunction asserts that the distance exceeds $c$ in either direction. The solution forms a disjoint union of two infinite rays:

(-\infty, x_{\text{left}}) \cup (x_{\text{right}}, \infty)

Comprehensive Step-by-Step Algebraic Solving Pipeline

To solve any general absolute value inequality $|ax + b| \le c$ or $|ax + b| \ge c$:

  1. Isolate the Absolute Value:

    Ensure the absolute value bracket stands alone on one side of the inequality before splitting.

  2. Apply the Splitting Rule:

    If $\le$, rewrite as $-c \le ax + b \le c$. If $\ge$, rewrite as $ax + b \le -c$ or $ax + b \ge c$.

  3. Subtract $b$:

    Subtract constant $b$ from all parts of the inequality to isolate the term $ax$.

  4. Divide by $a$ (Sign Inversion Warning):

    Divide by $a$. If $a < 0$, reverse all inequality signs ($≤ \to ≥$ and $< \to >$).

  5. Express in Interval and Set Notation:

    Use square brackets $[ \dots ]$ for $\le, \ge$ and round parentheses $( \dots )$ for $<, >$.

Critical Boundary Cases: Negative Right-Hand Side ($c \le 0$)

Standard splitting rules assume $c > 0$. When $c \le 0$, intuitive splitting fails, and the fundamental non-negativity property $|u| \ge 0$ must be invoked:

Inequality Form Condition on $c$ Solution Set Mathematical Justification
|ax + b| < c c ≤ 0 ∅ (Empty Set) Absolute value can never be negative or strictly less than 0.
|ax + b| ≤ 0 c = 0 { -b/a } (Single Point) Equality $|ax + b| = 0$ holds only at the root $x = -b/a$.
|ax + b| > c c < 0 (-\infty, \infty) (\mathbb{R}) Since $|u| \ge 0$, it is always strictly greater than any negative number.
|ax + b| > 0 c = 0 \mathbb{R} \setminus { -b/a } All real numbers satisfy $|u| > 0$ except where $u = 0$.

Dual Graphical Interpretation: 1D Number Line vs. 2D Coordinate System

Our calculator simultaneously renders the solution across two distinct geometric paradigms:

1D Real Number Line

Directly graphs the 1D solution set on the real axis $\mathbb{R}$. Critical roots appear as solid circles ($\bullet$) for included boundaries or open circles ($\circ$) for excluded boundaries. A green shaded bar illustrates the feasible zone.

2D Cartesian Coordinate System

Graphs the system $y_1 = |ax + b|$ as a V-shaped piecewise function alongside the horizontal threshold $y_2 = c$. The solution region corresponds to the vertical projection strip where the V-curve satisfies the inequality condition relative to the horizontal line.

Step-by-Step Graded Numerical Examples

Example 1: Bounded Conjunction ($|3x - 6| < 12$)

Solve and write the solution in interval notation: $|3x - 6| < 12$.

1. Split into double inequality: $-12 < 3x - 6 < 12$.
2. Add 6 across all three parts: $-12 + 6 < 3x < 12 + 6 \implies -6 < 3x < 18$.
3. Divide by 3: $-2 < x < 6$.
4. Interval Notation: $(-2, 6)$ (Open interval, strictly excluded endpoints).
Example 2: Disjoint Union ($|-2x + 4| \ge 8$)

Solve and graph: $|-2x + 4| \ge 8$ with a negative leading coefficient.

1. Factor out negative: $|-1(2x - 4)| = |2x - 4| \ge 8$.
2. Split into disjunction: $2x - 4 \le -8$  OR  $2x - 4 \ge 8$.
3. Left branch: $2x \le -4 \implies x \le -2$.
4. Right branch: $2x \ge 12 \implies x \ge 6$.
5. Combined Union: $(-\infty, -2] \cup [6, \infty)$ with solid discs at $x = -2$ and $x = 6$.

Engineering Tolerances, Signal Processing & Error Bounds

  • Mechanical Manufacturing Tolerances: Precision engineering specifications for machined parts (e.g. crankshaft diameters, piston clearances) are formally expressed as absolute value inequalities: $|d - d_{\text{nominal}}| \le \delta$. A diameter specified as $50.00 \pm 0.05\text{ mm}$ translates to $|d - 50.00| \le 0.05$.
  • Signal Processing & Voltage Clamping: Operational amplifiers and limiters clip voltage spikes exceeding threshold $V_{\text{max}}$. Signals operate in the linear regime when $|V(t)| \le V_{\text{max}}$. For single variable linear inequality fundamentals, visit our Single-Variable Linear Inequality Grapher.
  • Quality Control & Statistical Process Control: In six-sigma manufacturing, acceptable production runs require sample averages to reside within three standard errors of the target: $|\bar{x} - \mu| \le 3\sigma / \sqrt{n}$.

Common Pitfalls & Examination Traps

  • Writing a Conjunction for a Greater-Than Sign: Never write $-c > ax + b > c$. Writing a single connected chain for a greater-than inequality is a severe logical error. Always use the word OR to connect two distinct statements: $ax + b \le -c$ or $ax + b \ge c$.
  • Blindly Splitting When $c < 0$: Students frequently split $|x - 4| < -2$ into $2 < x - 4 < -2$, which produces contradictory nonsense. Always inspect the right-hand constant first: if $c < 0$, $|u| < c$ has no solution immediately.
  • Forgetting to Invert Signs When Dividing by Negative $a$: In $-3x \le 9$, dividing by $-3$ yields $x \ge -3$. Failing to flip the inequality sign reverses the entire feasible interval.
Fact-Checked & Verified • Computational Accuracy Standards
Updated September 2026 • Editorial Policy
Authored By
Sanjay Samanta

Lead Developer & Founder of Basic Math Tools. Specializes in browser-native computational algorithms and applied mathematics.

Reviewed & Verified By
Academic Review Board

Mathematics & curriculum specialists. Audited against standard algebraic and arithmetic principles.

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Frequently Asked Questions

How do you solve an absolute value inequality with a less-than sign (|u| < c)?
An inequality of the form |u| < c represents a conjunction (an 'AND' statement). It splits into the continuous double inequality -c < u < c. Geometrically, this means the distance between u and zero is strictly less than c, bounding the solution within an open interval.
How do you solve an absolute value inequality with a greater-than sign (|u| > c)?
An inequality of the form |u| > c represents a disjunction (an 'OR' statement). It splits into two disjoint rays: u < -c OR u > c. Geometrically, this means the distance between u and zero is strictly greater than c, resulting in two unbounded intervals extending to -∞ and +∞.
What happens if the boundary constant c is negative (|ax + b| < -3)?
Because the absolute value of any real number is always non-negative (|u| ≥ 0), it is impossible for |u| to be strictly less than or equal to a negative number. Thus, |ax + b| < -c (with c > 0) has no solution (empty set ∅). Conversely, |ax + b| > -c is satisfied for all real numbers (-∞, ∞).
When do you use open circles versus closed circles on the number line graph?
Use open circles (○) for strict inequalities (< and >) because the boundary points are excluded from the solution set. Use closed solid discs (●) for inclusive inequalities (≤ and ≥) because the boundary points satisfy equality and are included in the solution set.
Why does the direction of an inequality sign flip when dividing by a negative number?
Multiplying or dividing by a negative number reverses the order relation on the real number line. For example, 2 < 5, but multiplying by -1 yields -2 > -5. When isolating x in an absolute value inequality with a negative coefficient (e.g. -2x ≤ 6), dividing by -2 flips the sign to x ≥ -3.
How do you represent an absolute value inequality on a 2D Cartesian graph?
Plot the left-hand side y₁ = |ax + b| as a V-shaped piecewise linear function with vertex at (-b/a, 0), and plot the right-hand side y₂ = c as a horizontal line. The solution to |ax + b| ≤ c corresponds to the x-values where the V-curve lies on or below the horizontal line.