Learn on PengiOpenstax Prealgebre 2EChapter 2: The Language of Algebra

Lesson 3: Solving Equations Using the Subtraction and Addition Properties of Equality

In this lesson from OpenStax Prealgebra 2e, Chapter 2, students learn how to solve equations using the Subtraction and Addition Properties of Equality, including how to determine whether a number is a solution to an equation by substituting values for variables. Students also practice translating word phrases into algebraic equations and solving real-world problems using these foundational equation-solving techniques.

Section 1

πŸ“˜ Solving Equations Using the Subtraction and Addition Properties of Equality

New Concept

This lesson introduces the foundation of solving algebraic equations: isolating the variable. You'll learn how the Addition and Subtraction Properties of Equality allow you to 'undo' operations and find the value that makes an equation true.

What’s next

Next, you'll put these properties into action with interactive examples, practice problems, and challenge questions to master solving equations.

Section 2

Solution of an equation

Property

A solution to an equation is a value of a variable that makes a true statement when substituted into the equation. The process of finding the solution to an equation is called solving the equation.

Examples

  • Is x=6x=6 a solution to x+9=15x+9=15? We substitute 6 for xx to get 6+9=156+9=15. Since 15=1515=15 is a true statement, 6 is a solution.
  • Is d=2d=2 a solution to 18dβˆ’9=2718d-9=27? We substitute 2 for dd to get 18(2)βˆ’9=36βˆ’9=2718(2)-9 = 36-9=27. Since 27=2727=27 is true, 2 is a solution.

Section 3

Subtraction property of equality

Property

For any numbers aa, bb, and cc, if a=ba=b, then aβˆ’c=bβˆ’ca-c=b-c. When we subtract the same quantity from both sides of an equation, we still have equality. The goal is to isolate the variable.

Examples

  • To solve x+8=17x+8=17, we use the Subtraction Property of Equality to isolate xx. Subtracting 8 from both sides gives x+8βˆ’8=17βˆ’8x+8-8=17-8, which simplifies to x=9x=9.
  • To solve a+2=18a+2=18, we get aa by itself by subtracting 2 from both sides. This gives a+2βˆ’2=18βˆ’2a+2-2=18-2, which simplifies to a=16a=16.

Section 4

Addition property of equality

Property

For any numbers aa, bb, and cc, if a=ba=b, then a+c=b+ca+c=b+c. We can add the same number to both sides of the equation without changing the equality.

Examples

  • To solve xβˆ’5=8x-5=8, we use the Addition Property of Equality to isolate xx. Adding 5 to both sides gives xβˆ’5+5=8+5x-5+5=8+5, which simplifies to x=13x=13.
  • To solve yβˆ’3=19y-3=19, we get yy by itself by adding 3 to both sides. This gives yβˆ’3+3=19+3y-3+3=19+3, which simplifies to y=22y=22.

Section 5

Translate word phrases to equations

Property

An equation has an equal sign between two algebraic expressions. To translate a sentence into an equation, look for clue words that mean 'equals' (is, gives, was, etc.). Then, translate the phrases on each side of the 'equals' word into expressions.

Examples

  • The sentence 'The sum of 6 and 12 is equal to 18' translates to the equation 6+12=186+12=18.
  • The sentence 'The product of 4 and 7 is equal to 28' translates to the equation 4β‹…7=284 \cdot 7=28.

Section 6

Translate to an equation and solve

Property

First, translate a word sentence into an algebraic equation. Then, solve the equation by using the Subtraction and Addition Properties of Equality to isolate the variable.

Examples

  • 'Five more than pp is equal to 21' translates to p+5=21p+5=21. Subtracting 5 from both sides gives p=16p=16.
  • 'The difference of yy and 9 is 20' translates to yβˆ’9=20y-9=20. Adding 9 to both sides gives y=29y=29.

Book overview

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

Continue this chapter

Chapter 2: The Language of Algebra

  1. Lesson 1

    Lesson 1: Use the Language of Algebra

  2. Lesson 2

    Lesson 2: Evaluate, Simplify, and Translate Expressions

  3. Lesson 3Current

    Lesson 3: Solving Equations Using the Subtraction and Addition Properties of Equality

  4. Lesson 4

    Lesson 4: Find Multiples and Factors

  5. Lesson 5

    Lesson 5: Prime Factorization and the Least Common Multiple

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

πŸ“˜ Solving Equations Using the Subtraction and Addition Properties of Equality

New Concept

This lesson introduces the foundation of solving algebraic equations: isolating the variable. You'll learn how the Addition and Subtraction Properties of Equality allow you to 'undo' operations and find the value that makes an equation true.

What’s next

Next, you'll put these properties into action with interactive examples, practice problems, and challenge questions to master solving equations.

Section 2

Solution of an equation

Property

A solution to an equation is a value of a variable that makes a true statement when substituted into the equation. The process of finding the solution to an equation is called solving the equation.

Examples

  • Is x=6x=6 a solution to x+9=15x+9=15? We substitute 6 for xx to get 6+9=156+9=15. Since 15=1515=15 is a true statement, 6 is a solution.
  • Is d=2d=2 a solution to 18dβˆ’9=2718d-9=27? We substitute 2 for dd to get 18(2)βˆ’9=36βˆ’9=2718(2)-9 = 36-9=27. Since 27=2727=27 is true, 2 is a solution.

Section 3

Subtraction property of equality

Property

For any numbers aa, bb, and cc, if a=ba=b, then aβˆ’c=bβˆ’ca-c=b-c. When we subtract the same quantity from both sides of an equation, we still have equality. The goal is to isolate the variable.

Examples

  • To solve x+8=17x+8=17, we use the Subtraction Property of Equality to isolate xx. Subtracting 8 from both sides gives x+8βˆ’8=17βˆ’8x+8-8=17-8, which simplifies to x=9x=9.
  • To solve a+2=18a+2=18, we get aa by itself by subtracting 2 from both sides. This gives a+2βˆ’2=18βˆ’2a+2-2=18-2, which simplifies to a=16a=16.

Section 4

Addition property of equality

Property

For any numbers aa, bb, and cc, if a=ba=b, then a+c=b+ca+c=b+c. We can add the same number to both sides of the equation without changing the equality.

Examples

  • To solve xβˆ’5=8x-5=8, we use the Addition Property of Equality to isolate xx. Adding 5 to both sides gives xβˆ’5+5=8+5x-5+5=8+5, which simplifies to x=13x=13.
  • To solve yβˆ’3=19y-3=19, we get yy by itself by adding 3 to both sides. This gives yβˆ’3+3=19+3y-3+3=19+3, which simplifies to y=22y=22.

Section 5

Translate word phrases to equations

Property

An equation has an equal sign between two algebraic expressions. To translate a sentence into an equation, look for clue words that mean 'equals' (is, gives, was, etc.). Then, translate the phrases on each side of the 'equals' word into expressions.

Examples

  • The sentence 'The sum of 6 and 12 is equal to 18' translates to the equation 6+12=186+12=18.
  • The sentence 'The product of 4 and 7 is equal to 28' translates to the equation 4β‹…7=284 \cdot 7=28.

Section 6

Translate to an equation and solve

Property

First, translate a word sentence into an algebraic equation. Then, solve the equation by using the Subtraction and Addition Properties of Equality to isolate the variable.

Examples

  • 'Five more than pp is equal to 21' translates to p+5=21p+5=21. Subtracting 5 from both sides gives p=16p=16.
  • 'The difference of yy and 9 is 20' translates to yβˆ’9=20y-9=20. Adding 9 to both sides gives y=29y=29.

Book overview

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

Continue this chapter

Chapter 2: The Language of Algebra

  1. Lesson 1

    Lesson 1: Use the Language of Algebra

  2. Lesson 2

    Lesson 2: Evaluate, Simplify, and Translate Expressions

  3. Lesson 3Current

    Lesson 3: Solving Equations Using the Subtraction and Addition Properties of Equality

  4. Lesson 4

    Lesson 4: Find Multiples and Factors

  5. Lesson 5

    Lesson 5: Prime Factorization and the Least Common Multiple