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Absolute value

Absolute value inequalities

Absolute value inequalities are easiest to read as distance questions: is the point closer, farther away, or exactly on the boundary?

|x| < a
|x - a| > r
Solution intervals

Inequalities of the form |x| around zero

Assume a > 0. The expression |x| is the distance from x to zero, so < and <= choose the middle interval, while > and >= choose the two outside parts of the line.

|x| < a

Geometric interpretation

We want points whose distance from zero is less than a. They lie strictly between -a and a.

Algebraic solution

By the distance interpretation, |x| < a is equivalent to the double inequality -a < x < a.

Example

For |x| < 5, the solution is x ∈ (-5, 5).

Solution set

x ∈ (-a, a)

Step by step

1.Make sure a > 0.

2.Mark -a and a on the number line.

3.Choose the points between them, without endpoints.

4.Write the open interval.

Geometric interpretation

Choose a variant

Inequality

|x| < 2

Interval

x ∈ (-2, 2)

-2
0
2

We want points whose distance from zero is less than a. They lie strictly between -a and a.

|x| <= a

Geometric interpretation

We want points at most a units away from zero. The points -a and a are included.

Algebraic solution

The inequality |x| <= a gives -a <= x <= a.

Example

For |x| <= 3, the solution is x ∈ [-3, 3].

Solution set

x ∈ [-a, a]

Step by step

1.Make sure a > 0.

2.Mark the boundaries -a and a.

3.Because the sign is <=, include both boundaries.

4.Write the closed interval.

Geometric interpretation

Choose a variant

Inequality

|x| <= 2

Interval

x ∈ [-2, 2]

-2
0
2

We want points at most a units away from zero. The points -a and a are included.

|x| > a

Geometric interpretation

We want points farther than a units from zero. They lie left of -a or right of a.

Algebraic solution

The inequality |x| > a splits into two conditions: x < -a or x > a.

Example

For |x| > 4, the solution is x ∈ (-∞, -4) ∪ (4, ∞).

Solution set

x ∈ (-∞, -a) ∪ (a, ∞)

Step by step

1.Make sure a > 0.

2.Mark -a and a.

3.Choose the outside parts of the number line.

4.Do not include the boundaries because the inequality is strict.

Geometric interpretation

Choose a variant

Inequality

|x| > 2

Interval

x ∈ (-∞, -2) ∪ (2, ∞)

-2
0
2

We want points farther than a units from zero. They lie left of -a or right of a.

|x| >= a

Geometric interpretation

We want points at least a units away from zero. The boundaries -a and a are included.

Algebraic solution

The inequality |x| >= a gives x <= -a or x >= a.

Example

For |x| >= 2, the solution is x ∈ (-∞, -2] ∪ [2, ∞).

Solution set

x ∈ (-∞, -a] ∪ [a, ∞)

Step by step

1.Make sure a > 0.

2.Mark the boundary points -a and a.

3.Choose the outside parts of the line.

4.Include the boundaries because equality is allowed.

Geometric interpretation

Choose a variant

Inequality

|x| >= 2

Interval

x ∈ (-∞, -2] ∪ [2, ∞)

-2
0
2

We want points at least a units away from zero. The boundaries -a and a are included.

Moving the center: |x - a|

The expression |x - a| measures the distance from x to the point a. The radius r determines two boundary points: a - r and a + r.

Inequalities of the form |x - a| with radius r

Assume r > 0. These are the same four patterns, but the center moves from 0 to a.

|x - a| < r

Geometric interpretation

x is closer than r units to the point a.

Algebraic solution

Write -r < x - a < r, then add a to every part.

Example

For |x - 2| < 3, the solution is x ∈ (-1, 5).

Solution set

x ∈ (a - r, a + r)

Step by step

1.Check that r > 0.

2.Turn the absolute value into a double inequality: -r < x - a < r.

3.Add a on the left, in the middle, and on the right.

4.You get an open interval.

Geometric interpretation

Choose a variant

Inequality

|x - 2| < 3

Interval

x ∈ (-1, 5)

-1
2
5

x is closer than r units to the point a.

|x - a| <= r

Geometric interpretation

x is at most r units away from the point a.

Algebraic solution

Write -r <= x - a <= r and add a to every part.

Example

For |x + 1| <= 4, the solution is x ∈ [-5, 3].

Solution set

x ∈ [a - r, a + r]

Step by step

1.Check that r > 0.

2.Write -r <= x - a <= r.

3.Add a to every side of the double inequality.

4.Include both endpoints.

Geometric interpretation

Choose a variant

Inequality

|x - 2| <= 3

Interval

x ∈ [-1, 5]

-1
2
5

x is at most r units away from the point a.

|x - a| > r

Geometric interpretation

x is farther than r units from the point a, so it lies outside the boundary segment.

Algebraic solution

Split into x - a < -r or x - a > r, then add a.

Example

For |x - 3| > 2, the solution is x ∈ (-∞, 1) ∪ (5, ∞).

Solution set

x ∈ (-∞, a - r) ∪ (a + r, ∞)

Step by step

1.Check that r > 0.

2.Write two cases: x - a < -r or x - a > r.

3.Add a in both inequalities.

4.Join the two outside intervals.

Geometric interpretation

Choose a variant

Inequality

|x - 2| > 3

Interval

x ∈ (-∞, -1) ∪ (5, ∞)

-1
2
5

x is farther than r units from the point a, so it lies outside the boundary segment.

|x - a| >= r

Geometric interpretation

x is at least r units away from the point a, so the boundaries also satisfy the condition.

Algebraic solution

Split into x - a <= -r or x - a >= r, then add a.

Example

For |x + 2| >= 5, the solution is x ∈ (-∞, -7] ∪ [3, ∞).

Solution set

x ∈ (-∞, a - r] ∪ [a + r, ∞)

Step by step

1.Check that r > 0.

2.Write x - a <= -r or x - a >= r.

3.Add a in both parts.

4.Include the endpoints of the outside intervals.

Geometric interpretation

Choose a variant

Inequality

|x - 2| >= 3

Interval

x ∈ (-∞, -1] ∪ [5, ∞)

-1
2
5

x is at least r units away from the point a, so the boundaries also satisfy the condition.

Inequalities that require cases

If the right-hand side contains x, or the absolute value is not a clean distance from a fixed point, the safest method is to consider the sign of the expression inside the absolute value.

|x - 2| <= x + 4

One absolute value and an expression on the right

Geometric interpretation

The left-hand side is the distance from 2, while the right-hand side changes with x. You cannot treat it as a fixed radius.

Algebraic solution

The split point is x = 2. The right-hand side also has to be nonnegative, but the case solution naturally enforces that.

Step by step

  1. 1.

    Consider the case x < 2.

    |x - 2| = -x + 2

    -x + 2 <= x + 4

  2. 2.

    Solve the inequality and intersect with the case condition.

    -x + 2 <= x + 4

    -2 <= 2x

    x >= -1

    x ∈ [-1, 2)

  3. 3.

    Consider the case x >= 2.

    |x - 2| = x - 2

    x - 2 <= x + 4

  4. 4.

    Check whether the inequality holds on the whole interval.

    x - 2 <= x + 4

    -2 <= 4

    x ∈ [2, ∞)

  5. 5.

    Combine the results from both cases.

    [-1, 2) ∪ [2, ∞) = [-1, ∞)

Solution set

x ∈ [-1, ∞)

-1
2

x ∈ [-1, ∞)

Inequalities with many absolute values

With many absolute values, each one can change its formula at a different point. Split the number line by all zeros of the expressions inside absolute values.

|x - 1| + |x + 2| < 7

Two absolute values and a sum of distances

Geometric interpretation

The sum of distances from 1 and -2 must be less than 7.

Algebraic solution

The split points are x = -2 and x = 1. Solve the inequality separately on the three intervals.

Step by step

  1. 1.

    Consider the interval x < -2.

    |x - 1| + |x + 2| = -x + 1 - x - 2

    -2x - 1 < 7

    x > -4

    x ∈ (-4, -2)

  2. 2.

    Consider the interval -2 <= x < 1.

    |x - 1| + |x + 2| = -x + 1 + x + 2

    3 < 7

    x ∈ [-2, 1)

  3. 3.

    Consider the interval x >= 1.

    |x - 1| + |x + 2| = x - 1 + x + 2

    2x + 1 < 7

    x < 3

    x ∈ [1, 3)

  4. 4.

    Combine the three solution parts.

    (-4, -2) ∪ [-2, 1) ∪ [1, 3) = (-4, 3)

Solution set

x ∈ (-4, 3)

-4
-2
1
3

x ∈ (-4, 3)

Exam-style examples

In exam tasks, the key is to recognize the inequality type quickly, mark endpoints correctly, and check the conditions from each case.

|2x - 6| <= 8

Closed interval example

Geometric interpretation

Since |2x - 6| = 2|x - 3|, the distance from x to 3 is at most 4.

Algebraic solution

Write -8 <= 2x - 6 <= 8 and solve the double inequality.

Step by step

  1. 1.

    Write the double inequality and add 6.

    -8 <= 2x - 6 <= 8

    -2 <= 2x <= 14

  2. 2.

    Divide each part by 2.

    -1 <= x <= 7

  3. 3.

    Include the closed endpoints.

    x ∈ [-1, 7]

Solution set

x ∈ [-1, 7]

-1
3
7

x ∈ [-1, 7]

|3 - x| > 4

Two-interval example

Geometric interpretation

|3 - x| = |x - 3|, so x must be farther than 4 units from 3.

Algebraic solution

Split into x - 3 < -4 or x - 3 > 4.

Step by step

  1. 1.

    Solve the left case.

    x - 3 < -4

    x < -1

  2. 2.

    Solve the right case.

    x - 3 > 4

    x > 7

  3. 3.

    Combine the outside intervals without endpoints.

    x ∈ (-∞, -1) ∪ (7, ∞)

Solution set

x ∈ (-∞, -1) ∪ (7, ∞)

-1
3
7

x ∈ (-∞, -1) ∪ (7, ∞)

|x + 1| > 2x - 1

Example requiring cases

Geometric interpretation

The left-hand side is distance from -1, but the right-hand side depends on x, so it is not a fixed radius.

Algebraic solution

Split the line at x = -1 and solve the inequality on each interval.

Step by step

  1. 1.

    Consider the case x < -1.

    |x + 1| = -x - 1

    -x - 1 > 2x - 1

    x < 0

    x ∈ (-∞, -1)

  2. 2.

    Consider the case x >= -1.

    |x + 1| = x + 1

    x + 1 > 2x - 1

    x < 2

    x ∈ [-1, 2)

  3. 3.

    Combine the results.

    (-∞, -1) ∪ [-1, 2) = (-∞, 2)

Solution set

x ∈ (-∞, 2)

-1
2

x ∈ (-∞, 2)

Common mistakes

Most errors come from applying formulas automatically without checking the inequality sign and whether endpoints should be included.

Confusing center and radius

In |x - a| < r, the center is a and the radius is r. The endpoints are a - r and a + r.

Turning a union into an intersection

For |x| > a, the answer is two outside intervals joined by or.

Losing equality

The signs <= and >= mean closed endpoints. The signs < and > mean open endpoints.

Not checking the case

If you solve on an interval, the result must belong to that interval.

How to recognize the inequality type quickly

First check whether the absolute value is compared with a positive constant. If yes, use geometry. If the right-hand side contains x or there are many absolute values, use cases.

One distance from zero

|x| < a or |x| > a. The center is 0 and the boundaries are -a and a.

One distance from a point

|x - a| < r or |x - a| > r. The center is a and the boundaries are a - r and a + r.

Right-hand side contains x

Do not treat it as a fixed radius. Consider cases or check nonnegativity conditions.

Many absolute values

Find all zeros of the expressions inside absolute values and split the number line there.

Summary table

In the table, assume a > 0 and r > 0. These are the core patterns behind most simple absolute value inequalities.

TypeConditionGeometrySolution set
|x| < aa > 0closer than a to zero(-a, a)
|x| <= aa > 0at most a from zero[-a, a]
|x| > aa > 0farther than a from zero(-∞, -a) ∪ (a, ∞)
|x| >= aa > 0at least a from zero(-∞, -a] ∪ [a, ∞)
|x - a| < rr > 0closer than r to a(a - r, a + r)
|x - a| <= rr > 0at most r from a[a - r, a + r]
|x - a| > rr > 0farther than r from a(-∞, a - r) ∪ (a + r, ∞)
|x - a| >= rr > 0at least r from a(-∞, a - r] ∪ [a + r, ∞)