Inequalities and Absolute Value

Understand solution sets, intervals, constraints, and absolute value as distance.

Learning objectives

After this lesson you should be able to:

  • solve linear inequalities and compound inequalities;
  • write solution sets using interval notation;
  • explain why multiplying an inequality by a negative number reverses the sign;
  • interpret absolute value as distance;
  • solve absolute value equations and inequalities;
  • identify basic domain restrictions from denominators and radicals.

Motivation

Equations often ask for exact values. Inequalities describe allowable regions. That makes them essential for optimization, probability, statistics, and modeling.

Examples:

  • A portfolio weight must satisfy 0 <= w <= 1.
  • A variance must be nonnegative.
  • A logarithm input must be positive.
  • A feasible region in optimization is defined by inequalities.

An inequality is not a weaker equation. It is a statement about a set of values.

Inequality symbols

x < 3     x is less than 3
x <= 3    x is less than or equal to 3
x > 3     x is greater than 3
x >= 3    x is greater than or equal to 3

The solution to an inequality is usually an interval, not a single point.

x < 3

means all real numbers to the left of 3.

Interval notation

Intervals summarize solution sets.

Inequality Interval notation Meaning
x < 3 (-∞, 3) all numbers less than 3
x <= 3 (-∞, 3] all numbers less than or equal to 3
x > -2 (-2, ∞) all numbers greater than -2
-2 <= x < 3 [-2, 3) from -2 included to 3 excluded

Parentheses mean an endpoint is not included. Brackets mean it is included.

Solving linear inequalities

Solve:

3x - 5 < 7

Add 5 to both sides:

3x < 12

Divide by 3:

x < 4

Solution set:

(-∞, 4)

Why the sign flips

When you multiply or divide by a negative number, the inequality direction reverses.

Example:

-2x < 8

Divide by -2:

x > -4

Why? Try one value. If x = -5, then -2x = 10, and 10 < 8 is false. If x = -3, then -2x = 6, and 6 < 8 is true. The correct side is x > -4.

Compound inequalities

A compound inequality combines two constraints:

-2 <= 3x + 1 < 10

Subtract 1 from all three parts:

-3 <= 3x < 9

Divide by 3:

-1 <= x < 3

Interval notation:

[-1, 3)

Absolute value as distance

The expression |x - a| is the distance from x to a.

For example:

|x - 2| <= 5

means "the distance from x to 2 is at most 5." Therefore:

-5 <= x - 2 <= 5

Add 2:

-3 <= x <= 7

So the solution is:

[-3, 7]

Visualizing absolute value

The graph of |x - a| forms a V-shape centered at x = a. A horizontal cutoff line tells you which points are close enough to the center.

The solution is where the V-shaped graph is at or below the cutoff line.

Domain restrictions as inequalities

Many algebraic expressions only make sense for some inputs.

Denominators

1/(x + 1)

requires:

x + 1 != 0

so:

x != -1

Square roots over the real numbers

sqrt(x - 3)

requires:

x - 3 >= 0

so:

x >= 3

Logarithms

log(x - 4)

requires:

x - 4 > 0

so:

x > 4

Worked examples

Example 1: solve a negative-coefficient inequality

Solve:

5 - 2x >= 17

Subtract 5:

-2x >= 12

Divide by -2 and reverse the sign:

x <= -6

Example 2: solve an absolute value equation

Solve:

|x + 3| = 4

The distance from x to -3 is 4, so:

x + 3 = 4  or  x + 3 = -4

Therefore:

x = 1  or  x = -7

Example 3: find a domain

Find the real domain of:

sqrt(x - 2)/(x + 1)

The square root requires x - 2 >= 0, so x >= 2.

The denominator requires x + 1 != 0, so x != -1.

Because x >= 2 already excludes -1, the domain is:

[2, ∞)

Common mistakes

Forgetting to reverse the inequality

Dividing by -2 changes < into > and >= into <=.

Treating absolute value like parentheses

|x - 2| < 5 is not the same as x - 2 < 5. It means the distance from x to 2 is less than 5, which creates a two-sided inequality.

Ignoring endpoint types

x < 3 uses 3) in interval notation, while x <= 3 uses 3].

Practice

  1. Solve 2x + 1 <= 9.
  2. Solve -3x + 4 > 16.
  3. Solve -1 < 2x + 3 <= 7.
  4. Write (-4, 2] as inequalities.
  5. Solve |x - 5| < 2.
  6. Solve |x + 1| >= 3.
  7. Find the real domain of sqrt(x + 4).
  8. Find the real domain of 1/(x^2 - 9).
  9. Find the real domain of log(2x - 6).
  10. Challenge: explain why |x - a| <= r has solution [a-r, a+r] when r >= 0.

Solutions

2x + 1 <= 9 gives 2x <= 8, so x <= 4.

-3x + 4 > 16 gives -3x > 12. Divide by -3 and reverse the sign: x < -4.

Subtract 3: -4 < 2x <= 4. Divide by 2: -2 < x <= 2. Interval notation: (-2, 2].

-4 < x <= 2.

|x - 5| < 2 means -2 < x - 5 < 2. Add 5: 3 < x < 7.

|x + 1| >= 3 means the distance from x to -1 is at least 3. So x + 1 <= -3 or x + 1 >= 3. Therefore x <= -4 or x >= 2.

sqrt(x + 4) requires x + 4 >= 0, so x >= -4.

The denominator cannot be zero. x^2 - 9 = (x - 3)(x + 3), so exclude x = 3 and x = -3.

The logarithm input must be positive: 2x - 6 > 0, so x > 3.

|x - a| <= r means the distance between x and a is at most r. That means x can be no more than r units left of a and no more than r units right of a, so a - r <= x <= a + r.

Why this matters later

  • Optimization: constraints are inequalities that define feasible regions.
  • Probability: random variables often have restricted support such as x >= 0.
  • Calculus: domains determine where limits, derivatives, and integrals make sense.
  • Statistics: confidence intervals and hypothesis-test rejection regions are sets defined by inequalities.

Subtopic guided practice and checkpoints

Inequality work is about preserving the direction of comparison. Guess first, solve carefully, then test a value from your answer set.

Lab 1: interval notation and endpoints

Guess first. Does [-2, 5) include -2? Does it include 5?

Guided exercise.

  1. A square bracket means the endpoint is included.
  2. A parenthesis means the endpoint is not included.
  3. Therefore [-2, 5) includes -2, all numbers between -2 and 5, but not 5.

Checkpoint. Write x > -3 and x <= 4 in interval notation. Then draw the number-line endpoint types in your notes.

Lab 2: solve a linear inequality

Guess first. For -2x + 5 < 13, will the final inequality point left or right?

Guided exercise.

-2x + 5 < 13
-2x < 8                subtract 5
x > -4                 divide by -2 and reverse the inequality

Test x = 0, which is greater than -4:

-2(0) + 5 < 13
5 < 13                 true

Checkpoint. Solve 7 - 3x >= 16. Explain exactly when and why the sign changes direction.

Lab 3: solve a compound inequality

Guess first. Does -1 <= 2x + 3 < 9 produce one interval or two separate intervals?

Guided exercise.

-1 <= 2x + 3 < 9
-4 <= 2x < 6           subtract 3 from all three parts
-2 <= x < 3            divide all three parts by 2

The answer is one connected interval: [-2, 3).

Checkpoint. Solve 4 < 3x - 2 <= 13 and write the answer in interval notation.

Lab 4: absolute value as distance

Guess first. If |x - 4| = 6, should there be one solution or two?

Guided exercise.

|x - 4| = 6 means “the distance from x to 4 is 6.”

x - 4 = 6      or      x - 4 = -6
x = 10                 x = -2

For inequalities, translate distance language:

|x - 4| < 6  means  -6 < x - 4 < 6

Checkpoint. Solve |2x + 1| <= 7. Then test one number inside the interval and one number outside it.

Guessing game checkpoint

Before moving to the Python/SymPy appendix, practice the core moves from this lesson. This short quiz may show an expression, graph, equation, solution set, or description and ask you to choose the matching mathematical object.

Using this lesson with edumath and SymPy

SymPy can solve inequalities and show interval-style answers. Use it to verify your work after you solve the inequality by hand.

import sympy as sp

x = sp.Symbol("x", real=True)
sp.solve_univariate_inequality(-2 <= 3*x + 1, x)

\(\displaystyle -1 \leq x\)

sp.solve_univariate_inequality(sp.Abs(x - 2) <= 5, x)

\(\displaystyle -3 \leq x \wedge x \leq 7\)

Further reading