Learn on PengiOpenstax Intermediate Algebra 2EChapter 2: Solving Linear Equations

Lesson 2.5: Solve Linear Inequalities

In this lesson from OpenStax Intermediate Algebra 2E, students learn to graph linear inequalities on the number line and express solution sets using interval notation, including open and closed endpoints represented by parentheses and brackets. The lesson covers solving linear inequalities, translating word problems into inequality statements, and applying these skills to real-world scenarios. This intermediate algebra content builds foundational skills for working with solution sets, infinity symbols, and compound inequalities.

Section 1

πŸ“˜ Solve Linear Inequalities

New Concept

A linear inequality is a statement comparing linear expressions, like ax+b≀cax + b \leq c. In this lesson, we'll master solving these inequalities, graphing their solutions on a number line, translating words into inequalities, and solving real-world applications.

What’s next

First, you'll practice graphing inequalities on the number line. Then, you'll solve them step-by-step through a series of interactive examples and challenge problems.

Section 2

Graphing Inequalities on the Number Line

Property

We show all the solutions to an inequality on the number line by shading. An open parenthesis shows an endpoint is not a solution; a bracket shows an endpoint is a solution. In interval notation, we use parentheses or brackets to show the interval of solutions. The symbol ∞\infty is read as β€œinfinity,” and βˆ’βˆž-\infty is read as β€œnegative infinity.”

Examples

  • The inequality x>5x > 5 is written in interval notation as (5,∞)(5, \infty). On a number line, an open parenthesis is placed at 5, and the line is shaded to the right.
  • The inequality yβ‰€βˆ’1y \leq -1 is written in interval notation as (βˆ’βˆž,βˆ’1](-\infty, -1]. On a number line, a bracket is placed at -1, and the line is shaded to the left.
  • The inequality βˆ’4≀z<2-4 \leq z < 2 is written as [βˆ’4,2)[-4, 2). On a number line, there is a bracket at -4, a parenthesis at 2, and the line between them is shaded.

Explanation

Think of graphing an inequality as drawing a map of all its solutions. A parenthesis '(' or ')' means the endpoint is not included, while a bracket '[' or ']' means it is. Interval notation is just a shorthand for this map.

Section 3

Addition and Subtraction Property of Inequality

Property

For any numbers aa, bb, and cc, if a<ba < b, then

a+c<b+caβˆ’c<bβˆ’ca + c < b + c \qquad a - c < b - c

For any numbers aa, bb, and cc, if a>ba > b, then

a+c>b+caβˆ’c>bβˆ’ca + c > b + c \qquad a - c > b - c

We can add or subtract the same quantity from both sides of an inequality and still keep the inequality.

Examples

  • To solve x+7<15x + 7 < 15, subtract 7 from both sides: x+7βˆ’7<15βˆ’7x + 7 - 7 < 15 - 7, which simplifies to x<8x < 8.
  • To solve yβˆ’4β‰₯βˆ’2y - 4 \geq -2, add 4 to both sides: yβˆ’4+4β‰₯βˆ’2+4y - 4 + 4 \geq -2 + 4, which simplifies to yβ‰₯2y \geq 2.
  • Given 12>z+512 > z + 5, subtract 5 from both sides: 12βˆ’5>z+5βˆ’512 - 5 > z + 5 - 5, so 7>z7 > z, which means z<7z < 7.

Explanation

This property is just like it is for equations. You can add or subtract the same number on both sides of an inequality, and the relationship between the two sides stays the same. The inequality sign does not change.

Section 4

Multiplication and Division Property of Inequality

Property

For any numbers aa, bb, and cc,
multiply or divide by a positive:
if a<ba < b and c>0c > 0, then ac<bcac < bc and ac<bc\frac{a}{c} < \frac{b}{c}.
if a>ba > b and c>0c > 0, then ac>bcac > bc and ac>bc\frac{a}{c} > \frac{b}{c}.
multiply or divide by a negative:
if a<ba < b and c<0c < 0, then ac>bcac > bc and ac>bc\frac{a}{c} > \frac{b}{c}.
if a>ba > b and c<0c < 0, then ac<bcac < bc and ac<bc\frac{a}{c} < \frac{b}{c}.
When we divide or multiply an inequality by a negative number, the inequality sign reverses.

Examples

  • To solve 3x>213x > 21, divide by 3 (a positive number): 3x3>213\frac{3x}{3} > \frac{21}{3}, so x>7x > 7. The sign stays the same.
  • To solve βˆ’4yβ‰₯20-4y \geq 20, divide by -4 (a negative number): βˆ’4yβˆ’4≀20βˆ’4\frac{-4y}{-4} \leq \frac{20}{-4}, so yβ‰€βˆ’5y \leq -5. The sign reverses.
  • To solve zβˆ’2<5\frac{z}{-2} < 5, multiply by -2 (a negative number): βˆ’2(zβˆ’2)>βˆ’2(5)-2(\frac{z}{-2}) > -2(5), so z>βˆ’10z > -10. The sign reverses.

Explanation

The game-changing rule! When you multiply or divide both sides by a positive number, nothing changes. But if you use a negative number, you MUST flip the inequality sign. For example, > becomes <.

Section 5

Solving Multi-Step Linear Inequalities

Property

A linear inequality is an inequality in one variable that can be written in one of the following forms where aa, bb, and cc are real numbers and a≠0a \neq 0:

ax+b<c,ax+b≀c,ax+b>c,ax+bβ‰₯c.ax + b < c, \quad ax + b \leq c, \quad ax + b > c, \quad ax + b \geq c.

To solve, follow the same steps as for linear equations, but reverse the inequality sign when multiplying or dividing by a negative number.

Examples

  • To solve 7xβˆ’5>3x+117x - 5 > 3x + 11, subtract 3x3x from both sides to get 4xβˆ’5>114x - 5 > 11. Then add 5 to get 4x>164x > 16. Finally, divide by 4 to get x>4x > 4.
  • To solve 3(kβˆ’2)≀123(k - 2) \leq 12, distribute to get 3kβˆ’6≀123k - 6 \leq 12. Add 6 to both sides to get 3k≀183k \leq 18. Divide by 3 to get k≀6k \leq 6.
  • To solve 2yβ‰₯9y+352y \geq 9y + 35, subtract 9y9y from both sides to get βˆ’7yβ‰₯35-7y \geq 35. Divide by -7 and reverse the sign to get yβ‰€βˆ’5y \leq -5.

Explanation

Combine everything you know! Simplify each side, collect variables on one side, and then isolate the variable. Just keep an eye out for that one tricky step: multiplying or dividing by a negative number flips the inequality sign.

Section 6

Identities and Contradictions in Inequalities

Property

When solving an inequality, if the variable is eliminated and the result is a true statement (like βˆ’10<36-10 < 36), the inequality is an identity and the solution is all real numbers, (βˆ’βˆž,∞)(-\infty, \infty). If the result is a false statement (like 0>180 > 18), the inequality is a contradiction and there is no solution.

Examples

  • The inequality 5(x+2)>5x+35(x+2) > 5x+3 simplifies to 5x+10>5x+35x+10 > 5x+3, then 10>310 > 3. This is true, so the solution is all real numbers, (βˆ’βˆž,∞)(-\infty, \infty).
  • The inequality 2(yβˆ’3)≀2yβˆ’102(y-3) \leq 2y-10 simplifies to 2yβˆ’6≀2yβˆ’102y-6 \leq 2y-10, then βˆ’6β‰€βˆ’10-6 \leq -10. This is false, so there is no solution.
  • The inequality 4zβˆ’1βˆ’4z<βˆ’14z - 1 - 4z < -1 simplifies to βˆ’1<βˆ’1-1 < -1. This is false because -1 is not less than itself, so there is no solution.

Explanation

Sometimes the variables completely cancel out! If you are left with a true statement like 5<105 < 10, the answer is all real numbers. If you are left with a false statement like 5>105 > 10, there is no solution.

Section 7

Translate Words to an Inequality

Property

To translate English sentences into inequalities, we need to recognize the phrases that indicate the inequality. Some common phrases are 'is greater than' (>>), 'is at least' (β‰₯\geq), 'is less than' (<<), and 'is at most' (≀\leq).

Examples

  • 'Thirty less than a number nn is at least 50' translates to nβˆ’30β‰₯50n - 30 \geq 50.
  • 'Four times a number yy is no more than 24' translates to 4y≀244y \leq 24.
  • 'A number pp increased by 10 exceeds 25' translates to p+10>25p + 10 > 25.

Explanation

Math has its own language. Learning keywords helps you translate from English to an inequality. 'Exceeds' means 'greater than,' while 'at most' means 'less than or equal to.' Pay close attention to these phrases.

Section 8

Applications with Linear Inequalities

Property

To solve applications with linear inequalities, first identify what you are looking for and assign a variable. Translate the problem into an inequality by representing the situation. Solve the inequality, and ensure the answer makes sense in the context of the problem, rounding if necessary.

Examples

  • A grant is 3,000 dollars for science kits. Each kit costs 75 dollars. What is the maximum number of kits, kk, you can buy? The inequality is 75k≀300075k \leq 3000. Solving gives k≀40k \leq 40. You can buy a maximum of 40 kits.
  • Your phone bill is 25 dollars plus 0.15 dollars per text, tt. To keep your bill no more than 40 dollars, the inequality is 25+0.15t≀4025 + 0.15t \leq 40. Solving gives t≀100t \leq 100. You can send at most 100 texts.
  • To earn a profit of at least 500 dollars, revenue must exceed costs by that amount. If you sell items for 10 dollars each and expenses are 400 dollars, how many items, bb, must you sell? The inequality is 10bβˆ’400β‰₯50010b - 400 \geq 500. Solving gives bβ‰₯90b \geq 90 items.

Explanation

Use inequalities to solve real-world problems about budgets, profits, or limits. Set up an expression for the total cost or amount and compare it to the limit using an inequality sign. Then, solve for your variable.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 2: Solving Linear Equations

  1. Lesson 1

    Lesson 2.1: Use a General Strategy to Solve Linear Equations

  2. Lesson 2

    Lesson 2.2: Use a Problem Solving Strategy

  3. Lesson 3

    Lesson 3: 2.3 Solve a Formula for a Specific Variable

  4. Lesson 4

    Lesson 2.4: Solve Mixture and Uniform Motion Applications

  5. Lesson 5Current

    Lesson 2.5: Solve Linear Inequalities

  6. Lesson 6

    Lesson 2.6: Solve Compound Inequalities

  7. Lesson 7

    Lesson 2.7: Solve Absolute Value Inequalities

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

πŸ“˜ Solve Linear Inequalities

New Concept

A linear inequality is a statement comparing linear expressions, like ax+b≀cax + b \leq c. In this lesson, we'll master solving these inequalities, graphing their solutions on a number line, translating words into inequalities, and solving real-world applications.

What’s next

First, you'll practice graphing inequalities on the number line. Then, you'll solve them step-by-step through a series of interactive examples and challenge problems.

Section 2

Graphing Inequalities on the Number Line

Property

We show all the solutions to an inequality on the number line by shading. An open parenthesis shows an endpoint is not a solution; a bracket shows an endpoint is a solution. In interval notation, we use parentheses or brackets to show the interval of solutions. The symbol ∞\infty is read as β€œinfinity,” and βˆ’βˆž-\infty is read as β€œnegative infinity.”

Examples

  • The inequality x>5x > 5 is written in interval notation as (5,∞)(5, \infty). On a number line, an open parenthesis is placed at 5, and the line is shaded to the right.
  • The inequality yβ‰€βˆ’1y \leq -1 is written in interval notation as (βˆ’βˆž,βˆ’1](-\infty, -1]. On a number line, a bracket is placed at -1, and the line is shaded to the left.
  • The inequality βˆ’4≀z<2-4 \leq z < 2 is written as [βˆ’4,2)[-4, 2). On a number line, there is a bracket at -4, a parenthesis at 2, and the line between them is shaded.

Explanation

Think of graphing an inequality as drawing a map of all its solutions. A parenthesis '(' or ')' means the endpoint is not included, while a bracket '[' or ']' means it is. Interval notation is just a shorthand for this map.

Section 3

Addition and Subtraction Property of Inequality

Property

For any numbers aa, bb, and cc, if a<ba < b, then

a+c<b+caβˆ’c<bβˆ’ca + c < b + c \qquad a - c < b - c

For any numbers aa, bb, and cc, if a>ba > b, then

a+c>b+caβˆ’c>bβˆ’ca + c > b + c \qquad a - c > b - c

We can add or subtract the same quantity from both sides of an inequality and still keep the inequality.

Examples

  • To solve x+7<15x + 7 < 15, subtract 7 from both sides: x+7βˆ’7<15βˆ’7x + 7 - 7 < 15 - 7, which simplifies to x<8x < 8.
  • To solve yβˆ’4β‰₯βˆ’2y - 4 \geq -2, add 4 to both sides: yβˆ’4+4β‰₯βˆ’2+4y - 4 + 4 \geq -2 + 4, which simplifies to yβ‰₯2y \geq 2.
  • Given 12>z+512 > z + 5, subtract 5 from both sides: 12βˆ’5>z+5βˆ’512 - 5 > z + 5 - 5, so 7>z7 > z, which means z<7z < 7.

Explanation

This property is just like it is for equations. You can add or subtract the same number on both sides of an inequality, and the relationship between the two sides stays the same. The inequality sign does not change.

Section 4

Multiplication and Division Property of Inequality

Property

For any numbers aa, bb, and cc,
multiply or divide by a positive:
if a<ba < b and c>0c > 0, then ac<bcac < bc and ac<bc\frac{a}{c} < \frac{b}{c}.
if a>ba > b and c>0c > 0, then ac>bcac > bc and ac>bc\frac{a}{c} > \frac{b}{c}.
multiply or divide by a negative:
if a<ba < b and c<0c < 0, then ac>bcac > bc and ac>bc\frac{a}{c} > \frac{b}{c}.
if a>ba > b and c<0c < 0, then ac<bcac < bc and ac<bc\frac{a}{c} < \frac{b}{c}.
When we divide or multiply an inequality by a negative number, the inequality sign reverses.

Examples

  • To solve 3x>213x > 21, divide by 3 (a positive number): 3x3>213\frac{3x}{3} > \frac{21}{3}, so x>7x > 7. The sign stays the same.
  • To solve βˆ’4yβ‰₯20-4y \geq 20, divide by -4 (a negative number): βˆ’4yβˆ’4≀20βˆ’4\frac{-4y}{-4} \leq \frac{20}{-4}, so yβ‰€βˆ’5y \leq -5. The sign reverses.
  • To solve zβˆ’2<5\frac{z}{-2} < 5, multiply by -2 (a negative number): βˆ’2(zβˆ’2)>βˆ’2(5)-2(\frac{z}{-2}) > -2(5), so z>βˆ’10z > -10. The sign reverses.

Explanation

The game-changing rule! When you multiply or divide both sides by a positive number, nothing changes. But if you use a negative number, you MUST flip the inequality sign. For example, > becomes <.

Section 5

Solving Multi-Step Linear Inequalities

Property

A linear inequality is an inequality in one variable that can be written in one of the following forms where aa, bb, and cc are real numbers and a≠0a \neq 0:

ax+b<c,ax+b≀c,ax+b>c,ax+bβ‰₯c.ax + b < c, \quad ax + b \leq c, \quad ax + b > c, \quad ax + b \geq c.

To solve, follow the same steps as for linear equations, but reverse the inequality sign when multiplying or dividing by a negative number.

Examples

  • To solve 7xβˆ’5>3x+117x - 5 > 3x + 11, subtract 3x3x from both sides to get 4xβˆ’5>114x - 5 > 11. Then add 5 to get 4x>164x > 16. Finally, divide by 4 to get x>4x > 4.
  • To solve 3(kβˆ’2)≀123(k - 2) \leq 12, distribute to get 3kβˆ’6≀123k - 6 \leq 12. Add 6 to both sides to get 3k≀183k \leq 18. Divide by 3 to get k≀6k \leq 6.
  • To solve 2yβ‰₯9y+352y \geq 9y + 35, subtract 9y9y from both sides to get βˆ’7yβ‰₯35-7y \geq 35. Divide by -7 and reverse the sign to get yβ‰€βˆ’5y \leq -5.

Explanation

Combine everything you know! Simplify each side, collect variables on one side, and then isolate the variable. Just keep an eye out for that one tricky step: multiplying or dividing by a negative number flips the inequality sign.

Section 6

Identities and Contradictions in Inequalities

Property

When solving an inequality, if the variable is eliminated and the result is a true statement (like βˆ’10<36-10 < 36), the inequality is an identity and the solution is all real numbers, (βˆ’βˆž,∞)(-\infty, \infty). If the result is a false statement (like 0>180 > 18), the inequality is a contradiction and there is no solution.

Examples

  • The inequality 5(x+2)>5x+35(x+2) > 5x+3 simplifies to 5x+10>5x+35x+10 > 5x+3, then 10>310 > 3. This is true, so the solution is all real numbers, (βˆ’βˆž,∞)(-\infty, \infty).
  • The inequality 2(yβˆ’3)≀2yβˆ’102(y-3) \leq 2y-10 simplifies to 2yβˆ’6≀2yβˆ’102y-6 \leq 2y-10, then βˆ’6β‰€βˆ’10-6 \leq -10. This is false, so there is no solution.
  • The inequality 4zβˆ’1βˆ’4z<βˆ’14z - 1 - 4z < -1 simplifies to βˆ’1<βˆ’1-1 < -1. This is false because -1 is not less than itself, so there is no solution.

Explanation

Sometimes the variables completely cancel out! If you are left with a true statement like 5<105 < 10, the answer is all real numbers. If you are left with a false statement like 5>105 > 10, there is no solution.

Section 7

Translate Words to an Inequality

Property

To translate English sentences into inequalities, we need to recognize the phrases that indicate the inequality. Some common phrases are 'is greater than' (>>), 'is at least' (β‰₯\geq), 'is less than' (<<), and 'is at most' (≀\leq).

Examples

  • 'Thirty less than a number nn is at least 50' translates to nβˆ’30β‰₯50n - 30 \geq 50.
  • 'Four times a number yy is no more than 24' translates to 4y≀244y \leq 24.
  • 'A number pp increased by 10 exceeds 25' translates to p+10>25p + 10 > 25.

Explanation

Math has its own language. Learning keywords helps you translate from English to an inequality. 'Exceeds' means 'greater than,' while 'at most' means 'less than or equal to.' Pay close attention to these phrases.

Section 8

Applications with Linear Inequalities

Property

To solve applications with linear inequalities, first identify what you are looking for and assign a variable. Translate the problem into an inequality by representing the situation. Solve the inequality, and ensure the answer makes sense in the context of the problem, rounding if necessary.

Examples

  • A grant is 3,000 dollars for science kits. Each kit costs 75 dollars. What is the maximum number of kits, kk, you can buy? The inequality is 75k≀300075k \leq 3000. Solving gives k≀40k \leq 40. You can buy a maximum of 40 kits.
  • Your phone bill is 25 dollars plus 0.15 dollars per text, tt. To keep your bill no more than 40 dollars, the inequality is 25+0.15t≀4025 + 0.15t \leq 40. Solving gives t≀100t \leq 100. You can send at most 100 texts.
  • To earn a profit of at least 500 dollars, revenue must exceed costs by that amount. If you sell items for 10 dollars each and expenses are 400 dollars, how many items, bb, must you sell? The inequality is 10bβˆ’400β‰₯50010b - 400 \geq 500. Solving gives bβ‰₯90b \geq 90 items.

Explanation

Use inequalities to solve real-world problems about budgets, profits, or limits. Set up an expression for the total cost or amount and compare it to the limit using an inequality sign. Then, solve for your variable.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 2: Solving Linear Equations

  1. Lesson 1

    Lesson 2.1: Use a General Strategy to Solve Linear Equations

  2. Lesson 2

    Lesson 2.2: Use a Problem Solving Strategy

  3. Lesson 3

    Lesson 3: 2.3 Solve a Formula for a Specific Variable

  4. Lesson 4

    Lesson 2.4: Solve Mixture and Uniform Motion Applications

  5. Lesson 5Current

    Lesson 2.5: Solve Linear Inequalities

  6. Lesson 6

    Lesson 2.6: Solve Compound Inequalities

  7. Lesson 7

    Lesson 2.7: Solve Absolute Value Inequalities