Learn on PengiBig Ideas Math, Course 2Chapter 3: Expressions and Equations

Lesson 4: Solving Equations Using Multiplication or Division

In this Grade 7 lesson from Big Ideas Math, Course 2, students learn to solve one-step equations using the Multiplication Property of Equality and the Division Property of Equality, including equations with fractions solved by multiplying by the reciprocal. The lesson covers equations of the form x/a = b and ax = b with integers, decimals, and fractions, and applies these skills to real-life problems. Part of Chapter 3: Expressions and Equations, it aligns with Florida standard MAFS.7.EE.2.4a.

Section 1

Multiplication and Division as Inverse Operations

Property

Multiplication and division are opposite or inverse operations, because each operation undoes the effects of the other.

Examples

Section 2

Solving with multiplication and division

Property

The Division Property of Equality: For any numbers $a$ , $b$ , and $c$ , and $c \neq 0$ , if $a = b$ , then $\frac{a}{c} = \frac{b}{c}$ .

The Multiplication Property of Equality: For any numbers $a$ , $b$ , and $c$ , if $a = b$ , then $ac = bc$ .

Use these properties to isolate the variable by performing the inverse operation on both sides of the equation.

Section 3

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 $\frac{a}{b}x = c$ , you multiply by $\frac{b}{a}$ .

Examples

To solve $\frac{2}{3}x = 18$ , multiply both sides by the reciprocal of $\frac{2}{3}$ , which is $\frac{3}{2}$ . This gives $\frac{3}{2} \cdot \frac{2}{3}x = \frac{3}{2} \cdot 18$ , so $x = 27$ .

To solve $\frac{3}{4}r = 15$ , multiply both sides by $\frac{4}{3}$ . You get $\frac{4}{3} \cdot \frac{3}{4}r = \frac{4}{3} \cdot 15$ , which simplifies to $r = 20$ .

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Multiplication and Division as Inverse Operations

Property

Multiplication and division are opposite or inverse operations, because each operation undoes the effects of the other.

Examples

Section 2

Solving with multiplication and division

Property

The Division Property of Equality: For any numbers $a$ , $b$ , and $c$ , and $c \neq 0$ , if $a = b$ , then $\frac{a}{c} = \frac{b}{c}$ .

The Multiplication Property of Equality: For any numbers $a$ , $b$ , and $c$ , if $a = b$ , then $ac = bc$ .

Use these properties to isolate the variable by performing the inverse operation on both sides of the equation.

Section 3

Solving equations with fractional coefficients

Property

Examples

To solve $\frac{2}{3}x = 18$ , multiply both sides by the reciprocal of $\frac{2}{3}$ , which is $\frac{3}{2}$ . This gives $\frac{3}{2} \cdot \frac{2}{3}x = \frac{3}{2} \cdot 18$ , so $x = 27$ .

To solve $\frac{3}{4}r = 15$ , multiply both sides by $\frac{4}{3}$ . You get $\frac{4}{3} \cdot \frac{3}{4}r = \frac{4}{3} \cdot 15$ , which simplifies to $r = 20$ .

Overview

Lesson overview

Review the lesson summary and core properties without leaving the current page.

Overview

Section 1

Multiplication and Division as Inverse Operations

Property

Multiplication and division are opposite or inverse operations, because each operation undoes the effects of the other.

Examples

Section 2

Solving with multiplication and division

Property

The Division Property of Equality: For any numbers $a$ , $b$ , and $c$ , and $c \neq 0$ , if $a = b$ , then $\frac{a}{c} = \frac{b}{c}$ .

The Multiplication Property of Equality: For any numbers $a$ , $b$ , and $c$ , if $a = b$ , then $ac = bc$ .

Use these properties to isolate the variable by performing the inverse operation on both sides of the equation.

Section 3

Solving equations with fractional coefficients

Property

Examples

To solve $\frac{2}{3}x = 18$ , multiply both sides by the reciprocal of $\frac{2}{3}$ , which is $\frac{3}{2}$ . This gives $\frac{3}{2} \cdot \frac{2}{3}x = \frac{3}{2} \cdot 18$ , so $x = 27$ .

To solve $\frac{3}{4}r = 15$ , multiply both sides by $\frac{4}{3}$ . You get $\frac{4}{3} \cdot \frac{3}{4}r = \frac{4}{3} \cdot 15$ , which simplifies to $r = 20$ .