Absolute Value Inequalities Calculator

Solve any absolute value inequality with detailed step-by-step solutions

Expression inside absolute value (e.g., 2x + 3)

Select inequality type

How to Use This Calculator

This calculator makes solving absolute value inequalities straightforward. Here’s what you need to do:

Enter the expression that goes inside the absolute value bars. For example, if you’re solving |2x + 3| < 7, type “2x + 3” in the expression field.
Click on the inequality symbol you need: less than (<), greater than (>), less than or equal to (≤), or greater than or equal to (≥).
Enter the number on the right side of your inequality.
Hit the “Solve Inequality” button and watch as the calculator shows you the solution with complete steps.

The calculator handles all types of expressions, including those with variables, coefficients, and constants. You’ll get interval notation, number line visualization, and detailed explanation of each step.

What Are Absolute Value Inequalities?

Think of absolute value as the distance from zero on a number line. When you write |x|, you’re asking “how far is x from zero?” That distance is always positive or zero, never negative.

An absolute value inequality combines this distance concept with comparison. When you see |x – 5| < 3, you’re really asking: “what values of x are less than 3 units away from 5?” The answer forms a range of values, not just a single number.

Real-World Example

Imagine a thermostat set to 68°F with a tolerance of ±2°F. This situation can be written as |T – 68| ≤ 2, which means the actual temperature T must stay within 2 degrees of 68. The solution is 66 ≤ T ≤ 70.

Solving Different Types

Less Than Inequalities (|ax + b| < c)

When you have an absolute value less than a number, you’re looking for values trapped between two boundaries. The expression |ax + b| < c splits into: -c < ax + b < c. This creates a bounded interval.

Example: |x – 2| < 4

This means x must be less than 4 units from 2. Split it: -4 < x – 2 < 4. Add 2 throughout: -2 < x < 6. The solution is all numbers between -2 and 6.

Greater Than Inequalities (|ax + b| > c)

When absolute value exceeds a number, you’re looking for values far from a center point. The expression |ax + b| > c splits into two separate conditions: ax + b < -c OR ax + b > c. This creates two separate regions.

Example: |2x + 1| > 5

Split into: 2x + 1 < -5 OR 2x + 1 > 5. Solve each: x < -3 OR x > 2. The solution includes everything less than -3 and everything greater than 2.

Special Cases You Should Know

Sometimes absolute value inequalities have unusual solutions:

If |expression| < negative number: No solution exists. Absolute values can’t be negative, so nothing satisfies this.
If |expression| > negative number: All real numbers work. Since absolute values are never negative, they’re always greater than any negative number.
If |expression| < 0: No solution. The only way absolute value equals zero is if the expression inside equals zero.
If |expression| ≥ 0: All real numbers satisfy this since absolute value is never negative.

Inequality Type	How to Split	Solution Form	Number Line
\|x\| < a	-a < x < a	Single interval	Bounded region
\|x\| > a	x < -a OR x > a	Two intervals	Two separate regions
\|x\| ≤ a	-a ≤ x ≤ a	Closed interval	Bounded region with endpoints
\|x\| ≥ a	x ≤ -a OR x ≥ a	Two rays	Two regions with endpoints

Common Mistakes and How to Avoid Them

Mistake: Forgetting to split the inequality into two cases

When solving |2x – 1| > 3, students sometimes only solve 2x – 1 > 3 and forget about 2x – 1 < -3. Remember: greater than inequalities always create two separate regions. Always write both cases.

Mistake: Using OR instead of AND for less than inequalities

For |x + 2| < 5, the solution is -5 < x + 2 AND x + 2 < 5, which combines into -7 < x < -3. Don’t write it as two separate conditions with OR. The less than symbol creates a single bounded region.

Mistake: Incorrectly handling negative numbers on the right side

If you see |x – 3| < -2, stop right there. This has no solution because absolute value is never negative. Similarly, |x – 3| > -2 is true for all real numbers.

Mistake: Forgetting to flip the inequality when dividing by negatives

When solving -2x < 6 (which might come from splitting an absolute value inequality), you must flip the sign when dividing by -2, giving you x > -3. This rule applies to all inequalities, not just those with absolute values.

Mistake: Mixing up open and closed circles on number lines

Use open circles for < and > (strict inequalities). Use closed circles for ≤ and ≥ (inequalities that include the endpoint). For |x| ≤ 3, the points -3 and 3 are included, so use closed circles.

Frequently Asked Questions

What’s the difference between absolute value equations and inequalities?

Equations like |x| = 5 have specific solutions (x = 5 or x = -5). Inequalities like |x| < 5 have ranges of solutions (-5 < x < 5). Equations give you points; inequalities give you intervals.

Can an absolute value inequality have no solution?

Yes, definitely. Any inequality like |expression| < negative number has no solution. For example, |x + 1| < -3 is impossible because absolute values are never negative.

How do I write my answer in interval notation?

For bounded solutions like -2 < x < 3, write (-2, 3). Use parentheses for < and >, brackets for ≤ and ≥. For two separate regions like x < -1 or x > 4, write (-∞, -1) ∪ (4, ∞). The union symbol ∪ means “or.”

Why do greater than inequalities create two separate solutions?

Think about |x| > 3. You need values more than 3 units from zero. That happens in two ways: go far right (x > 3) or go far left (x < -3). You can’t get from one region to the other without passing through values that don’t work.

Can I have absolute value on both sides of an inequality?

Yes, but it gets trickier. For |2x – 1| < |x + 3|, you need to consider multiple cases based on where the expressions inside change sign. It’s best to solve these by testing intervals or squaring both sides (when appropriate).

What if the coefficient of the variable is negative inside the absolute value?

It works the same way. For |-2x + 5| < 7, split it as -7 < -2x + 5 < 7, then solve. Remember to flip the inequality when dividing by -2. You’ll get -1 < x < 6.

How do I check if my solution is correct?

Pick test values from your solution regions and plug them into the original inequality. Also test values outside your solution to confirm they don’t work. If you got -2 < x < 5, try x = 0 (should work) and x = 6 (shouldn’t work).

Step-by-Step Solution Process

Here’s your roadmap for solving any absolute value inequality:

Step 1: Isolate the Absolute Value

Make sure the absolute value expression stands alone on one side. If you see 2|x – 3| + 5 < 15, first subtract 5 from both sides, then divide by 2 to get |x – 3| < 5.

Step 2: Check the Right Side

Look at the number you’re comparing to. Is it positive, negative, or zero? This determines whether you have solutions, no solutions, or all numbers as solutions.

Step 3: Split According to Inequality Type

For less than: Create one compound inequality with the expression sandwiched between negative and positive versions of the right side. For greater than: Create two separate inequalities with OR between them.

Step 4: Solve Each Piece

Work through the algebra carefully. Remember to flip inequality signs when multiplying or dividing by negative numbers.

Step 5: Write in Interval Notation

Convert your solution to interval notation. Use ∪ to combine separate regions for greater than inequalities.

Step 6: Verify Your Answer

Test a value from your solution in the original inequality. If it works, you’re done. If not, check your algebra.