MGU Mathematics Series — Algebra I
Part II: Equations and Inequalities
Symbols · Solution Sets · Graphing · Compound Inequalities · Applications
Chapter 4 of 14
MGU Mathematics Series  ·  Algebra I  ·  Library Textbook Edition

Chapter 4:
Linear Inequalities
in One Variable

Inequality symbols, solution sets, number line graphs, the reversal rule, compound inequalities, and real-world applications — the language of range and constraint

■ Math Level: Algebra I ■

■ 14 Sections  •  Part II of 5 ■

Follows Chapter 3: Linear Equations  ·  Prepares students for Chapter 5 through Chapter 8

Chapter 4 — Front Matter

Purpose and Overview

Chapter 3 introduced equations as precise mathematical statements with one solution, or in special cases, no solution or infinitely many solutions. In each of those situations, the goal was to identify an exact value, or to recognize that no such value existed. Chapter 4 extends that reasoning into new territory.

An inequality is not a statement of equality. It is a statement of comparison. It says that one quantity is greater than, less than, greater than or equal to, or less than or equal to another. When you solve an inequality, you are not usually looking for a single answer. You are looking for a range of values — a set of numbers that all satisfy the condition described.

This shift from single-answer thinking to range-of-values thinking is one of the most important conceptual steps in Algebra I. It appears everywhere in mathematics and in real life. A speed limit is an inequality. A minimum wage requirement is an inequality. A budget constraint is an inequality. A temperature range for safe food storage is an inequality. The language of comparison and range is the language of constraints, and algebra gives us precise tools for working with that language.

This chapter builds directly from Chapter 3. You will use the same inverse-operation strategies you learned for equations. The algebra looks almost identical, with one important exception that deserves special attention: when you multiply or divide both sides of an inequality by a negative number, the direction of the inequality symbol must reverse. This rule has a logical reason behind it, and this chapter explains that reason carefully.

Opening Question

A part-time worker earns $14.50 per hour. She needs to earn at least $400 this week to cover her expenses. How many hours does she need to work? Notice the question does not ask for the exact number of hours — it asks for the minimum. She could work 28 hours, 30 hours, or 35 hours, and all of those satisfy the goal. The answer is not one number; it is a range of numbers. Every value of h satisfying 14.50h ≥ 400 is a valid answer. Finding that solution set is the goal of this chapter.

14Sections
9Objectives
9Key Terms
15+Worked Examples

Learning Objectives

  1. 1Interpret the meaning of an inequality statement in words and symbols.
  2. 2Identify solutions and solution sets of inequalities.
  3. 3Graph inequality solution sets on a number line.
  4. 4Solve one-step, two-step, and multi-step linear inequalities.
  5. 5Solve inequalities with variables on both sides.
  6. 6Explain and apply the rule for reversing the inequality symbol.
  7. 7Solve compound inequalities involving "and" and "or."
  8. 8Write and solve inequalities from word problems.
  9. 9Recognize and correct common inequality errors.
Chapter 4 — Contents

Sections at a Glance

Sections 4.1 – 4.4
What Inequalities Mean: Symbols, Solutions, and Graphs
What an Inequality Says · Symbols · Solutions · Number Line Graphs
Section 4.1

What an Inequality Says

An equation says two quantities are equal. An inequality says one quantity is greater than, less than, greater than or equal to, or less than or equal to another. Consider the difference:

StatementTypeMeaningNumber of Solutions
x = 5Equationx is exactly 5One
x > 5Inequalityx is some number greater than 5Infinitely many
x ≤ 5Inequalityx is 5 or any number less than 5Infinitely many

Inequalities appear naturally wherever you describe a constraint, a range, a limit, a requirement, or a threshold. Here are four real-world examples:

Real SituationInequality
A roller coaster requires riders to be at least 48 inches tall.h ≥ 48
A recipe calls for less than 2 cups of flour.f < 2
The speed limit is at most 65 miles per hour.s ≤ 65
The temperature must be above 20 degrees Celsius.T > 20
Key Idea

An inequality does not mean something is wrong with the equation form. It means the situation calls for a range of possibilities rather than one exact answer. Many of the most important real-world problems are inequalities, not equations.

Section 4.2

Inequality Symbols and Meaning

There are four inequality symbols used in algebra. Each has a precise meaning. The small, pointed end of the symbol always faces the smaller quantity.

SymbolMeaningExampleRead As
<less thanx < 3"x is less than 3"
>greater thanx > 3"x is greater than 3"
≤less than or equal tox ≤ 3"x is less than or equal to 3"
≥greater than or equal tox ≥ 3"x is greater than or equal to 3"

Translating Words to Symbols

PhraseSymbol
is less than / is fewer than<
is greater than / is more than>
is at most / no more than / does not exceed≤
is at least / no less than / minimum of≥
Example 4.2.1
Translating phrases into inequality symbols
  • ProblemWrite an inequality for each statement. (a) The weight w of a package is at most 50 pounds. (b) A student's score s must be more than 70 to pass. (c) Guests g cannot exceed 200. (d) Temperature T is at least −10 degrees.
  • ThinkMatch each phrase to an inequality symbol. Check whether the boundary value is included or excluded.
  • Solve (a) "At most 50" — 50 is allowed, nothing above: w ≤ 50 (b) "More than 70" — 70 is not included: s > 70 (c) "Cannot exceed 200" — 200 is maximum: g ≤ 200 (d) "At least −10" — −10 is allowed, values above: T ≥ −10
  • CheckReread each phrase with the symbol. "The weight is less than or equal to 50 pounds." Matches "at most 50." ✓
Answers: (a) w ≤ 50   (b) s > 70   (c) g ≤ 200   (d) T ≥ −10
Section 4.3

Solutions and Solution Sets

A solution of an inequality is any value of the variable that makes the inequality a true statement. To test whether a value is a solution, substitute it into the inequality and evaluate. The solution set is the complete collection of all values that make the inequality true. For most linear inequalities, this is an infinite set forming a continuous ray on the number line.

Example 4.3.1
Testing whether a value is a solution
  • ProblemDetermine whether each value is a solution of 3x − 1 < 8. (a) x = 2   (b) x = 3   (c) x = 4
  • ThinkSubstitute each value and simplify. Check whether the resulting comparison is true.
  • Solve (a) x = 2: 3(2) − 1 = 5. Is 5 < 8? Yes. ✓ Solution. (b) x = 3: 3(3) − 1 = 8. Is 8 < 8? No. Not a solution. (c) x = 4: 3(4) − 1 = 11. Is 11 < 8? No. Not a solution.
  • CheckValues below 3 should work; x = 2 is below 3. Values at or above 3 should not; x = 3, 4 are at or above 3. Consistent. ✓
Answer: Only x = 2 is a solution.
Solution Set Notation

For the inequality 3x − 1 < 8, the solution is all real numbers less than 3.

Set notation: { x | x < 3 } — "the set of all x such that x is less than 3"

Interval notation: (−∞, 3) — the parenthesis means 3 is not included

Section 4.4

Graphing Inequalities on a Number Line

A number line graph gives an immediate visual picture of what values are included in a solution set. Two rules govern every graph.

The Two Graphing Rules

Open circle ○ — boundary NOT included. Use with < or >.

Closed circle ● — boundary IS included. Use with ≤ or ≥.

After placing the circle, draw an arrow toward the solution set: left for < or ≤, right for > or ≥.

Example 4.4.1
Graphing four inequalities on a number line
  • ProblemGraph: (a) x > 2   (b) x ≤ −1   (c) x ≥ 0   (d) x < 4
  • ThinkFor each, identify the boundary value, whether it is included, and which direction to shade.
  • Solve (a) x > 2: Open circle at 2, arrow pointing right → (b) x ≤ −1: Closed circle at −1, arrow pointing left ← (c) x ≥ 0: Closed circle at 0, arrow pointing right → (d) x < 4: Open circle at 4, arrow pointing left ←

(a) x > 2 — open circle, arrow right

-1 0 1 2 3

(b) x ≤ −1 — closed circle, arrow left

-3 -2 -1 0 1
Open circle = boundary excluded. Closed circle = boundary included. Arrow shows the direction of all solutions.
Student Tip

The open and closed circles carry mathematical meaning — they are not decorative. Ask yourself: "Is the boundary value a solution?" If yes, use a closed circle. If no, use an open circle. Never skip this detail.

Practice 4.1 – 4.4Concept Check
  1. Write an inequality for each: (a) The number of items n must be less than 15. (b) A car's speed s cannot exceed 70 mph. (c) A student needs at least 80 points for a B. (d) A monthly payment p is at most $250.
  2. Test each value in −3x + 5 > 2. Is it a solution? (a) x = 0   (b) x = 1   (c) x = 2
  3. Graph each inequality on a number line. (a) x > −3   (b) x ≤ 5   (c) x < 0   (d) x ≥ −7
  4. Write the inequality shown by each description: (a) Closed circle at 3, arrow pointing right.   (b) Open circle at −2, arrow pointing left.   (c) Open circle at −5, open circle at 3, bold segment between them.
Sections 4.5 – 4.9
Solving Inequalities: One Step to Variables on Both Sides
One-Step · Two-Step · Multi-Step · Variables on Both Sides · Reversal Rule
Section 4.5

One-Step Inequalities

Solving a one-step inequality means applying one inverse operation to isolate the variable. The approach mirrors one-step equations: whatever you do to one side, you must do to the other. This preserves the relationship between the two sides. The critical difference from equations — multiplying or dividing by a negative — is treated fully in Section 4.9.

Example 4.5.1
One-step inequality — subtraction
  • ProblemSolve and graph: x + 7 < 10
  • ThinkThe variable has 7 added to it. Subtract 7 from both sides.
  • Solve x + 7 < 10 x + 7 − 7 < 10 − 7 x < 3
  • CheckTest x = 0: 0 + 7 = 7 < 10. True. ✓   Test x = 4: 4 + 7 = 11 < 10. False. ✓
Answer: x < 3 — Open circle at 3, arrow pointing left.
Example 4.5.2
One-step inequality — division by positive
  • ProblemSolve and graph: 4x ≥ 20
  • ThinkThe variable is multiplied by positive 4. Divide both sides by 4.
  • Solve 4x ≥ 20 4x / 4 ≥ 20 / 4 x ≥ 5
  • CheckTest x = 5: 4(5) = 20 ≥ 20. True. ✓   Test x = 3: 4(3) = 12 ≥ 20. False. ✓
Answer: x ≥ 5 — Closed circle at 5, arrow pointing right.
Example 4.5.3
One-step inequality — division by negative (reversal preview)
  • ProblemSolve: −3x > 12
  • ThinkDivide both sides by −3. Because we divide by a negative, the symbol reverses. (Full explanation in Section 4.9.)
  • Solve −3x > 12 −3x / −3 < 12 / −3 ← symbol reverses x < −4
  • CheckTest x = −5: −3(−5) = 15 > 12. True. ✓   Test x = −3: −3(−3) = 9 > 12. False. ✓
Answer: x < −4 — Open circle at −4, arrow pointing left.
Section 4.6

Two-Step Inequalities

Two-step inequalities require two inverse operations. The strategy: undo addition or subtraction first, then undo multiplication or division. Apply the reversal rule if you multiply or divide by a negative.

Example 4.6.1
Two-step inequality
  • ProblemSolve and graph: 2x − 5 > 9
  • ThinkFirst undo the subtraction of 5 by adding 5. Then divide by 2.
  • Solve 2x − 5 > 9 2x − 5 + 5 > 9 + 5 2x > 14 2x / 2 > 14 / 2 x > 7
  • CheckTest x = 8: 2(8) − 5 = 11 > 9. True. ✓   Test x = 7: 2(7) − 5 = 9 > 9. False. ✓ (7 is excluded.)
Answer: x > 7 — Open circle at 7, arrow pointing right.
Example 4.6.2
Two-step inequality with negative coefficient
  • ProblemSolve and graph: −3x + 4 ≤ 13
  • ThinkSubtract 4 first. Then divide by −3. Dividing by negative → symbol reverses.
  • Solve −3x + 4 ≤ 13 −3x + 4 − 4 ≤ 13 − 4 −3x ≤ 9 −3x / −3 ≥ 9 / −3 ← symbol reverses x ≥ −3
  • CheckTest x = 0: −3(0) + 4 = 4 ≤ 13. True. ✓   Test x = −5: −3(−5) + 4 = 19 ≤ 13. False. ✓
Answer: x ≥ −3 — Closed circle at −3, arrow pointing right.
Section 4.7

Multi-Step Inequalities

Some inequalities require three or more steps. Always simplify each side first — distribute, combine like terms — then isolate the variable. The reversal rule applies whenever you multiply or divide by a negative number, regardless of how many steps the problem takes.

Example 4.7.1
Multi-step inequality with distribution
  • ProblemSolve and graph: 3(x − 2) + 4 < 13
  • ThinkDistribute first, then combine like terms, then solve as a two-step inequality.
  • Solve 3(x − 2) + 4 < 13 3x − 6 + 4 < 13 3x − 2 < 13 3x − 2 + 2 < 13 + 2 3x < 15 x < 5
  • CheckTest x = 4: 3(4 − 2) + 4 = 10 < 13. True. ✓   Test x = 6: 3(6 − 2) + 4 = 16 < 13. False. ✓
Answer: x < 5 — Open circle at 5, arrow pointing left.
Example 4.7.2
Multi-step inequality with combining like terms
  • ProblemSolve and graph: 5x − 3 + 2x ≥ 4x + 9
  • ThinkCombine like terms on the left first, then gather variable terms on one side.
  • Solve 5x − 3 + 2x ≥ 4x + 9 7x − 3 ≥ 4x + 9 7x − 4x − 3 ≥ 9 3x − 3 ≥ 9 3x ≥ 12 x ≥ 4
  • CheckTest x = 5: left side = 32; right side = 29. 32 ≥ 29. True. ✓
Answer: x ≥ 4 — Closed circle at 4, arrow pointing right.
Section 4.8

Inequalities with Variables on Both Sides

When a variable appears on both sides of an inequality, collect all variable terms on one side and all constant terms on the other — exactly as you did with equations in Chapter 3. After simplifying, isolate and solve.

Example 4.8.1
Variables on both sides
  • ProblemSolve and graph: 6x − 2 > 2x + 10
  • ThinkSubtract 2x from both sides to gather variable terms on the left.
  • Solve 6x − 2 > 2x + 10 6x − 2x − 2 > 10 4x − 2 > 10 4x > 12 x > 3
  • CheckTest x = 5: 28 > 20. True. ✓   Test x = 2: 10 > 14. False. ✓
Answer: x > 3 — Open circle at 3, arrow pointing right.
Example 4.8.2
Variables on both sides — variable on the right
  • ProblemSolve and graph: 3x + 7 ≤ 5x − 1
  • ThinkSubtract 3x from both sides to avoid a negative coefficient on the variable term.
  • Solve 3x + 7 ≤ 5x − 1 7 ≤ 5x − 3x − 1 7 ≤ 2x − 1 8 ≤ 2x 4 ≤ x Rewrite: x ≥ 4
  • CheckTest x = 4: 19 ≤ 19. True. ✓   Test x = 2: 13 ≤ 9. False. ✓
Answer: x ≥ 4 — Closed circle at 4, arrow pointing right.
Student Tip

When the variable lands on the right side, rewrite the inequality with the variable on the left. The statements 4 ≤ x and x ≥ 4 say exactly the same thing. Writing the variable first makes the solution easier to read and graph.

Section 4.9

Reversing the Inequality Symbol

This section explains one of the most important and most frequently forgotten rules in inequality algebra: when you multiply or divide both sides of an inequality by a negative number, you must reverse the direction of the inequality symbol.

Why does this rule exist? Consider the true statement 3 < 7. Multiply both sides by −1 and ask: is −3 < −7? No. On the number line, −3 lies to the right of −7, meaning −3 > −7. Multiplication by −1 reversed the ordering. To preserve truth, the symbol must flip: 3 < 7 implies −3 > −7. This is not a special case — it is true for all real numbers. The rule applies only to multiplication and division by negatives. Adding or subtracting any number, positive or negative, never reverses the symbol.

The Reversal Rule

When you multiply or divide both sides of an inequality by a negative number, reverse the inequality symbol.

Adding or subtracting — even a negative — does not trigger a reversal.

Example 4.9.1
Demonstrating the reversal with numbers
  • StartTrue inequality: 2 < 5
  • MultiplyMultiply both sides by −3: left gives −6, right gives −15.
  • CheckIs −6 < −15? No. −6 is to the right of −15 on the number line, so −6 > −15. The reversed symbol is correct: −6 > −15. ✓
  • LessonForgetting the reversal gives the false statement −6 < −15, which is clearly wrong.
Conclusion: 2 < 5 implies −6 > −15 after multiplying by −3.
Example 4.9.2
Solving an inequality requiring a reversal
  • ProblemSolve and graph: −5x + 3 < 18
  • ThinkSubtract 3 first. Then divide by −5. Dividing by negative → reverse the symbol.
  • Solve −5x + 3 < 18 −5x + 3 − 3 < 18 − 3 −5x < 15 −5x / −5 > 15 / −5 ← symbol reverses x > −3
  • CheckTest x = 0: −5(0) + 3 = 3 < 18. True. ✓   Test x = −4: −5(−4) + 3 = 23 < 18. False. ✓
Answer: x > −3 — Open circle at −3, arrow pointing right.
Critical Student Warning

Forgetting to reverse the inequality symbol when dividing by a negative is the single most common error in this chapter. After every solution involving negative division, test a value from both sides of your boundary. The test value will immediately reveal whether the direction is correct. Never skip this check.

Example 4.9.3
Catching a missing reversal with a test value
  • ProblemA student solved −2x ≥ 8 and got x ≥ −4. Is the student correct?
  • ThinkTest x = 0, which is in the student's claimed solution set since 0 ≥ −4.
  • Check −2(0) = 0 ≥ 8? False. Zero should not be a solution, but the student's answer includes it. The answer is wrong.
  • Correct −2x ≥ 8 −2x / −2 ≤ 8 / −2 ← symbol reverses x ≤ −4 Check: x = −6: −2(−6) = 12 ≥ 8. True. ✓
Answer: x ≤ −4. The student forgot to reverse the symbol.
Practice 4.5 – 4.9Solving Inequalities

Solve each inequality. Write your answer and note whether the graph uses an open or closed circle.

  1. x − 6 > 4
  2. 3 + x ≤ 10
  3. 5x ≥ 35
  4. x/4 < −2
  5. −2x > 10
  6. −x/3 ≤ 4
  7. 4x − 3 < 9
  8. 2x + 7 ≥ 3
  9. −5x + 2 ≤ 17
  10. 3(x + 1) > 12
  11. 2(x − 4) + 6 ≤ 10
  12. 6x − 2 > 3x + 7
  13. 5x + 4 ≤ 2x − 8
  14. −4(x − 1) ≥ 12
  15. 7 − 3x < 4x + 7
Sections 4.10 – 4.12
Compound Inequalities and Applications
Compound Inequalities · "And" and "Or" · Inequality Word Problems
Sections 4.10 – 4.11

Compound Inequalities: "And" and "Or"

A compound inequality combines two separate inequalities. There are two types, and they produce fundamentally different solution sets.

An "and" compound inequality requires the variable to satisfy both conditions simultaneously. The solution is the intersection of the two sets — the values that work for both. "And" inequalities are often written as a three-part expression: a < x < b, which means x > a and x < b at the same time. The solution is a bounded interval between two values.

An "or" compound inequality requires the variable to satisfy at least one of the two conditions. The solution is the union of the two sets — the values that work for at least one. "Or" solutions typically produce two separate rays extending in opposite directions, leaving a gap in the middle.

Example 4.11.1
Solving a three-part "and" inequality
  • ProblemSolve and graph: −2 ≤ 3x − 1 < 8
  • ThinkApply the same operations to all three parts simultaneously. First add 1, then divide by 3.
  • Solve −2 ≤ 3x − 1 < 8 −2 + 1 ≤ 3x − 1 + 1 < 8 + 1 −1 ≤ 3x < 9 −1/3 ≤ x < 3
  • CheckTest x = 0: −2 ≤ −1 < 8. True. ✓   Test x = −1: −2 ≤ −4? No. False. ✓   Test x = 3: 3(3) − 1 = 8 < 8? No. False. ✓ (boundary excluded)
Answer: −1/3 ≤ x < 3 — Closed circle at −1/3, open circle at 3, bold segment between them.
-1 0 1 2 3 -⅓
Example 4.11.2
Solving an "or" compound inequality
  • ProblemSolve and graph: 2x + 1 > 7 or x − 3 < −5
  • ThinkSolve each inequality separately. Then combine the solution sets. Any value satisfying either inequality is in the solution.
  • Solve Inequality 1: 2x + 1 > 7 2x > 6 x > 3 Inequality 2: x − 3 < −5 x < −2 Combined: x < −2 OR x > 3
  • CheckTest x = 5: 2(5) + 1 = 11 > 7. True. ✓   Test x = −4: −4 − 3 = −7 < −5. True. ✓   Test x = 0: neither condition met. False. ✓
Answer: x < −2 or x > 3 — Two open circles with arrows pointing away from each other; gap between −2 and 3.
-2 -1 0 1 2 3
Key Difference — "And" vs. "Or"

An "and" solution is a bounded segment — a finite interval. An "or" solution is two rays extending away from each other with a gap in the middle. Recognizing the shape helps you verify your answer.

Section 4.12

Inequality Word Problems

The Read → Represent → Solve → Check → Explain method works for inequalities just as it did for equations. The key additional step is identifying whether the situation calls for a strict inequality (< or >) or an inclusive one (≤ or ≥), and whether the answer should be a continuous range or a whole-number answer.

Example 4.12.1
Earning goal problem
  • ReadMarcus earns $12.75 per hour. He wants to earn at least $350 this week. Find the minimum hours.
  • RepresentLet h = number of hours worked.   12.75h ≥ 350
  • Solve 12.75h ≥ 350 h ≥ 350 / 12.75 h ≥ 27.45...
  • Check12.75 × 27.5 = 350.625 ≥ 350. True. ✓
  • ExplainMarcus needs to work at least approximately 27.45 hours. If hours must be whole numbers, he needs to work at least 28 hours.
Answer: h ≥ 27.45 (at least 28 whole hours).
Example 4.12.2
Maximum cost problem
  • ReadAlicia has $60 total. Her backpack cost $22. Each composition notebook costs $2.75. How many notebooks can she buy?
  • RepresentLet n = number of notebooks.   22 + 2.75n ≤ 60
  • Solve 22 + 2.75n ≤ 60 2.75n ≤ 38 n ≤ 38 / 2.75 n ≤ 13.81...
  • Check22 + 2.75(13) = 57.75 ≤ 60. True. ✓   22 + 2.75(14) = 60.50 ≤ 60. False. ✓
  • ExplainSince notebooks come in whole units, Alicia can buy at most 13 notebooks.
Answer: n ≤ 13 notebooks.
Example 4.12.3
Temperature range problem
  • ReadA biology experiment requires lab temperature between 18°C and 26°C, inclusive. Write and interpret the inequality.
  • RepresentLet T = lab temperature in degrees Celsius.   18 ≤ T ≤ 26
  • CheckTest T = 22: 18 ≤ 22 ≤ 26. True. ✓   Test T = 10: 18 ≤ 10? No. False. ✓
  • ExplainAny temperature from 18°C to 26°C, including both endpoints, satisfies the requirement. The solution set is a closed interval.
Answer: 18 ≤ T ≤ 26
Practice 4.10 – 4.12Compound Inequalities and Word Problems
  1. Solve and graph: −3 < 2x + 1 < 9
  2. Solve and graph: −1 ≤ 4x − 3 ≤ 13
  3. Solve and graph: 3x + 2 > 8 or 3x + 2 < −4
  4. Solve and graph: x + 5 < 3 or 2x − 1 > 11
  5. A telephone plan charges $25/month plus $0.08 per text. A customer's bill must stay under $45. What is the maximum number of texts?
  6. A student scored 71, 84, 78, and 66 on four tests. What score must she earn on the fifth test to have an average of at least 75?
  7. A pilot must maintain altitude between 1,000 and 10,000 feet. Write the compound inequality for allowable altitude a.
  8. Two friends are saving money. Maya has $150 and saves $20 per week. Jordan has $90 and saves $30 per week. After how many weeks will Jordan have at least as much as Maya?
Sections 4.13 – 4.14
Common Mistakes and the Road Ahead
Seven Recurring Errors · Bridge to Graphing, Systems, and Functions
Section 4.13

Common Inequality Mistakes

Mistake 1 — Forgetting to reverse the inequality symbol
−4x > 20 → x > −5

Dividing by −4 requires a symbol reversal. Correct: x < −5. After every negative division, test a value from both sides of the boundary. If the original is false for a value in your solution set, the symbol direction is wrong.

Mistake 2 — Reversing the symbol when adding or subtracting a negative
x − (−3) > 5 → symbol flipped unnecessarily

Adding or subtracting any number, positive or negative, never reverses the symbol. Only multiplication and division by a negative trigger the reversal.

Mistake 3 — Using the wrong circle type on a number line
x ≤ 4 graphed with an open circle at 4

Ask yourself: "Is the boundary value a solution?" For ≤ or ≥, the boundary is included — use a closed circle. For < or >, the boundary is excluded — use an open circle.

Mistake 4 — Shading the middle for an "or" inequality
For x < −2 or x > 3, the segment from −2 to 3 is shaded

The "or" solution has two rays pointing away from each other. The middle region is not in the solution set. Shading the middle would describe an "and" solution, not an "or."

Mistake 5 — Writing a compound inequality backwards
5 ≥ x ≥ −2 (confusing to read)

Write compound inequalities left to right in number-line order: −2 ≤ x ≤ 5. The smaller boundary on the left, the larger on the right.

Mistake 6 — Confusing "at least" and "at most"
"At least 5" written as x ≤ 5

"At least 5" means 5 or more: x ≥ 5. "At most 5" means 5 or fewer: x ≤ 5. If you reverse these in a word problem, your inequality describes the wrong situation entirely.

Mistake 7 — Checking only one test value
Testing only one value inside the solution region

For full confidence, test one value inside your solution region and one value outside. Both tests together confirm the correct boundary and direction.

Section 4.14

Preparing for Linear Graphs and Regions

Chapter 4 closes the study of inequalities in one variable. The ideas you have developed here appear throughout the rest of this course and continue into Algebra II.

In Chapter 6, you will return to the coordinate plane and deepen your understanding of slope and graphing. The number line work here is a one-dimensional preview of the two-dimensional reasoning you will develop there.

In Chapter 7, you will study linear equations and their graphs. The moment you ask which side of a line satisfies an inequality condition, you are thinking in two-variable inequality terms.

In Chapter 8, you will extend Chapter 4's work into linear inequalities in two variables and systems of linear inequalities. You will shade regions of the plane, find their intersections, and use those regions to model real constraints. Every skill practiced here supports that work directly.

In Algebra II, inequalities return in the context of polynomial functions, rational functions, and absolute value. The sign-analysis and test-interval reasoning you practiced here becomes the foundation for those topics.

The transition from "find the answer" to "find the range of answers" is one of the defining moves of algebraic reasoning. You have now made that transition.

MGU Library Connection
  • Inequality Graphing Reference — Appendix F of this textbook
  • Algebra Symbol Reference — Appendix D of this textbook
  • Chapter 8 of this text: Systems of Linear Inequalities and Feasible Regions
  • Algebra II, Chapter 1: Domain restrictions using inequality notation
  • MGU Dictionary entries: inequality, solution set, compound inequality, intersection, union
Chapter 4 — Back Matter

Chapter Summary

Key Terms Review

inequality
A mathematical statement comparing two quantities using <, >, ≤, or ≥. May have infinitely many solutions.
inequality symbol
One of the four comparison symbols: less than, greater than, less than or equal to, greater than or equal to.
solution
Any value of the variable that makes an inequality a true statement when substituted.
solution set
The set of all values satisfying an inequality. For linear inequalities in one variable, usually a ray or bounded interval.
open circle
Symbol on a number line indicating the endpoint is NOT included. Used with < or >.
closed circle
Symbol on a number line indicating the endpoint IS included. Used with ≤ or ≥.
reversal rule
When both sides of an inequality are multiplied or divided by a negative number, the inequality symbol must be flipped.
compound inequality
Two inequalities combined with "and" or "or." Produces an intersection or a union as the solution set.
intersection
The set of values satisfying both conditions in an "and" compound inequality. Usually a bounded segment.
union
The set of values satisfying at least one condition in an "or" compound inequality. Usually two rays.

Concept Review Questions

  1. Explain the difference between an equation and an inequality. Give an example of each.
  2. What does the solution set of an inequality mean? How is it different from the solution of an equation?
  3. Explain when you use an open circle versus a closed circle on a number line graph.
  4. State the reversal rule in your own words. Why is it necessary? Use a numerical example to illustrate.
  5. What is the difference between an "and" compound inequality and an "or" compound inequality? Describe what each solution set looks like on a number line.
  6. Explain the difference between "at most 10" and "at least 10." Write each as an inequality using the variable x.
  7. When solving an inequality, you test a value in the solution region and the result is false. What does this tell you?
  8. When solving an inequality with variables on both sides, a student ends with 5 ≤ x. Rewrite this more naturally. Are the two forms equivalent?

Skill Practice

Part AWriting and Reading Inequalities
  1. n must be less than 15
  2. s cannot exceed 70 mph
  3. Needs at least 80 points
  4. Height more than 20 feet
  5. Payment at most $250
  6. Between 5 and 12, inclusive
Part BOne- and Two-Step Inequalities

Solve and graph each. Note open or closed circle.

  1. x + 9 > 4
  2. x − 3 ≤ 7
  3. 6x < −18
  4. −x/2 ≥ 5
  5. 3x + 5 < 17
  6. −4x − 1 ≥ 11
  7. x/3 + 2 > 4
  8. 7 − 2x ≤ 1
Part CMulti-Step and Variables on Both Sides
  1. 2(x + 4) − 3 > 9
  2. 4(x − 1) + 2 ≤ 14
  3. −3(x + 2) ≥ −9
  4. 5x + 3 > 2x + 12
  5. 7x − 4 ≤ 3x + 8
  6. 4 − x > 2x − 5
  7. 6x − 1 ≥ 2(3x − 1)
  8. 3(2x + 1) < 4x + 11
Part DCompound Inequalities
  1. −4 < 3x + 2 ≤ 11
  2. −10 < −2x + 4 ≤ 10
  3. −6 ≤ −2x + 4 ≤ 8
  4. 2x − 1 > 5 or x + 4 < 1
  5. x + 3 < 1 or 2x − 1 > 9
  6. −1 < (x+1)/2 ≤ 3

Word Problems

Define the variable, write the inequality, solve, and explain the answer in context.

  1. A delivery truck can carry at most 2,000 pounds. It holds 1,340 pounds. How much additional weight can be loaded?
  2. The perimeter of a rectangle must be no more than 60 cm. The length is 18 cm. What values are possible for the width?
  3. A car travels at 55 mph. Write an expression for distance in t hours. How long must the trip be to cover at least 200 miles?
  4. Maya has $150 and saves $20 per week. Jordan has $90 and saves $30 per week. After how many weeks will Jordan have at least as much as Maya?
  5. A nutritionist recommends between 1,500 and 2,200 calories per day. A person has eaten 840 calories so far. Write a compound inequality for the remaining calories.
  6. A part-time employee can work between 15 and 32 hours per week. Write the compound inequality. Find the possible weekly earnings at $16.40 per hour.
  7. A student scored 74, 88, 81, and 77 on four quizzes. What score on the fifth quiz gives an average of at least 80?
  8. A fish must be between 12 and 18 inches to be legally kept. Write a compound inequality and describe the legal catch range.

Real-World Applications

  1. Finance. A student has a balance of $215. She writes a check for $185. Write and solve an inequality describing how much additional money she can spend and keep a positive balance.
  2. Science. Safe drinking water must contain fewer than 0.010 mg/L of lead. A sample has 0.004 mg/L naturally. How much additional contamination c can be present before the water becomes unsafe?
  3. Construction. A beam supports at most 8,500 pounds. Three identical loads are placed on it. Write and solve an inequality for the maximum weight w of each load.
  4. Athletics. A sprinter needs to run 400 m in less than 55 seconds to qualify. She ran the first 200 m in 26.4 seconds. Write and solve an inequality for the maximum time t for the second 200 m.

Error Analysis

Each problem shows a student's work that contains at least one error. Identify the error, explain what went wrong, and show the correct solution.

Problem 1
−6x ≤ 24 → x ≤ −4

What is wrong? What is the correct answer?

Problem 2
3x + 7 < 10 → 3x < 17 → x < 17/3

What is wrong? What is the correct answer?

Problem 3
2(x + 3) > 10 → 2x + 3 > 10 → 2x > 7 → x > 3.5

What is wrong? What is the correct answer?

Problem 4
x > 4 or x < −1: student graphs a segment from −1 to 4 with open circles

What is wrong? What should the correct graph look like?

Problem 5
"The package weighs at most 30 pounds" → w > 30

What is wrong? What is the correct inequality?

Problem 6
−1 < 2x + 3 < 9 → −1 − 3 < 2x < 9 → −4 < 2x < 9 → −2 < x < 4.5

Identify the error in the middle step. What is the correct solution?

Chapter 4 Checkpoint

Chapter 4 Checkpoint
Complete without assistance — covers all key skills and concepts from Chapter 4
Part I — Foundations
  1. Write an inequality: "A tank must hold more than 40 gallons but no more than 120 gallons."
  2. Is x = −3 a solution of −2x + 5 ≥ 10? Show your check.
  3. Graph x < 2 on a number line. State whether 2 is in the solution set.
  4. Write the inequality shown by: closed circle at 1, open circle at 7, bold segment between them.
Part II — Solving Inequalities
  1. Solve: x − 9 ≥ −4
  2. Solve: 4x < −20
  3. Solve: −x/5 > 3
  4. Solve: 3x + 8 ≤ 20
  5. Solve: −2x − 7 > 3
  6. Solve: 5(x − 2) − 3 ≥ 12
  7. Solve: 4x + 9 < 7x − 3
  8. Solve: 6 − 2(x + 4) ≤ 0
Part III — Compound Inequalities
  1. Solve and graph: −4 < 3x + 2 ≤ 11
  2. Solve and graph: 2x − 1 > 5 or x + 4 < 1
  3. Solve: −6 ≤ −2x + 4 ≤ 8
Part IV — Word Problems
  1. A gym membership costs $35 per month. A new member budgets at most $500 for the first year. Write and solve an inequality for the maximum months the membership can run.
  2. Maria's quiz scores are 74, 88, 81, and 77. What score on the fifth quiz gives an average of at least 80?
  3. A container holds between 3 and 8 liters, inclusive. Water is added at 0.5 liters per minute. Write and solve a compound inequality for the number of minutes m of filling within the acceptable range. Assume the container starts empty.
Part V — Reasoning
  1. Explain in your own words why dividing both sides of an inequality by a negative number requires reversing the symbol. Use a specific numerical example to support your explanation.
  2. A student claims that the solution to |x| < 5 is the same as saying −5 < x < 5. Test three values and determine whether you agree. (Note: Absolute value inequalities are developed fully in Algebra II. This problem previews that connection.)
Looking Ahead
Chapter 5: Introduction to Functions

Chapter 4 closes Part II of this text. You can now write, solve, graph, and interpret linear inequalities in one variable — including compound forms — and apply them to real-world constraints. Chapter 5 opens Part III by introducing one of the most important ideas in all of mathematics: the function. A function is a rule that assigns exactly one output to every input. The expressions and equations you practiced in Chapters 1 through 4 become function rules. The domain of a function is often described using inequalities. Every skill from this chapter carries forward, now applied to inputs, outputs, and graphs of functions across all major function families.