Learn on PengiBig Ideas Math, Advanced 1Chapter 7: Equations and Inequalities

Lesson 3: Solving Equations Using Multiplication or Division

In this Grade 6 lesson from Big Ideas Math, Advanced 1, students learn to solve one-variable equations using the Multiplication Property of Equality and the Division Property of Equality. The lesson covers techniques such as multiplying both sides by a reciprocal to isolate a variable and dividing both sides by a coefficient, with practice using substitution to check solutions. Real-world word problems involving shared costs and unit conversions help students apply these equation-solving strategies in context.

Section 1

Multiplication and Division Properties of Equality

Property

Multiplication Property of Equality: Multiplying both sides of an equation by the same non-zero number maintains the equality. If a=ba=b and c0c \neq 0, then ac=bcac = bc.

Division Property of Equality: Dividing both sides of an equation by the same non-zero number maintains the equality. If a=ba=b and c0c \neq 0, then ac=bc\frac{a}{c} = \frac{b}{c}.

Section 2

Reciprocals and Multiplicative Inverse

Property

For multiplication, the “opposite” of a number aa is the solution of the equation ax=1ax = 1, denoted 1/a1/a.
It is the “multiplicative inverse” of aa. This means that a×(1/a)=1a \times (1/a) = 1. Division by aa undoes multiplication by aa.
Since 0×x=00 \times x = 0 for every number xx, there is no solution to the equation 0×x=10 \times x = 1. For this reason, we cannot divide by zero.

Examples

  • Calculate (21÷7)(-21 \div 7). Since we know that (3)×7=21(-3) \times 7 = -21, the answer must be 3-3.
  • Calculate 30÷(5)30 \div (-5). We are looking for a number xx that solves 5x=30-5x = 30. Since (5)×(6)=30(-5) \times (-6) = 30, the answer is 6-6.
  • Calculate (45)÷(9)(-45) \div (-9). The number which, when multiplied by 9-9, gives 45-45 must be positive. Since 5×(9)=455 \times (-9) = -45, the answer is 5. No, wait. Since (9)×5=45(-9) \times 5 = -45, the answer is 5. No, wait. A negative times a positive is negative. To get a negative product (45-45) from a negative factor (9-9), the other factor must be positive. The answer is 5, since (9)×5=45(-9) \times 5 = -45. No, wait. The product of two negatives is a positive. The number must be positive. The answer is 5, since (9)×(5)=45(-9) \times (-5) = 45. No, (9)×5=45(-9) \times 5 = -45. The correct answer is 5.

Explanation

Division is simply the reverse of multiplication. Dividing by a number is the same as multiplying by its inverse (like 5 and 15\frac{1}{5}). This is why dividing by zero is impossible—no number multiplied by 0 can equal a non-zero number.

Section 3

Solving Equations Using Multiplication or Division

Property

To solve equations involving multiplication or division, we apply the multiplication and division properties of equality to isolate the variable.
For equations in the form ax=bax = b, we divide both sides by aa (where a0a \neq 0) to get x=bax = \frac{b}{a}.
For equations in the form xa=b\frac{x}{a} = b, we multiply both sides by aa to get x=abx = ab.

Examples

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Chapter 7: Equations and Inequalities

  1. Lesson 1

    Lesson 1: Writing Equations in One Variable

  2. Lesson 2

    Lesson 2: Solving Equations Using Addition or Subtraction

  3. Lesson 3Current

    Lesson 3: Solving Equations Using Multiplication or Division

  4. Lesson 4

    Lesson 4: Writing Equations in Two Variables

  5. Lesson 5

    Lesson 5: Writing and Graphing Inequalities

  6. Lesson 6

    Lesson 6: Solving Inequalities Using Addition or Subtraction

  7. Lesson 7

    Lesson 7: Solving Inequalities Using Multiplication or Division

Lesson overview

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Section 1

Multiplication and Division Properties of Equality

Property

Multiplication Property of Equality: Multiplying both sides of an equation by the same non-zero number maintains the equality. If a=ba=b and c0c \neq 0, then ac=bcac = bc.

Division Property of Equality: Dividing both sides of an equation by the same non-zero number maintains the equality. If a=ba=b and c0c \neq 0, then ac=bc\frac{a}{c} = \frac{b}{c}.

Section 2

Reciprocals and Multiplicative Inverse

Property

For multiplication, the “opposite” of a number aa is the solution of the equation ax=1ax = 1, denoted 1/a1/a.
It is the “multiplicative inverse” of aa. This means that a×(1/a)=1a \times (1/a) = 1. Division by aa undoes multiplication by aa.
Since 0×x=00 \times x = 0 for every number xx, there is no solution to the equation 0×x=10 \times x = 1. For this reason, we cannot divide by zero.

Examples

  • Calculate (21÷7)(-21 \div 7). Since we know that (3)×7=21(-3) \times 7 = -21, the answer must be 3-3.
  • Calculate 30÷(5)30 \div (-5). We are looking for a number xx that solves 5x=30-5x = 30. Since (5)×(6)=30(-5) \times (-6) = 30, the answer is 6-6.
  • Calculate (45)÷(9)(-45) \div (-9). The number which, when multiplied by 9-9, gives 45-45 must be positive. Since 5×(9)=455 \times (-9) = -45, the answer is 5. No, wait. Since (9)×5=45(-9) \times 5 = -45, the answer is 5. No, wait. A negative times a positive is negative. To get a negative product (45-45) from a negative factor (9-9), the other factor must be positive. The answer is 5, since (9)×5=45(-9) \times 5 = -45. No, wait. The product of two negatives is a positive. The number must be positive. The answer is 5, since (9)×(5)=45(-9) \times (-5) = 45. No, (9)×5=45(-9) \times 5 = -45. The correct answer is 5.

Explanation

Division is simply the reverse of multiplication. Dividing by a number is the same as multiplying by its inverse (like 5 and 15\frac{1}{5}). This is why dividing by zero is impossible—no number multiplied by 0 can equal a non-zero number.

Section 3

Solving Equations Using Multiplication or Division

Property

To solve equations involving multiplication or division, we apply the multiplication and division properties of equality to isolate the variable.
For equations in the form ax=bax = b, we divide both sides by aa (where a0a \neq 0) to get x=bax = \frac{b}{a}.
For equations in the form xa=b\frac{x}{a} = b, we multiply both sides by aa to get x=abx = ab.

Examples

Book overview

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

Continue this chapter

Chapter 7: Equations and Inequalities

  1. Lesson 1

    Lesson 1: Writing Equations in One Variable

  2. Lesson 2

    Lesson 2: Solving Equations Using Addition or Subtraction

  3. Lesson 3Current

    Lesson 3: Solving Equations Using Multiplication or Division

  4. Lesson 4

    Lesson 4: Writing Equations in Two Variables

  5. Lesson 5

    Lesson 5: Writing and Graphing Inequalities

  6. Lesson 6

    Lesson 6: Solving Inequalities Using Addition or Subtraction

  7. Lesson 7

    Lesson 7: Solving Inequalities Using Multiplication or Division