Learn on PengiOpenstax Elementary Algebra 2EChapter 2: Solving Linear Equations and Inequalities

Lesson 2.4: Use a General Strategy to Solve Linear Equations

In this lesson from OpenStax Elementary Algebra 2E, students learn a five-step general strategy for solving any linear equation, including applying the Distributive Property, combining like terms, and using the Addition, Subtraction, Multiplication, and Division Properties of Equality. The lesson covers equations with constants on both sides, variables on both sides, and combinations of both, as well as how to classify equations. Students practice checking their solutions by substituting back into the original equation to verify a true statement.

Section 1

πŸ“˜ Use a General Strategy to Solve Linear Equations

New Concept

This lesson introduces a powerful, multi-step strategy for solving any linear equation. You'll learn to simplify complex expressions, isolate variables, and classify equations as conditional, identities, or contradictions based on their solutions.

What’s next

Next, we'll break down this strategy step-by-step and apply it to various equations through a series of interactive examples and practice problems.

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 βˆ’(a+5)=15-(a+5) = 15, first distribute the negative sign to get βˆ’aβˆ’5=15-a - 5 = 15. Add 5 to both sides to get βˆ’a=20-a = 20. Finally, divide by βˆ’1-1 to find a=βˆ’20a = -20.
  • To solve 12(4xβˆ’6)=5βˆ’x\frac{1}{2}(4x-6) = 5-x, distribute 12\frac{1}{2} to get 2xβˆ’3=5βˆ’x2x - 3 = 5 - x. Add xx to both sides to get 3xβˆ’3=53x - 3 = 5. Add 3 to both sides to get 3x=83x=8. The solution is x=83x = \frac{8}{3}.

Section 3

Conditional equation

Property

An equation that is true for one or more values of the variable and false for all other values of the variable is a conditional equation.

Examples

  • The equation 3x+5=143x + 5 = 14 is a conditional equation. It is only true when x=3x = 3, because 3(3)+5=143(3) + 5 = 14.
  • Solve 5(yβˆ’2)=155(y - 2) = 15. Distributing gives 5yβˆ’10=155y - 10 = 15. Adding 10 to both sides gives 5y=255y = 25, so y=5y=5. The equation is true only for y=5y=5.

Section 4

Identity

Property

An equation that is true for any value of the variable is called an identity. The solution of an identity is all real numbers.

Examples

  • The equation 3(x+2)=3x+63(x+2) = 3x+6 is an identity. If we distribute on the left, we get 3x+6=3x+63x+6 = 3x+6. This simplifies to 6=66=6, which is always true. The solution is all real numbers.
  • Consider 7+4(mβˆ’1)=4m+37 + 4(m-1) = 4m + 3. Simplifying the left side gives 7+4mβˆ’4=4m+37+4m-4 = 4m+3, which becomes 4m+3=4m+34m+3 = 4m+3. This is an identity, and the solution is all real numbers.

Section 5

Contradiction

Property

An equation that is false for all values of the variable is called a contradiction. A contradiction has no solution.

Examples

  • The equation 3xβˆ’5=3x+23x - 5 = 3x + 2 is a contradiction. If we subtract 3x3x from both sides, we are left with the false statement βˆ’5=2-5 = 2. Therefore, there is no solution.
  • Solve 4(y+2)=4yβˆ’14(y+2) = 4y - 1. Distributing gives 4y+8=4yβˆ’14y+8 = 4y-1. Subtracting 4y4y from both sides results in 8=βˆ’18 = -1. This is false, so the equation has no solution.

Book overview

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Chapter 2: Solving Linear Equations and Inequalities

  1. Lesson 1

    Lesson 2.1: Solve Equations Using the Subtraction and Addition Properties of Equality

  2. Lesson 2

    Lesson 2.2: Solve Equations using the Division and Multiplication Properties of Equality

  3. Lesson 3

    Lesson 2.3: Solve Equations with Variables and Constants on Both Sides

  4. Lesson 4Current

    Lesson 2.4: Use a General Strategy to Solve Linear Equations

  5. Lesson 5

    Lesson 2.5: Solve Equations with Fractions or Decimals

  6. Lesson 6

    Lesson 2.6: Solve a Formula for a Specific Variable

  7. Lesson 7

    Lesson 2.7: Solve Linear Inequalities

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

πŸ“˜ Use a General Strategy to Solve Linear Equations

New Concept

This lesson introduces a powerful, multi-step strategy for solving any linear equation. You'll learn to simplify complex expressions, isolate variables, and classify equations as conditional, identities, or contradictions based on their solutions.

What’s next

Next, we'll break down this strategy step-by-step and apply it to various equations through a series of interactive examples and practice problems.

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 βˆ’(a+5)=15-(a+5) = 15, first distribute the negative sign to get βˆ’aβˆ’5=15-a - 5 = 15. Add 5 to both sides to get βˆ’a=20-a = 20. Finally, divide by βˆ’1-1 to find a=βˆ’20a = -20.
  • To solve 12(4xβˆ’6)=5βˆ’x\frac{1}{2}(4x-6) = 5-x, distribute 12\frac{1}{2} to get 2xβˆ’3=5βˆ’x2x - 3 = 5 - x. Add xx to both sides to get 3xβˆ’3=53x - 3 = 5. Add 3 to both sides to get 3x=83x=8. The solution is x=83x = \frac{8}{3}.

Section 3

Conditional equation

Property

An equation that is true for one or more values of the variable and false for all other values of the variable is a conditional equation.

Examples

  • The equation 3x+5=143x + 5 = 14 is a conditional equation. It is only true when x=3x = 3, because 3(3)+5=143(3) + 5 = 14.
  • Solve 5(yβˆ’2)=155(y - 2) = 15. Distributing gives 5yβˆ’10=155y - 10 = 15. Adding 10 to both sides gives 5y=255y = 25, so y=5y=5. The equation is true only for y=5y=5.

Section 4

Identity

Property

An equation that is true for any value of the variable is called an identity. The solution of an identity is all real numbers.

Examples

  • The equation 3(x+2)=3x+63(x+2) = 3x+6 is an identity. If we distribute on the left, we get 3x+6=3x+63x+6 = 3x+6. This simplifies to 6=66=6, which is always true. The solution is all real numbers.
  • Consider 7+4(mβˆ’1)=4m+37 + 4(m-1) = 4m + 3. Simplifying the left side gives 7+4mβˆ’4=4m+37+4m-4 = 4m+3, which becomes 4m+3=4m+34m+3 = 4m+3. This is an identity, and the solution is all real numbers.

Section 5

Contradiction

Property

An equation that is false for all values of the variable is called a contradiction. A contradiction has no solution.

Examples

  • The equation 3xβˆ’5=3x+23x - 5 = 3x + 2 is a contradiction. If we subtract 3x3x from both sides, we are left with the false statement βˆ’5=2-5 = 2. Therefore, there is no solution.
  • Solve 4(y+2)=4yβˆ’14(y+2) = 4y - 1. Distributing gives 4y+8=4yβˆ’14y+8 = 4y-1. Subtracting 4y4y from both sides results in 8=βˆ’18 = -1. This is false, so the equation has no solution.

Book overview

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

Continue this chapter

Chapter 2: Solving Linear Equations and Inequalities

  1. Lesson 1

    Lesson 2.1: Solve Equations Using the Subtraction and Addition Properties of Equality

  2. Lesson 2

    Lesson 2.2: Solve Equations using the Division and Multiplication Properties of Equality

  3. Lesson 3

    Lesson 2.3: Solve Equations with Variables and Constants on Both Sides

  4. Lesson 4Current

    Lesson 2.4: Use a General Strategy to Solve Linear Equations

  5. Lesson 5

    Lesson 2.5: Solve Equations with Fractions or Decimals

  6. Lesson 6

    Lesson 2.6: Solve a Formula for a Specific Variable

  7. Lesson 7

    Lesson 2.7: Solve Linear Inequalities