Learn on PengiOpenstax Prealgebre 2EChapter 8: Solving Linear Equations

Lesson 2: Solve Equations Using the Division and Multiplication Properties of Equality

In this lesson from OpenStax Prealgebra 2E, Chapter 8, students learn to solve one-variable linear equations by applying the Division and Multiplication Properties of Equality, including cases where the variable is multiplied by a coefficient or divided by a number. Learners practice isolating the variable by performing inverse operations on both sides of the equation, then simplifying equations that require multiple steps before solving. The lesson also covers interpreting negative coefficients such as recognizing that negative r equals negative one times r.

Section 1

πŸ“˜ Solve Equations Using the Division and Multiplication Properties of Equality

New Concept

This lesson introduces two powerful tools: the Division and Multiplication Properties of Equality. You'll learn to use these inverse operations to isolate a variable and solve for its value, even in complex, multi-step equations.

What’s next

Now, let's see these properties in action. You'll work through interactive examples and then apply your skills on a series of practice cards.

Section 2

Division property of equality

Property

Division Property of Equality: For all real numbers aa, bb, cc, and c≠0c \neq 0, if a=ba = b, then ac=bc\frac{a}{c} = \frac{b}{c}.

When you divide both sides of an equation by the same non-zero quantity, you still have equality. The goal is to 'undo' the operation on the variable.

Examples

  • To solve 4x=βˆ’284x = -28, we 'undo' the multiplication by dividing both sides by 4. This gives 4x4=βˆ’284\frac{4x}{4} = \frac{-28}{4}, so x=βˆ’7x = -7.

Section 3

Multiplication property of equality

Property

Multiplication Property of Equality: For all real numbers aa, bb, cc, if a=ba = b, then ac=bcac = bc.

When you multiply both sides of an equation by the same quantity, you still have equality. This is used to 'undo' division in an equation.

Examples

  • To solve aβˆ’7=βˆ’42\frac{a}{-7} = -42, we multiply both sides by βˆ’7-7. This results in βˆ’7(aβˆ’7)=βˆ’7(βˆ’42)-7(\frac{a}{-7}) = -7(-42), which simplifies to a=294a = 294.

Section 4

Solving equations with fractional coefficients

Property

Since the product of a number and its reciprocal is 1, our strategy will be to isolate the variable by multiplying by the reciprocal of the fractional coefficient. For an equation like abx=c\frac{a}{b}x = c, you multiply by ba\frac{b}{a}.

Examples

  • To solve 23x=18\frac{2}{3}x = 18, multiply both sides by the reciprocal of 23\frac{2}{3}, which is 32\frac{3}{2}. This gives 32β‹…23x=32β‹…18\frac{3}{2} \cdot \frac{2}{3}x = \frac{3}{2} \cdot 18, so x=27x = 27.
  • To solve 34r=15\frac{3}{4}r = 15, multiply both sides by 43\frac{4}{3}. You get 43β‹…34r=43β‹…15\frac{4}{3} \cdot \frac{3}{4}r = \frac{4}{3} \cdot 15, which simplifies to r=20r = 20.

Section 5

Simplifying equations before solving

Property

Many equations start out more complicated than they need to be. Before applying the properties of equality, simplify both sides of the equation as much as possible by combining like terms or using the distributive property.

Examples

  • To solve 8x+9xβˆ’5x=βˆ’3+158x + 9x - 5x = -3 + 15, first combine like terms on each side to get 12x=1212x = 12. Then, divide by 12 to find x=1x = 1.
  • In βˆ’3(nβˆ’2)βˆ’6=21-3(n - 2) - 6 = 21, first distribute the βˆ’3-3 to get βˆ’3n+6βˆ’6=21-3n + 6 - 6 = 21. This simplifies to βˆ’3n=21-3n = 21. Dividing by βˆ’3-3 gives n=βˆ’7n = -7.

Book overview

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

Continue this chapter

Chapter 8: Solving Linear Equations

  1. Lesson 1

    Lesson 1: Solve Equations Using the Subtraction and Addition Properties of Equality

  2. Lesson 2Current

    Lesson 2: Solve Equations Using the Division and Multiplication Properties of Equality

  3. Lesson 3

    Lesson 3: Solve Equations with Variables and Constants on Both Sides

  4. Lesson 4

    Lesson 4: Solve Equations with Fraction or Decimal Coefficients

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

πŸ“˜ Solve Equations Using the Division and Multiplication Properties of Equality

New Concept

This lesson introduces two powerful tools: the Division and Multiplication Properties of Equality. You'll learn to use these inverse operations to isolate a variable and solve for its value, even in complex, multi-step equations.

What’s next

Now, let's see these properties in action. You'll work through interactive examples and then apply your skills on a series of practice cards.

Section 2

Division property of equality

Property

Division Property of Equality: For all real numbers aa, bb, cc, and c≠0c \neq 0, if a=ba = b, then ac=bc\frac{a}{c} = \frac{b}{c}.

When you divide both sides of an equation by the same non-zero quantity, you still have equality. The goal is to 'undo' the operation on the variable.

Examples

  • To solve 4x=βˆ’284x = -28, we 'undo' the multiplication by dividing both sides by 4. This gives 4x4=βˆ’284\frac{4x}{4} = \frac{-28}{4}, so x=βˆ’7x = -7.

Section 3

Multiplication property of equality

Property

Multiplication Property of Equality: For all real numbers aa, bb, cc, if a=ba = b, then ac=bcac = bc.

When you multiply both sides of an equation by the same quantity, you still have equality. This is used to 'undo' division in an equation.

Examples

  • To solve aβˆ’7=βˆ’42\frac{a}{-7} = -42, we multiply both sides by βˆ’7-7. This results in βˆ’7(aβˆ’7)=βˆ’7(βˆ’42)-7(\frac{a}{-7}) = -7(-42), which simplifies to a=294a = 294.

Section 4

Solving equations with fractional coefficients

Property

Since the product of a number and its reciprocal is 1, our strategy will be to isolate the variable by multiplying by the reciprocal of the fractional coefficient. For an equation like abx=c\frac{a}{b}x = c, you multiply by ba\frac{b}{a}.

Examples

  • To solve 23x=18\frac{2}{3}x = 18, multiply both sides by the reciprocal of 23\frac{2}{3}, which is 32\frac{3}{2}. This gives 32β‹…23x=32β‹…18\frac{3}{2} \cdot \frac{2}{3}x = \frac{3}{2} \cdot 18, so x=27x = 27.
  • To solve 34r=15\frac{3}{4}r = 15, multiply both sides by 43\frac{4}{3}. You get 43β‹…34r=43β‹…15\frac{4}{3} \cdot \frac{3}{4}r = \frac{4}{3} \cdot 15, which simplifies to r=20r = 20.

Section 5

Simplifying equations before solving

Property

Many equations start out more complicated than they need to be. Before applying the properties of equality, simplify both sides of the equation as much as possible by combining like terms or using the distributive property.

Examples

  • To solve 8x+9xβˆ’5x=βˆ’3+158x + 9x - 5x = -3 + 15, first combine like terms on each side to get 12x=1212x = 12. Then, divide by 12 to find x=1x = 1.
  • In βˆ’3(nβˆ’2)βˆ’6=21-3(n - 2) - 6 = 21, first distribute the βˆ’3-3 to get βˆ’3n+6βˆ’6=21-3n + 6 - 6 = 21. This simplifies to βˆ’3n=21-3n = 21. Dividing by βˆ’3-3 gives n=βˆ’7n = -7.

Book overview

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

Continue this chapter

Chapter 8: Solving Linear Equations

  1. Lesson 1

    Lesson 1: Solve Equations Using the Subtraction and Addition Properties of Equality

  2. Lesson 2Current

    Lesson 2: Solve Equations Using the Division and Multiplication Properties of Equality

  3. Lesson 3

    Lesson 3: Solve Equations with Variables and Constants on Both Sides

  4. Lesson 4

    Lesson 4: Solve Equations with Fraction or Decimal Coefficients