Learn on PengiAoPS: Introduction to Algebra (AMC 8 & 10)Chapter 3: One-Variable Linear Equations

Lesson 2: Solving Linear Equations II

In this Grade 4 lesson from AoPS: Introduction to Algebra, students learn to solve multi-step one-variable linear equations by combining like terms, moving variable terms to one side, and isolating the variable using inverse operations. The lesson covers techniques such as eliminating fractions by multiplying by the least common denominator and identifying equations with no solution or infinitely many solutions. Part of Chapter 3 in the AMC 8 and 10 curriculum, it also introduces translating word problems into algebraic equations and verifying solutions by substitution.

Section 1

Variables and constants on both sides

Property

When an equation has variables and constants on both sides, it may take several steps to solve. We need a clear and organized strategy. The strategy is to first collect the variable terms to one side of the equation, and then collect the constant terms to the other side.

Examples

Solve $8x + 4 = 7x + 9$ . First, subtract $7x$ from both sides to get $x + 4 = 9$ . Then, subtract 4 from both sides to find $x = 5$ .

Solve $10n - 5 = -3n + 21$ . First, add $3n$ to both sides to get $13n - 5 = 21$ . Then, add 5 to both sides to get $13n = 26$ . Finally, divide by 13 to find $n = 2$ .

Section 2

General strategy for solving linear equations

Property

Step 1. Simplify each side of the equation as much as possible.
Use the Distributive Property to remove any parentheses.
Combine like terms.
Step 2. Collect all the variable terms on one side of the equation.
Use the Addition or Subtraction Property of Equality.
Step 3. Collect all the constant terms on the other side of the equation.
Use the Addition or Subtraction Property of Equality.
Step 4. Make the coefficient of the variable term to equal to 1.
Use the Multiplication or Division Property of Equality.
State the solution to the equation.
Step 5. Check the solution.
Substitute the solution into the original equation to make sure the result is a true statement.

Examples

To solve $-(a+5) = 15$ , first distribute the negative sign to get $-a - 5 = 15$ . Add 5 to both sides to get $-a = 20$ . Finally, divide by $-1$ to find $a = -20$ .

To solve $\frac{1}{2}(4x-6) = 5-x$ , distribute $\frac{1}{2}$ to get $2x - 3 = 5 - x$ . Add $x$ to both sides to get $3x - 3 = 5$ . Add 3 to both sides to get $3x=8$ . The solution is $x = \frac{8}{3}$ .

Section 3

Classifying Equations

Property

Equations can be classified into three types based on their solutions:

A conditional equation is true for one or more specific values of the variable.
An identity is an equation that is true for all real numbers. Solving results in a true statement, such as $3=3$ .
A contradiction is an equation that has no solution. Solving results in a false statement, such as $5=-2$ .

Examples

Conditional: $4x-1=7$ . Solving this gives $4x=8$ , so $x=2$ . The equation is true only for this value.
Identity: $2(x+3) = 2x+6$ . Distributing gives $2x+6=2x+6$ . Subtracting $2x$ from both sides results in $6=6$ . This is an identity, and the solution is all real numbers.
Contradiction: $3y = 3(y-2)$ . Distributing gives $3y = 3y-6$ . Subtracting $3y$ from both sides results in $0=-6$ . This is a contradiction and has no solution.

Explanation

Not all equations have just one answer. If solving leads to a specific value like $x=2$ , it's conditional. If the variables cancel and you get a true statement like $5=5$ , it's an identity. If you get a false statement, it's a contradiction.

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Variables and constants on both sides

Property

Examples

Solve $8x + 4 = 7x + 9$ . First, subtract $7x$ from both sides to get $x + 4 = 9$ . Then, subtract 4 from both sides to find $x = 5$ .

Solve $10n - 5 = -3n + 21$ . First, add $3n$ to both sides to get $13n - 5 = 21$ . Then, add 5 to both sides to get $13n = 26$ . Finally, divide by 13 to find $n = 2$ .

Section 2

General strategy for solving linear equations

Property

Examples

To solve $-(a+5) = 15$ , first distribute the negative sign to get $-a - 5 = 15$ . Add 5 to both sides to get $-a = 20$ . Finally, divide by $-1$ to find $a = -20$ .

To solve $\frac{1}{2}(4x-6) = 5-x$ , distribute $\frac{1}{2}$ to get $2x - 3 = 5 - x$ . Add $x$ to both sides to get $3x - 3 = 5$ . Add 3 to both sides to get $3x=8$ . The solution is $x = \frac{8}{3}$ .

Section 3

Classifying Equations

Property

Equations can be classified into three types based on their solutions:

A conditional equation is true for one or more specific values of the variable.
An identity is an equation that is true for all real numbers. Solving results in a true statement, such as $3=3$ .
A contradiction is an equation that has no solution. Solving results in a false statement, such as $5=-2$ .

Examples

Conditional: $4x-1=7$ . Solving this gives $4x=8$ , so $x=2$ . The equation is true only for this value.
Identity: $2(x+3) = 2x+6$ . Distributing gives $2x+6=2x+6$ . Subtracting $2x$ from both sides results in $6=6$ . This is an identity, and the solution is all real numbers.
Contradiction: $3y = 3(y-2)$ . Distributing gives $3y = 3y-6$ . Subtracting $3y$ from both sides results in $0=-6$ . This is a contradiction and has no solution.

Explanation

Overview

Lesson overview

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

Overview

Section 1

Variables and constants on both sides

Property

Examples

Solve $8x + 4 = 7x + 9$ . First, subtract $7x$ from both sides to get $x + 4 = 9$ . Then, subtract 4 from both sides to find $x = 5$ .

Solve $10n - 5 = -3n + 21$ . First, add $3n$ to both sides to get $13n - 5 = 21$ . Then, add 5 to both sides to get $13n = 26$ . Finally, divide by 13 to find $n = 2$ .

Section 2

General strategy for solving linear equations

Property

Examples

To solve $-(a+5) = 15$ , first distribute the negative sign to get $-a - 5 = 15$ . Add 5 to both sides to get $-a = 20$ . Finally, divide by $-1$ to find $a = -20$ .

To solve $\frac{1}{2}(4x-6) = 5-x$ , distribute $\frac{1}{2}$ to get $2x - 3 = 5 - x$ . Add $x$ to both sides to get $3x - 3 = 5$ . Add 3 to both sides to get $3x=8$ . The solution is $x = \frac{8}{3}$ .

Section 3

Classifying Equations

Property

Equations can be classified into three types based on their solutions:

A conditional equation is true for one or more specific values of the variable.
An identity is an equation that is true for all real numbers. Solving results in a true statement, such as $3=3$ .
A contradiction is an equation that has no solution. Solving results in a false statement, such as $5=-2$ .

Examples

Conditional: $4x-1=7$ . Solving this gives $4x=8$ , so $x=2$ . The equation is true only for this value.
Identity: $2(x+3) = 2x+6$ . Distributing gives $2x+6=2x+6$ . Subtracting $2x$ from both sides results in $6=6$ . This is an identity, and the solution is all real numbers.
Contradiction: $3y = 3(y-2)$ . Distributing gives $3y = 3y-6$ . Subtracting $3y$ from both sides results in $0=-6$ . This is a contradiction and has no solution.