Absolute Value Inequality Calculator
Solve |x| inequalities step-by-step — and finally understand when to use AND vs OR.
Three supported modes
Absolute Value Solver
Built for absolute value inequalities with clear AND/OR rule guidance, number-line graphing, distance interpretation, interval notation, and value verification.
Supported Input Styles
|x-3|<5becomes an AND rule middle interval.abs(2x+1)>7is normalized into bars automatically.3|x-1|+2<8isolates the absolute value before solving.|x+5|<-2triggers the no-solution special case.|2x-4|>-3triggers the all-real-numbers special case.|x-1|>|x+3|uses squaring, factoring, and a sign chart.
Math Keyboard
Tap bars, centers, symbols, and constants for fast absolute-value input.
Result
The absolute-value solution set is shown below in notation and on the number line.
Both conditions must be true, so the answer becomes a bounded middle interval.
Original
Isolated
Center
x = 2
Radius
5
Step 1
Isolate the absolute value expression
The absolute-value expression is already isolated on one side of the inequality.
Before
After
Step 2
Check the right-hand side
A positive right-hand side means the inequality can be converted into linear branches.
Step 3
Apply the AND rule
For a strict less-than comparison, both linear bounds must hold and both endpoints stay open.
Before
After
Step 4
Upper bound: Move all terms to one side
Rewrite the inequality so the right-hand side is zero.
Before
After
Step 5
Upper bound: Isolate x
Add or subtract the constant term so that the x-term stands alone.
Before
After
Step 6
Upper bound: Divide by the coefficient of x
Dividing both sides by a positive number keeps the inequality direction the same.
Before
After
Step 7
Lower bound: Move all terms to one side
Rewrite the inequality so the right-hand side is zero.
Before
After
Step 8
Lower bound: Isolate x
Add or subtract the constant term so that the x-term stands alone.
Before
After
Step 9
Lower bound: Divide by the coefficient of x
Dividing both sides by a positive number keeps the inequality direction the same.
Before
After
Step 10
Keep only the overlap
For |A| <= k, the solution is the intersection of the lower and upper bounds.
Before
After
AND/OR Rule Visualizer
Less than stays between. Greater than moves beyond.
The calculator highlights this same decision after every solve. The default example |x - 2| < 5 is an AND case, while greater-than problems split into two outside branches.
|f(x)| < a -> AND
Betweenx must be between -a and a. Both bounds must be true at once, so the graph is one connected middle interval.
|f(x)| > a -> OR
Beyondx must be beyond -a or beyond a. Either outside branch makes the statement true, so the answer is a union.
Step-by-Step Breakdown
The route changes before the algebra starts
Positive constant
Less-than symbols use AND and create one middle interval. Greater-than symbols use OR and create two outer rays.
Negative or zero
The right side can decide the answer before branch solving: no solution, all real numbers, one point, or all reals except one point.
Two absolute values
The shortcut is not AND/OR. Square both sides, factor the difference of squares, then use the sign chart.
The most important rule on the page is simple to say and easy to forget under pressure: less-than absolute value inequalities become AND statements, while greater-than absolute value inequalities become OR statements. That is not a trick. It comes directly from the meaning of distance.
If |A| < c, then A must stay between -c and c. Both bounds have to hold at the same time, so the answer is an AND statement and the graph is a middle interval. If |A| > c, then A must fall to the left of -c or to the right of c. Only one outer branch needs to hold, so the answer is an OR statement and the graph becomes two outer rays.
This is why the number line picture matters so much. Less than means closer than c, so the answer stays inside a neighborhood around the center. Greater than means farther than c, so the answer moves outside that neighborhood. Once students connect the words closer and farther to the graph, the algebra becomes much easier to remember. For the same set logic without absolute-value bars, use the compound inequality calculator to compare AND as intersection and OR as union.
Examples
Three example types the calculator handles
Basic AND rule
|x - 2| < 5
Split the inequality into -5 < x - 2 < 5, then add 2 across the compound inequality to get -3 < x < 7.
Less than -> AND -> one connected interval.
Interval
Method
Use AND and keep the middle interval.
Negative right side
|x + 1| < -3
The calculator stops before branch solving because absolute value is never negative. The final answer is No Solution.
Absolute value is never negative, so no further solving is needed.
Interval
Method
No Solution
Two absolute values
|x - 1| > |x + 3|
Square both nonnegative sides, subtract, factor (x - 1)^2 - (x + 3)^2 into (-4)(2x + 2), then use the sign chart to keep x < -1.
Two absolute values become a sign-chart inequality.
Interval
Method
Square, factor, and solve with the sign chart.
FAQ
Frequently Asked Questions
When do you use AND vs OR when solving absolute value inequalities?
Use AND for < and <= because the expression must stay between two bounds at the same time. Use OR for > and >= because the expression can satisfy either outer branch.
What happens when an absolute value inequality is compared to a negative number?
If |A| is less than a negative number, there is no solution. If |A| is greater than a negative number, every real number works, because absolute value is always at least zero.
Why is |x| always greater than or equal to zero?
Absolute value measures distance from zero on the number line. Distance cannot be negative, so |x| is always zero or positive.
How do you solve an inequality with two absolute values?
For |f(x)| > |g(x)| or similar comparisons, square both nonnegative sides, move everything to one side, factor f(x)^2 - g(x)^2 as (f-g)(f+g), then solve the resulting sign chart.
What's the difference between absolute value equations and inequalities?
An absolute value equation usually asks for exact points at a fixed distance. An absolute value inequality asks for every point inside or outside a distance range, so the answer is often an interval or a union of intervals.
How do you write the solution to an absolute value inequality in interval notation?
A less-than absolute value inequality usually becomes one middle interval, such as (-3, 7). A greater-than inequality usually becomes two outer intervals joined by a union, such as (-infinity, -4) union (3, infinity). Open the interval notation calculator.