Algebra Practice

Absolute Value Inequalities Practice Test

Solve inside and outside absolute-value inequalities, interpret distance, and write solution sets correctly.

Absolute Value Inequalities Practice Test

20 mixed questions with interval notation and worked explanations.

Instant feedback · Worked explanations

After the test · absolute value distance zone

Absolute value inequalities describe how far an expression may be from zero or from a chosen center

The expression |x − a| represents the distance between x and a on a number line. That distance interpretation explains the two core patterns: a “less than” inequality keeps x inside a band around the center, while a “greater than” inequality sends x outside that band.

DistanceCenterRadiusInside range Outside rangeANDORInterval notation
Memorizing two formulas helps, but understanding “inside versus outside” makes the sign pattern much harder to confuse.
Expression
|x − 4| < 3
Center
4
Maximum distance
3 units
Meaning
x must stay within 3 units of 4.
Boundary points
4 − 3 = 1 and 4 + 3 = 7

Symmetry ruler

Absolute value creates two symmetric boundary points around a center

1
4
7
← 3 units →
← 3 units →
For |x − 4| < 3, the allowed values lie between the two boundary points: 1 < x < 7.
Less than · inside |x − a| < r
becomes
a − r < x < a + r

The distance from the center is smaller than r, so x must remain between two boundaries. This is an AND situation.

DISTANCE
LOGIC
Greater than · outside |x − a| > r
becomes
x < a − r OR x > a + r

The distance is larger than r, so x must lie outside the central band. This is an OR situation.

Absolute-value pattern converter

Convert the absolute value statement before solving the resulting linear inequality or inequalities

|u| < k
→
−k < u < k
Inside range · AND
|u| ≤ k
→
−k ≤ u ≤ k
Inside range including boundaries
|u| > k
→
u < −k OR u > k
Outside range · OR
|u| ≥ k
→
u ≤ −k OR u ≥ k
Outside range including boundaries
Inside-range example · |2x − 3| < 7
START
|2x − 3| < 7 The distance of 2x − 3 from zero is less than 7.
OPEN
−7 < 2x − 3 < 7 “Less than” becomes a three-part AND inequality.
+3
−4 < 2x < 10 Add 3 to all three parts.
÷2
−2 < x < 5 Divide all three parts by positive 2.
SET
(−2, 5) Both endpoints are open because the original inequality is strict.

Outside-range example · |3x + 1| ≥ 8

A greater-than absolute value inequality splits into two independent branches

Left branch

3x + 1 ≤ −8
3x ≤ −9
x ≤ −3

Values far enough to the left make the expression at most −8.

OR

Right branch

3x + 1 ≥ 8
3x ≥ 7
x ≥ 7/3

Values far enough to the right make the expression at least 8.

Solution: x ≤ −3 OR x ≥ 7/3, which is (−∞, −3] ∪ [7/3, ∞).

Inside versus outside on the number line

The graph should visibly match the distance condition

|x − 4| < 3

1 7

Shade the inside band: 1 < x < 7.

|x − 4| > 3

1 7

Shade outside the band: x < 1 or x > 7.

Special right-side cases

Before splitting the inequality, check whether the right side is positive, zero, or negative

|u| < negative
|x + 2| < −1
No solution. Absolute value cannot be negative.
|u| ≤ negative
|2x − 5| ≤ −3
No solution for the same reason.
|u| > negative
|x − 7| > −2
All real x, because every absolute value is at least 0.
|u| ≤ 0
|x − 6| ≤ 0
Only x = 6, because absolute value equals 0 only when its inside expression is 0.

Absolute-value interval notation

Inside solutions usually form one interval; outside solutions usually form a union of two intervals

|x − 4| < 3
(1, 7)
Strict inside band; both boundaries excluded.
|x − 4| ≤ 3
[1, 7]
Inclusive inside band; both boundaries included.
|x − 4| > 3
(−∞, 1) ∪ (7, ∞)
Two strict outside regions.
|x − 4| ≥ 3
(−∞, 1] ∪ [7, ∞)
Outside regions including both boundary points.

Tolerance model

Absolute value inequalities are natural for acceptable deviation from a target

Target mass
500 g
Allowed deviation
at most 8 g
Absolute value model
|m − 500| ≤ 8
Compound form
−8 ≤ m − 500 ≤ 8
Acceptable range
492 ≤ m ≤ 508

Absolute-value inequality error scan

Most mistakes come from choosing the wrong AND/OR structure or losing symmetry

Less-than turned into OR
|x| < 5 → x < −5 OR x > 5 −5 < x < 5

Distance less than 5 means x stays inside the two boundaries.

Greater-than turned into AND
|x| > 5 → −5 > x > 5 x < −5 OR x > 5

Distance greater than 5 means x lies outside the central band.

Only positive branch kept
|2x − 3| ≥ 7 → 2x − 3 ≥ 7 only 2x − 3 ≤ −7 OR 2x − 3 ≥ 7

Absolute value produces symmetric negative and positive boundary conditions.

Negative right side ignored
|x + 1| < −3 expanded mechanically. No solution.

An absolute value cannot be less than a negative number.

Distance-zone diagnostics

Classify missed questions by the exact absolute-value idea that failed

This keeps review focused on absolute value inequalities instead of treating them as ordinary compound inequalities.

Distance meaning Could you identify the center and distance boundary represented by the absolute value?
Inside vs outside Did less-than become an inside AND range and greater-than become outside OR branches?
Boundary handling Were strict and inclusive endpoints preserved in the graph and interval notation?
Special cases Did you check zero or negative right-hand values before splitting the inequality?