Enter an absolute value inequality like |2x - 3| ≤ 7 into the inequality expression field and choose your variable. The Absolute Value Inequalities Calculator solves for the solution set, returning the solution interval, inequality type, and a breakdown of both cases — all shown in standard interval notation.
Solution Interval
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Lower Bound (x ≥ or x >)
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Upper Bound (x ≤ or x <)
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Case 1: ax + b ≥ 0
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Case 2: ax + b < 0
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Solution Type
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An absolute value inequality is an inequality that contains an absolute value expression, such as |2x - 3| ≤ 7. The absolute value measures the distance of a number from zero, so |x| represents how far x is from 0 on the number line. Solving these inequalities means finding all values of x that satisfy the distance condition.
For a 'less than' type (|ax + b| < c or ≤ c), you split it into a compound inequality: -c < ax + b < c. Solve each part separately to find the range of x, which gives a bounded interval. For example, |2x - 3| ≤ 7 becomes -7 ≤ 2x - 3 ≤ 7, which solves to -2 ≤ x ≤ 5, written as [-2, 5].
For a 'greater than' type (|ax + b| > c or ≥ c), you split into two separate inequalities: ax + b > c OR ax + b < -c. Solve each independently and combine the results with a union (∪). This produces an unbounded solution — two rays going in opposite directions on the number line.
Interval notation describes the solution set using brackets and parentheses. A square bracket [ or ] means the endpoint is included (≤ or ≥), while a round bracket ( or ) means it is excluded (< or >). For example, [-2, 5] means -2 ≤ x ≤ 5, and (-∞, -1) ∪ (3, ∞) means x < -1 or x > 3.
If c < 0, then |ax + b| < c has no solution, because absolute values are always non-negative and can never be less than a negative number. Conversely, |ax + b| > c where c < 0 is always true for all real x, since any absolute value is greater than any negative number.
The difference is whether the boundary points are included in the solution. With ≤ (less than or equal to), the endpoints are part of the solution and shown with square brackets in interval notation. With < (strictly less than), the endpoints are excluded and shown with parentheses. On a number line, included endpoints are shown as filled circles and excluded ones as open circles.
Yes. |ax + b| < c where c ≤ 0 yields no solution. |ax + b| > c where c < 0 yields all real numbers (−∞, ∞). Also, if a = 0 and b is a constant, the expression simplifies to |b| compared to c, giving either no solution or all reals depending on the comparison.
First divide both sides by the outer coefficient to isolate the absolute value. For 3|x - 2| ≤ 9, divide by 3 to get |x - 2| ≤ 3. Then solve as normal: -3 ≤ x - 2 ≤ 3, giving -1 ≤ x ≤ 5. This calculator handles the outer coefficient automatically using the 'multiplier' field.