Absolute Value Inequalities Calculator

Understanding Absolute Value Inequalities

An absolute value inequality is a mathematical statement that compares an absolute value expression to a constant using an inequality symbol. The expression |ax + b| measures the distance of the linear expression ax + b from zero on the number line. Because distance is never negative, the direction of the inequality symbol determines whether the solution is a bounded interval or an unbounded union of two rays.

The Core Formula and Its Derivation

For a less-than inequality of the form |ax + b| < c, where c > 0 and a ≠ 0, the definition of absolute value transforms the single inequality into a compound inequality:

−c < ax + b < c

Subtracting b from all three parts gives −c − b < ax < c − b. Dividing through by a yields the closed-form solution interval:

(−c − b) / a < x < (c − b) / a

When a is negative, the division flips both inequality signs, so the left endpoint becomes the algebraically larger value. For a greater-than inequality |ax + b| > c, the expression ax + b must lie outside the interval [−c, c], producing two separate branches: ax + b < −c OR ax + b > c. Solving each branch independently yields a union of two rays.

Variable Reference

• a (Coefficient of x) — The nonzero multiplier of x inside the absolute value bars. Negative values reverse inequality directions during division; the calculator handles this automatically.
• b (Constant inside |·|) — The additive constant inside |ax + b|. It shifts the center of the solution interval left or right along the number line.
• c (Right-side value) — The non-negative threshold on the right-hand side. When c = 0, a strict less-than inequality has no solution. When c < 0, a less-than inequality is unsolvable; a greater-than inequality is satisfied by all real numbers.

Step-by-Step Solution Process

To solve |ax + b| < c manually, follow these steps: (1) Confirm that c ≥ 0; if not, the solution set is empty. (2) Write the compound inequality −c < ax + b < c. (3) Subtract b from all three parts: −c − b < ax < c − b. (4) Divide all parts by a, flipping the signs if a < 0. (5) State the solution as the open interval ((−c − b)/a, (c − b)/a). For > or ≥ inequalities, split into two separate branches and solve each before writing the union.

Worked Examples

Example 1: |2x + 3| < 7

With a = 2, b = 3, c = 7, the compound inequality becomes −7 < 2x + 3 < 7. Subtracting 3 gives −10 < 2x < 4. Dividing by 2 yields −5 < x < 2, an open interval of length 7 on the number line.

Example 2: |3x − 6| > 9

With a = 3, b = −6, c = 9, split into 3x − 6 < −9 OR 3x − 6 > 9. Adding 6 and dividing by 3 gives x < −1 OR x > 5, written in interval notation as (−∞, −1) ∪ (5, +∞).

Example 3: |−2x + 4| ≤ 8

With a = −2, b = 4, c = 8, expand to −8 ≤ −2x + 4 ≤ 8. Subtract 4: −12 ≤ −2x ≤ 4. Divide by −2 and flip signs: −2 ≤ x ≤ 6. The closed interval [−2, 6] is the complete solution set.

Special Cases and Edge Conditions

• |ax + b| < 0 with strict inequality: no real solution exists, since absolute values are always non-negative.
• c < 0: |ax + b| > c is true for all real x; |ax + b| < c has no solution.
• a = 0: the expression reduces to |b| compared to c, which is either always true or always false.

Real-World Applications

Manufacturing tolerances rely directly on this structure. A bolt specified at 25 mm ± 0.05 mm must satisfy |x − 25| ≤ 0.05, giving the acceptance range 24.95 mm to 25.05 mm. Voltage regulators, GPS positional accuracy windows, and statistical confidence intervals all use the same underlying inequality to bound acceptable variation around a target value.

Methodology and Sources

The algebraic methods applied by this calculator follow the two-case decomposition of absolute value expressions standard in precalculus curricula. The core procedure is documented in Khan Academy's introduction to absolute value inequalities and formalized in MCC Kansas City's Solving Absolute Value Equations and Inequalities (PDF). Graphical verification of solution intervals aligns with the approach presented in Richland Community College's Section 2.5 on solving inequalities algebraically and graphically.