Learn on PengiOpenstax Prealgebre 2EChapter 8: Solving Linear Equations

Lesson 1: Solve Equations Using the Subtraction and Addition Properties of Equality

In this OpenStax Prealgebra 2E lesson from Chapter 8, students learn to solve linear equations by applying the Subtraction and Addition Properties of Equality to isolate variables and find solutions. The lesson covers how to verify solutions by substitution, simplify equations before solving, and translate real-world word problems into algebraic equations. These foundational skills prepare students for more advanced equation-solving techniques introduced later in the chapter.

Section 1

πŸ“˜ Solve Equations Using the Subtraction and Addition Properties of Equality

New Concept

To solve an equation, you must keep it balanced. This lesson introduces the Subtraction and Addition Properties of Equalityβ€”your tools for adding or subtracting from both sides to isolate the variable and find its value.

What’s next

Next, you'll see these properties in action with worked examples, then apply them yourself in a series of interactive practice problems.

Section 2

Solution of an equation

Property

A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.

Examples

  • To check if x=9x=9 is a solution for xβˆ’2=7x-2=7, substitute 9 for xx. This gives 9βˆ’2=79-2=7, which simplifies to 7=77=7. Since this is a true statement, x=9x=9 is a solution.
  • Is y=10y=10 a solution for yβˆ’4=6y-4=6? Substitute 10 for yy to get 10βˆ’4=610-4=6. This simplifies to 6=66=6, which is true, so y=10y=10 is a solution.

Section 3

Subtraction and Addition Properties of Equality

Property

Subtraction Property of Equality
For all real numbers aa, bb, and cc, if a=ba = b, then aβˆ’c=bβˆ’ca - c = b - c.

Addition Property of Equality
For all real numbers aa, bb, and cc, if a=ba = b, then a+c=b+ca + c = b + c.
When you add or subtract the same quantity from both sides of an equation, you still have equality.

Examples

  • To solve pβˆ’7=12p - 7 = 12, we use the Addition Property. Add 7 to both sides: pβˆ’7+7=12+7p - 7 + 7 = 12 + 7, which simplifies to p=19p = 19.

Section 4

Translate word sentences to equations

Property

Step 1. Locate the 'equals' word(s). Translate to an equal sign.
Step 2. Translate the words to the left of the 'equals' word(s) into an algebraic expression.
Step 3. Translate the words to the right of the 'equals' word(s) into an algebraic expression.

Examples

  • 'Five more than xx is equal to 26' translates to the equation x+5=26x + 5 = 26. The words 'is equal to' become the == sign.
  • 'The difference of 5p5p and 4p4p is 23' translates to 5pβˆ’4p=235p - 4p = 23. The word 'is' becomes the == sign.

Section 5

Problem-solving strategy

Property

Step 1. Read the problem. Make sure you understand all the words and ideas.
Step 2. Identify what you are looking for.
Step 3. Name what you are looking for. Choose a variable to represent that quantity.
Step 4. Translate into an equation.
Step 5. Solve the equation using good algebra techniques.
Step 6. Check the answer in the problem and make sure it makes sense.
Step 7. Answer the question with a complete sentence.

Examples

  • Problem: A number increased by 9 is 21. Find the number. Identify 'the number' as nn. Translate: n+9=21n + 9 = 21. Solve: n=12n = 12. The number is 12.
  • Problem: A shirt costs 25 dollars after a 10 dollar discount. What was the original price? Identify 'original price' as pp. Translate: pβˆ’10=25p - 10 = 25. Solve: p=35p = 35. The original price was 35 dollars.

Book overview

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Chapter 8: Solving Linear Equations

  1. Lesson 1Current

    Lesson 1: Solve Equations Using the Subtraction and Addition Properties of Equality

  2. Lesson 2

    Lesson 2: Solve Equations Using the Division and Multiplication Properties of Equality

  3. Lesson 3

    Lesson 3: Solve Equations with Variables and Constants on Both Sides

  4. Lesson 4

    Lesson 4: Solve Equations with Fraction or Decimal Coefficients

Lesson overview

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Section 1

πŸ“˜ Solve Equations Using the Subtraction and Addition Properties of Equality

New Concept

To solve an equation, you must keep it balanced. This lesson introduces the Subtraction and Addition Properties of Equalityβ€”your tools for adding or subtracting from both sides to isolate the variable and find its value.

What’s next

Next, you'll see these properties in action with worked examples, then apply them yourself in a series of interactive practice problems.

Section 2

Solution of an equation

Property

A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.

Examples

  • To check if x=9x=9 is a solution for xβˆ’2=7x-2=7, substitute 9 for xx. This gives 9βˆ’2=79-2=7, which simplifies to 7=77=7. Since this is a true statement, x=9x=9 is a solution.
  • Is y=10y=10 a solution for yβˆ’4=6y-4=6? Substitute 10 for yy to get 10βˆ’4=610-4=6. This simplifies to 6=66=6, which is true, so y=10y=10 is a solution.

Section 3

Subtraction and Addition Properties of Equality

Property

Subtraction Property of Equality
For all real numbers aa, bb, and cc, if a=ba = b, then aβˆ’c=bβˆ’ca - c = b - c.

Addition Property of Equality
For all real numbers aa, bb, and cc, if a=ba = b, then a+c=b+ca + c = b + c.
When you add or subtract the same quantity from both sides of an equation, you still have equality.

Examples

  • To solve pβˆ’7=12p - 7 = 12, we use the Addition Property. Add 7 to both sides: pβˆ’7+7=12+7p - 7 + 7 = 12 + 7, which simplifies to p=19p = 19.

Section 4

Translate word sentences to equations

Property

Step 1. Locate the 'equals' word(s). Translate to an equal sign.
Step 2. Translate the words to the left of the 'equals' word(s) into an algebraic expression.
Step 3. Translate the words to the right of the 'equals' word(s) into an algebraic expression.

Examples

  • 'Five more than xx is equal to 26' translates to the equation x+5=26x + 5 = 26. The words 'is equal to' become the == sign.
  • 'The difference of 5p5p and 4p4p is 23' translates to 5pβˆ’4p=235p - 4p = 23. The word 'is' becomes the == sign.

Section 5

Problem-solving strategy

Property

Step 1. Read the problem. Make sure you understand all the words and ideas.
Step 2. Identify what you are looking for.
Step 3. Name what you are looking for. Choose a variable to represent that quantity.
Step 4. Translate into an equation.
Step 5. Solve the equation using good algebra techniques.
Step 6. Check the answer in the problem and make sure it makes sense.
Step 7. Answer the question with a complete sentence.

Examples

  • Problem: A number increased by 9 is 21. Find the number. Identify 'the number' as nn. Translate: n+9=21n + 9 = 21. Solve: n=12n = 12. The number is 12.
  • Problem: A shirt costs 25 dollars after a 10 dollar discount. What was the original price? Identify 'original price' as pp. Translate: pβˆ’10=25p - 10 = 25. Solve: p=35p = 35. The original price was 35 dollars.

Book overview

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

Continue this chapter

Chapter 8: Solving Linear Equations

  1. Lesson 1Current

    Lesson 1: Solve Equations Using the Subtraction and Addition Properties of Equality

  2. Lesson 2

    Lesson 2: Solve Equations Using the Division and Multiplication Properties of Equality

  3. Lesson 3

    Lesson 3: Solve Equations with Variables and Constants on Both Sides

  4. Lesson 4

    Lesson 4: Solve Equations with Fraction or Decimal Coefficients