Last updated October 12, 2026 · Report an error
An absolute value inequality like |2x+3| < 5 hides two separate conditions at once, since the expression inside the bars could be positive or negative. This tool splits it into the correct compound inequality automatically and shows the interval notation.
How it works
For a "less than" inequality (c > 0), it becomes -c < ax+b < c. For a "greater than" inequality, it splits into two separate rays: ax+b > c or ax+b < -c. Negative or zero right-hand sides are handled as special cases (no solution, or all real numbers).
- Enter a (coefficient of x).
- Enter b (constant inside).
- Enter inequality.
- Enter c (right side).
- Click Calculate to see your results.
Examples
A bounded solution
|2x+3| < 5 becomes -5 < 2x+3 < 5, which solves to -4 < x < 1.
A two-ray solution
|x-1| > 2 splits into x-1 > 2 or x-1 < -2, giving x > 3 or x < -1.
Who should use it
- Algebra 1 and Algebra 2 coursework on absolute value inequalities.
- Checking homework answers before a test.
Industry applications
- Mathematics education
Advantages
- Handles all four inequality types (<, ≤, >, ≥) and the negative/zero right-hand-side edge cases correctly.
- Shows both the compound inequality and interval notation.
Limitations
- Limited to a single absolute value term (|ax+b|) rather than more complex nested expressions.
Common mistakes to avoid
- Forgetting to flip the inequality sign appropriately when a is negative and dividing through by it.
Best practices
- Always check the sign of the right-hand side first — it determines whether you get a bounded interval, two rays, no solution, or all real numbers.
Tips
- Sketching the boundary points on a number line first makes it easy to double-check whether the solution should be the middle interval or the two outer rays.
Frequently asked questions
Since an absolute value can never be negative, |expression| < (negative number) has no solution, while |expression| > (negative number) is true for every real number.
An absolute value inequality like |x| > c means x is more than c units from zero in either direction, so the solution set naturally splits into two disconnected rays rather than one bounded interval.