Enter one absolute value comparison with a linear expression inside the bars. See why a small-distance condition gives a middle interval and a large-distance condition gives two outer rays.
Result
The expression remains within 7 units of zero from -8/3 through 2
−38≤x≤2⟺[−38,2]
Conditions
x is real and the expression inside the absolute value is linear.
The positive threshold 7 allows the inside expression to lie between -7 and 7.
Both endpoints are included because the comparison uses ≤.
Steps
Rewrite the distance condition At most 7 units from zero means the inside expression lies between -7 and 7.−7≤3x+1≤7
Subtract from all three parts Subtract 1 from both outer bounds and the middle expression.−8≤3x≤6
Divide by the positive coefficient Dividing all parts by 3 preserves both comparison directions.−38≤x≤2
✓
Independent check
At x = -8/3 and x = 2, |3x + 1| = 7. At x = 0 it is 1, so the middle passes; at x = 3 it is 10, so the outside fails.
Scope
What this absolute value inequality covers
When c is positive, |u| ≤ c means -c ≤ u ≤ c, while |u| ≥ c means u ≤ -c or u ≥ c. Strict comparisons exclude their boundary points. Zero and negative thresholds need separate checks before applying either pattern.
Find a bounded middle interval
For an absolute value less than a positive threshold, solve both bounds together and retain the correct endpoint types.
Examples: |3x + 1| ≤ 7, |x - 2| < 4
Find two outer rays
For an absolute value greater than a positive threshold, solve the negative-side and positive-side cases and join them with OR.
Examples: |2x + 1| ≥ 3, |x + 4| > 3
Handle zero or negative thresholds
Use the fact that absolute value is nonnegative before splitting; otherwise an impossible middle interval can appear plausible.
Examples: |x - 1| < 0, |x + 1| ≥ -2
Check distance and endpoints
Substitute a point in each region into the original absolute value comparison, then check any candidate endpoint for inclusion.
Examples: open circle for <, closed circle for ≤
How to use it
Enter enough information for one clear task
1
Enter one distance condition
Use absolute value bars or abs(...) around a linear expression in x, followed by one comparison to a real constant. Keep parentheses around the full inside expression.
2
Read the shape before the endpoints
Less than produces a middle region; greater than produces outer regions when the threshold is positive. The number line helps distinguish the shapes.
3
Check the original distance
Test one point inside and outside the reported set. At a boundary, see whether equality is permitted by the original symbol.
Worked inputs
Examples to try
Use these examples to recognize the method, compare equivalent forms, and check your own work.
Closed middle interval
Write -7 ≤ 3x + 1 ≤ 7 and solve all three parts together.
∣3x+1∣≤7
Expected result
[−38,2]
Open middle interval
The inside expression lies strictly between -4 and 4.
∣x−2∣<4
Expected result
(−2,6)
Two closed outer rays
Solve 2x + 1 ≤ -3 or 2x + 1 ≥ 3.
∣2x+1∣≥3
Expected result
(−∞,−2]∪[1,∞)
Two open outer rays
The distance exceeds 3 to the left of -7 or to the right of -1.
∣x+4∣>3
Expected result
(−∞,−7)∪(−1,∞)
Impossible strict negative distance
No absolute value can be less than zero.
∣x−1∣<0
Expected result
∅
Negative threshold accepts everything
Every absolute value is at least zero and therefore at least -2.
∣x+1∣≥−2
Expected result
x∈R
Zero threshold leaves one point
Nonnegative absolute value can be at most zero only when its inside is exactly zero.
∣2x−6∣≤0
Expected result
x=3
Complete example
Split a greater-than distance into two cases
A distance above 7 cannot lie in the middle. One case falls below -7; the other rises above 7.
∣3x+2∣>7
1
Write both outer cases
The strict comparison makes each case strict.
3x+2<−7or3x+2>7
2
Solve each case
Subtract 2 and divide by positive 3 in each branch.
x<−3orx>35
3
Join the outer intervals
The boundary points make the distance exactly 7, so neither is included.
(−∞,−3)∪(35,∞)
x<−3orx>35
Verification: At x = -3 and 5/3 the absolute value equals 7, so both are excluded. At x = -4 it is 10 and at x = 2 it is 8; both pass. At x = 0 it is 2 and fails.
Avoidable errors
Common mistakes and how to fix them
Using AND for a greater-than result
Problem: Write -7 > 3x + 2 > 7.
Why it matters: A value cannot lie below -7 and above 7 at the same time.
Better approach: Write two outer cases joined by OR, then combine their solution intervals.
Applying the positive-threshold rule to a negative number
Problem: For |x| < -2, split into 2 < x < -2.
Why it matters: Distance is nonnegative, so the original condition is already impossible.
Better approach: Check whether the threshold is negative or zero before splitting into cases.
Forgetting an endpoint's equality status
Problem: Show a closed point for |x - 2| < 4.
Why it matters: At either boundary the absolute value is exactly 4, not less than 4.
Better approach: Substitute each boundary into the original comparison, then mark it open or closed.
Dropping the other side of the distance
Problem: Solve only 3x + 1 ≤ 7 for |3x + 1| ≤ 7.
Why it matters: A large negative inside value has a large absolute value too.
Better approach: For a positive threshold and ≤, include both -7 ≤ 3x + 1 and 3x + 1 ≤ 7.
Trust the result for the right reasons
Checks, assumptions, and limits
How results are checked
Nonnegative absolute value is compared with the threshold before any case split.
Each case is solved as an ordinary linear inequality and combined as an intersection or union.
Endpoints are substituted into the original absolute value statement.
A point from each reported and rejected region is checked against the original distance.
When to stop and revise the input
The expression inside the absolute value must be linear in one real variable.
One absolute value compared with a finite real constant is supported; nested bars, sums of bars, and variable thresholds are not.
An inequality with an absolute value on both sides needs a different method and is not silently simplified.
An empty set or all-real result can be correct, especially when the threshold is zero or negative.
Common questions
Solve an absolute value inequality FAQ
Why does less than give a middle interval?
The absolute value measures distance from zero. A value whose distance is below a positive threshold must stay between the negative and positive threshold values.
Why does greater than give two intervals?
A distance above a positive threshold can occur on either side of zero: sufficiently negative or sufficiently positive. Those alternatives are joined by OR.
What happens if the comparison number is zero?
Because absolute value cannot be negative, |u| < 0 has no solutions, |u| ≤ 0 means u = 0, |u| > 0 excludes the zeros of u, and |u| ≥ 0 is true for all real inputs.
Are the boundary points included?
They are included only for ≤ or ≥ when the original absolute value equals the threshold there. Strict < and > comparisons exclude equality.
Can I enter a quadratic inside the absolute value?
Not in this calculator's supported input. It is designed for a single absolute value of a linear expression compared with a constant, so it can explain each case reliably.
Sources and curriculum alignment
This page follows standard introductory mathematics notation and learning sequences. Use these references to continue with a complete course treatment.
