Learn on PengiThe Art of Problem Solving: Prealgebra (AMC 8)Chapter 5: Equations and Inequalities

Lesson 3: Solving Linear Equations II

In this Grade 4 AMC 8 lesson from Art of Problem Solving: Prealgebra, students learn to solve multi-step linear equations by combining addition, subtraction, multiplication, and division in sequence. Key skills include collecting like terms, eliminating variables from both sides of an equation, applying the distributive property, and working with equations that contain fractions and mixed numbers. The lesson builds systematically on earlier equation-solving strategies to handle more complex forms such as 4(t−7) = 3(2t+3) and (2r−7)/9 = 3.

Section 1

Solve equations with variables and constants on both sides

Property

Step 1. Choose one side to be the variable side and then the other will be the constant side.
Step 2. Collect the variable terms to the variable side, using the Addition or Subtraction Property of Equality.
Step 3. Collect the constants to the other side, using the Addition or Subtraction Property of Equality.
Step 4. Make the coefficient of the variable 1, using the Multiplication or Division Property of Equality.
Step 5. Check the solution by substituting it into the original equation.

It is a good idea to make the variable side the one in which the variable has the larger coefficient. This usually makes the arithmetic easier.

Examples

Given $9x + 3 = 4x + 23$ , subtract $4x$ from both sides to get $5x + 3 = 23$ . Then subtract 3 to get $5x = 20$ , so $x = 4$ .

Section 2

Clearing an equation of fractions

Property

Solve equations with fraction coefficients by clearing the fractions.

Step 1. Find the least common denominator of all the fractions in the equation.
Step 2. Multiply both sides of the equation by that LCD. This clears the fractions.
Step 3. Solve using the General Strategy for Solving Linear Equations.

Examples

To solve $\frac{1}{6}x + \frac{1}{3} = \frac{1}{2}$ , the LCD is 6. Multiplying by 6 gives $6(\frac{1}{6}x) + 6(\frac{1}{3}) = 6(\frac{1}{2})$ , which simplifies to $x + 2 = 3$ , so $x = 1$ .

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Solve equations with variables and constants on both sides

Property

It is a good idea to make the variable side the one in which the variable has the larger coefficient. This usually makes the arithmetic easier.

Examples

Given $9x + 3 = 4x + 23$ , subtract $4x$ from both sides to get $5x + 3 = 23$ . Then subtract 3 to get $5x = 20$ , so $x = 4$ .

Section 2

Clearing an equation of fractions

Property

Solve equations with fraction coefficients by clearing the fractions.

Examples

To solve $\frac{1}{6}x + \frac{1}{3} = \frac{1}{2}$ , the LCD is 6. Multiplying by 6 gives $6(\frac{1}{6}x) + 6(\frac{1}{3}) = 6(\frac{1}{2})$ , which simplifies to $x + 2 = 3$ , so $x = 1$ .

Overview

Lesson overview

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

Overview

Section 1

Solve equations with variables and constants on both sides

Property

It is a good idea to make the variable side the one in which the variable has the larger coefficient. This usually makes the arithmetic easier.

Examples

Given $9x + 3 = 4x + 23$ , subtract $4x$ from both sides to get $5x + 3 = 23$ . Then subtract 3 to get $5x = 20$ , so $x = 4$ .

Section 2

Clearing an equation of fractions

Property

Solve equations with fraction coefficients by clearing the fractions.

Examples

To solve $\frac{1}{6}x + \frac{1}{3} = \frac{1}{2}$ , the LCD is 6. Multiplying by 6 gives $6(\frac{1}{6}x) + 6(\frac{1}{3}) = 6(\frac{1}{2})$ , which simplifies to $x + 2 = 3$ , so $x = 1$ .