Solve absolute value inequalities and find the solution set or interval.
Real-number inputs only (decimals or scientific notation); complex values are not supported here.
Absolute value inequalities involve comparing an absolute value expression to a number using inequality symbols (<, ≤, >, ≥). Unlike equations, inequalities describe a range or set of values that satisfy the condition rather than specific solutions.
The key insight is that absolute value represents distance from zero. For example, |x| < 5 means "all numbers within 5 units of zero," giving the interval -5 < x < 5. Meanwhile, |x| > 5 means "all numbers more than 5 units from zero," giving x < -5 or x > 5.
If |x| < c, then -c < x < c (AND compound inequality - interval solution)
If |x| > c, then x < -c OR x > c (OR compound inequality - union solution)
Because we divide by coefficient a to isolate the absolute value term, the solver automatically flips the inequality whenever a is negative so the math stays correct.
< means strictly less than (open interval, boundary excluded), while ≤ includes the boundary value (closed interval). For |x| < 3, we get (-3, 3). For |x| ≤ 3, we get [-3, 3] with parentheses vs brackets.
Use AND (intersection) for 'less than' inequalities (interval: -c < x < c). Use OR (union) for 'greater than' inequalities (x < -c or x > c). This reflects whether values are close to or far from zero.
Yes. For example, |x| < -2 has no solution because absolute values are always non-negative and cannot be less than a negative number.
Yes. For example, |x| > -1 is always true since absolute values are always non-negative, so every real number satisfies it.
Use parentheses ( ) for open intervals (boundary excluded) and brackets [ ] for closed intervals (boundary included). Use ∞ for infinity: (-∞, 5) means all x < 5.
If a < 0 in a|x+b| = c, divide both sides by a and flip the inequality sign. For example, -2|x| < 4 becomes |x| > -2 (always true).
For intervals like -4 < x < 10, mark -4 and 10 with open circles and shade between them. For unions like x ≤ -3 or x ≥ 2, use closed circles and shade two separate regions.
Yes! Graph both sides of the inequality: y₁ = |expression| and y₂ = constant. The solution is where the graphs satisfy the inequality relationship (above, below, or intersecting).
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