Learn on PengiSaxon Algebra 1Chapter 8: Advanced Factoring and Functions

Lesson 73: Solving Compound Inequalities

New Concept A compound inequality is two inequalities combined with the word AND or OR. What’s next Next, you’ll apply this concept to write, graph, and solve problems involving both 'AND' (conjunctions) and 'OR' (disjunctions) statements.

Section 1

📘 Solving Compound Inequalities

New Concept

A compound inequality is two inequalities combined with the word AND or OR.

What’s next

Next, you’ll apply this concept to write, graph, and solve problems involving both 'AND' (conjunctions) and 'OR' (disjunctions) statements.

Section 2

Compound Inequality

Property

A compound inequality consists of two inequalities joined by the word AND or OR. A conjunction is a compound inequality linked by AND. For instance, the statement x3x \ge -3 AND x5x \le 5 can be written more compactly as 3x5-3 \le x \le 5.

Explanation

Think of a conjunction as a VIP club with two rules: you must be old enough AND have an invitation. You have to satisfy both conditions to get in! The solution is the 'sweet spot' where both inequalities overlap, representing all the numbers that fit both criteria at the same time.

Examples

  • All real numbers greater than 2 and less than 7 are written as 2<x<72 < x < 7.
  • A recommended lasagna cooking time is from 16 to 20 minutes: 16t2016 \le t \le 20.
  • A graph showing a line segment between solid dots at -2 and 6 is represented by 2x6-2 \le x \le 6.

Section 3

Solving a Conjunction

Property

A conjunction can be solved as two inequalities connected with AND or as a single inequality with three parts, such as 3010+0.10x4030 \le 10 + 0.10x \le 40.

Explanation

Solving a three-part inequality is like a balancing act! To find your variable, you must perform the same operation on all three parts—the left, the middle, and the right. This ensures that you keep the relationship between the numbers fair and balanced while you work to isolate the variable in the center.

Examples

  • A phone bill between 30 and 40 dollars, with a 10 dollars fee and 0.10 dollars per minute, xx, is 3010+0.10x4030 \le 10 + 0.10x \le 40.
  • First, subtract 10 from all three parts: 200.10x3020 \le 0.10x \le 30.
  • Next, divide all three parts by 0.10 to find the range of minutes: 200x300200 \le x \le 300.

Section 4

Example Card: Solving a Conjunction (Phone Bill)

Let's see how compound inequalities can help you stay within your monthly phone budget. This example uses a conjunction, where two conditions must be true.

Example Problem: A phone plan costs 15 dollars plus $0.20 per minute. The monthly bill is between 45 dollars and 65 dollars. Find the possible minutes used.

Step-by-Step:

  1. Let xx be the number of minutes. The total cost is 15+0.20x15 + 0.20x. The bill is between 45 dollars and 65 dollars, so we write the conjunction: 4515+0.20x6545 \le 15 + 0.20x \le 65.
  2. We solve this as one inequality. First, subtract 15 from all three parts.
451515+0.20x15651545 - 15 \le 15 + 0.20x - 15 \le 65 - 15
300.20x5030 \le 0.20x \le 50
  1. Now, divide each part by 0.200.20 to isolate xx.
300.200.20x0.20500.20\frac{30}{0.20} \le \frac{0.20x}{0.20} \le \frac{50}{0.20}
150x250150 \le x \le 250

Section 5

Disjunction

Property

A disjunction is a compound inequality that uses the word OR. A solution to a disjunction is any number that makes either of the inequalities true. These must be written as two separate inequalities connected by OR.

Explanation

Think of a disjunction as a choice at a theme park: you can ride the roller coaster OR the Ferris wheel. You don't have to do both! Any number that satisfies at least one of the conditions is part of the solution, which is why the graphs often shoot off in opposite directions.

Examples

  • All real numbers greater than 9 or less than 3 is written as x<3x < 3 OR x>9x > 9.
  • All real numbers no more than 7 or no less than 11 is written as x7x \le 7 OR x11x \ge 11.
  • The solution to x3<5x - 3 < -5 OR 2x<6-2x < -6 is x<2x < -2 OR x>3x > 3.

Section 6

Example Card: Solving a Disjunction

What if only one of two conditions needs to be true? This example uses a disjunction, connected by 'OR'.

Example Problem: Solve the disjunction x+2<4x + 2 < -4 OR 3x<12-3x < -12.

Step-by-Step:

  1. Solve each inequality separately. For the first one, subtract 2 from both sides.
x+2<4x + 2 < -4
x<6x < -6
  1. For the second inequality, divide by 3-3. Remember to reverse the inequality sign when dividing by a negative number.
3x<12-3x < -12
3x3>123\frac{-3x}{-3} > \frac{-12}{-3}
x>4x > 4
  1. Combine the two results with "OR". The solution is any number that satisfies either condition: x<6x < -6 OR x>4x > 4.

Book overview

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

Continue this chapter

Chapter 8: Advanced Factoring and Functions

  1. Lesson 1

    Lesson 71: Making and Analyzing Scatter Plots

  2. Lesson 2

    Lesson 72: Factoring Trinomials: x² + bx + c

  3. Lesson 3Current

    Lesson 73: Solving Compound Inequalities

  4. Lesson 4

    Lesson 74: Solving Absolute-Value Equations

  5. Lesson 5

    Lesson 75: Factoring Trinomials: ax² + bx + c

  6. Lesson 6

    Lesson 76: Multiplying Radical Expressions

  7. Lesson 7

    Lesson 77: Solving Two-Step and Multi-Step Inequalities

  8. Lesson 8

    Lesson 78: Graphing Rational Functions

  9. Lesson 9

    Lesson 79: Factoring Trinomials by Using the GCF

  10. Lesson 10

    Lesson 80: Calculating Frequency Distributions

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

📘 Solving Compound Inequalities

New Concept

A compound inequality is two inequalities combined with the word AND or OR.

What’s next

Next, you’ll apply this concept to write, graph, and solve problems involving both 'AND' (conjunctions) and 'OR' (disjunctions) statements.

Section 2

Compound Inequality

Property

A compound inequality consists of two inequalities joined by the word AND or OR. A conjunction is a compound inequality linked by AND. For instance, the statement x3x \ge -3 AND x5x \le 5 can be written more compactly as 3x5-3 \le x \le 5.

Explanation

Think of a conjunction as a VIP club with two rules: you must be old enough AND have an invitation. You have to satisfy both conditions to get in! The solution is the 'sweet spot' where both inequalities overlap, representing all the numbers that fit both criteria at the same time.

Examples

  • All real numbers greater than 2 and less than 7 are written as 2<x<72 < x < 7.
  • A recommended lasagna cooking time is from 16 to 20 minutes: 16t2016 \le t \le 20.
  • A graph showing a line segment between solid dots at -2 and 6 is represented by 2x6-2 \le x \le 6.

Section 3

Solving a Conjunction

Property

A conjunction can be solved as two inequalities connected with AND or as a single inequality with three parts, such as 3010+0.10x4030 \le 10 + 0.10x \le 40.

Explanation

Solving a three-part inequality is like a balancing act! To find your variable, you must perform the same operation on all three parts—the left, the middle, and the right. This ensures that you keep the relationship between the numbers fair and balanced while you work to isolate the variable in the center.

Examples

  • A phone bill between 30 and 40 dollars, with a 10 dollars fee and 0.10 dollars per minute, xx, is 3010+0.10x4030 \le 10 + 0.10x \le 40.
  • First, subtract 10 from all three parts: 200.10x3020 \le 0.10x \le 30.
  • Next, divide all three parts by 0.10 to find the range of minutes: 200x300200 \le x \le 300.

Section 4

Example Card: Solving a Conjunction (Phone Bill)

Let's see how compound inequalities can help you stay within your monthly phone budget. This example uses a conjunction, where two conditions must be true.

Example Problem: A phone plan costs 15 dollars plus $0.20 per minute. The monthly bill is between 45 dollars and 65 dollars. Find the possible minutes used.

Step-by-Step:

  1. Let xx be the number of minutes. The total cost is 15+0.20x15 + 0.20x. The bill is between 45 dollars and 65 dollars, so we write the conjunction: 4515+0.20x6545 \le 15 + 0.20x \le 65.
  2. We solve this as one inequality. First, subtract 15 from all three parts.
451515+0.20x15651545 - 15 \le 15 + 0.20x - 15 \le 65 - 15
300.20x5030 \le 0.20x \le 50
  1. Now, divide each part by 0.200.20 to isolate xx.
300.200.20x0.20500.20\frac{30}{0.20} \le \frac{0.20x}{0.20} \le \frac{50}{0.20}
150x250150 \le x \le 250

Section 5

Disjunction

Property

A disjunction is a compound inequality that uses the word OR. A solution to a disjunction is any number that makes either of the inequalities true. These must be written as two separate inequalities connected by OR.

Explanation

Think of a disjunction as a choice at a theme park: you can ride the roller coaster OR the Ferris wheel. You don't have to do both! Any number that satisfies at least one of the conditions is part of the solution, which is why the graphs often shoot off in opposite directions.

Examples

  • All real numbers greater than 9 or less than 3 is written as x<3x < 3 OR x>9x > 9.
  • All real numbers no more than 7 or no less than 11 is written as x7x \le 7 OR x11x \ge 11.
  • The solution to x3<5x - 3 < -5 OR 2x<6-2x < -6 is x<2x < -2 OR x>3x > 3.

Section 6

Example Card: Solving a Disjunction

What if only one of two conditions needs to be true? This example uses a disjunction, connected by 'OR'.

Example Problem: Solve the disjunction x+2<4x + 2 < -4 OR 3x<12-3x < -12.

Step-by-Step:

  1. Solve each inequality separately. For the first one, subtract 2 from both sides.
x+2<4x + 2 < -4
x<6x < -6
  1. For the second inequality, divide by 3-3. Remember to reverse the inequality sign when dividing by a negative number.
3x<12-3x < -12
3x3>123\frac{-3x}{-3} > \frac{-12}{-3}
x>4x > 4
  1. Combine the two results with "OR". The solution is any number that satisfies either condition: x<6x < -6 OR x>4x > 4.

Book overview

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

Continue this chapter

Chapter 8: Advanced Factoring and Functions

  1. Lesson 1

    Lesson 71: Making and Analyzing Scatter Plots

  2. Lesson 2

    Lesson 72: Factoring Trinomials: x² + bx + c

  3. Lesson 3Current

    Lesson 73: Solving Compound Inequalities

  4. Lesson 4

    Lesson 74: Solving Absolute-Value Equations

  5. Lesson 5

    Lesson 75: Factoring Trinomials: ax² + bx + c

  6. Lesson 6

    Lesson 76: Multiplying Radical Expressions

  7. Lesson 7

    Lesson 77: Solving Two-Step and Multi-Step Inequalities

  8. Lesson 8

    Lesson 78: Graphing Rational Functions

  9. Lesson 9

    Lesson 79: Factoring Trinomials by Using the GCF

  10. Lesson 10

    Lesson 80: Calculating Frequency Distributions