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

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

In this lesson from OpenStax Elementary Algebra 2E, students learn how to verify solutions of equations and apply the Subtraction and Addition Properties of Equality to solve one-variable linear equations. The lesson covers substituting values to check whether a number is a solution, isolating the variable by adding or subtracting the same quantity on both sides of an equation, and translating real-world problems into equations to solve. It also includes solving equations that require simplification before applying these properties.

Section 1

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

New Concept

This lesson introduces the core principle of balancing equations. You'll use the Addition and Subtraction Properties of Equality to isolate a variable and find its value, a foundational skill for solving more complex algebraic problems.

What’s next

Soon, you'll walk through interactive examples and solve practice problems using these properties to find the value of the unknown variable.

Section 2

Verify a 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.

To determine whether a number is a solution to an equation.
Step 1. Substitute the number in for the variable in the equation.
Step 2. Simplify the expressions on both sides of the equation.
Step 3. Determine whether the resulting equation is true. If it is true, the number is a solution. If it is not true, the number is not a solution.

Examples

  • Is y=4y=4 a solution to 5yβˆ’3=175y - 3 = 17? Substitute y=4y=4: 5(4)βˆ’3=20βˆ’3=175(4) - 3 = 20 - 3 = 17. Since 17=1717=17, yes, it is a solution.
  • Is x=βˆ’3x=-3 a solution to 2x+8=x+42x + 8 = x+4? Substitute x=βˆ’3x=-3: 2(βˆ’3)+8=22(-3) + 8 = 2 and βˆ’3+4=1-3+4=1. Since 2β‰ 12 \neq 1, it is not a solution.
  • Is a=12a = \frac{1}{2} a solution to 8aβˆ’1=38a - 1 = 3? Substitute a=12a=\frac{1}{2}: 8(12)βˆ’1=4βˆ’1=38(\frac{1}{2}) - 1 = 4 - 1 = 3. Since 3=33=3, yes, it is a solution.

Section 3

Subtraction Property of Equality

Property

For any numbers aa, bb, and cc,

Ifa=b,thenaβˆ’c=bβˆ’c\begin{array}{ll} \text{If} & a = b, \\ \text{then} & a - c = b - c \end{array}

When you subtract the same quantity from both sides of an equation, you still have equality.

Examples

  • To solve x+10=25x + 10 = 25, subtract 10 from both sides: x+10βˆ’10=25βˆ’10x + 10 - 10 = 25 - 10, which simplifies to x=15x = 15.
  • To solve y+4.2=8.7y + 4.2 = 8.7, subtract 4.2 from both sides: y+4.2βˆ’4.2=8.7βˆ’4.2y + 4.2 - 4.2 = 8.7 - 4.2, which simplifies to y=4.5y = 4.5.
  • To solve z+18=5z + 18 = 5, subtract 18 from both sides: z+18βˆ’18=5βˆ’18z + 18 - 18 = 5 - 18, which simplifies to z=βˆ’13z = -13.

Explanation

This property is like removing the same weight from both sides of a balanced scale. The scale stays balanced. We use this to 'undo' addition and isolate the variable to find its value.

Section 4

Addition Property of Equality

Property

For any numbers aa, bb, and cc,

Ifa=b,thena+c=b+c\begin{array}{ll} \text{If} & a = b, \\ \text{then} & a + c = b + c \end{array}

When you add the same quantity to both sides of an equation, you still have equality.

Examples

  • To solve pβˆ’8=14p - 8 = 14, add 8 to both sides: pβˆ’8+8=14+8p - 8 + 8 = 14 + 8, which simplifies to p=22p = 22.
  • To solve qβˆ’15=35q - \frac{1}{5} = \frac{3}{5}, add 15\frac{1}{5} to both sides: qβˆ’15+15=35+15q - \frac{1}{5} + \frac{1}{5} = \frac{3}{5} + \frac{1}{5}, which simplifies to q=45q = \frac{4}{5}.
  • To solve rβˆ’5.5=βˆ’10r - 5.5 = -10, add 5.5 to both sides: rβˆ’5.5+5.5=βˆ’10+5.5r - 5.5 + 5.5 = -10 + 5.5, which simplifies to r=βˆ’4.5r = -4.5.

Explanation

This property helps you 'undo' subtraction in an equation. By adding the same number to both sides, you keep the equation balanced and get the variable all by itself.

Section 5

Equations requiring simplification

Property

Usually, we will need to simplify one or both sides of an equation before using the Subtraction or Addition Properties of Equality. You should always simplify as much as possible before you try to isolate the variable. Simplify one side of the equation at a time.

Examples

  • Solve 10xβˆ’3βˆ’9x=1210x - 3 - 9x = 12. First, combine like terms on the left: (10xβˆ’9x)βˆ’3=12(10x - 9x) - 3 = 12, which gives xβˆ’3=12x - 3 = 12. Then add 3 to both sides to get x=15x = 15.
  • Solve 4(z+2)βˆ’3z=114(z + 2) - 3z = 11. First, distribute the 4: 4z+8βˆ’3z=114z + 8 - 3z = 11. Combine like terms: z+8=11z + 8 = 11. Then subtract 8 from both sides to get z=3z = 3.
  • Solve 2y+5+3y=4y+92y + 5 + 3y = 4y + 9. Combine terms on the left: 5y+5=4y+95y + 5 = 4y + 9. Subtract 4y4y from both sides: y+5=9y + 5 = 9. Then subtract 5 to get y=4y = 4.

Explanation

It's like tidying up before solving a puzzle. Combine all like terms on the left side, then do the same on the right. This makes the equation much simpler and easier to solve correctly.

Section 6

Translate to an equation and solve

Property

To translate an English sentence to an algebraic equation:
Step 1. Locate the β€œequals” word(s). Translate to an equals sign (=).
Step 2. Translate the words to the left of the β€œequals” word(s) into an algebraic expression.
Step 3. Translate the words to the right of the β€œequals” word(s) into an algebraic expression.

Examples

  • Translate and solve: "Twenty more than a number nn is 35." This translates to n+20=35n + 20 = 35. Subtracting 20 from both sides gives n=15n = 15.
  • Translate and solve: "The difference of 5y5y and 4y4y is -8." This translates to 5yβˆ’4y=βˆ’85y - 4y = -8. Simplifying the left side gives y=βˆ’8y = -8.
  • Translate and solve: "A number kk decreased by 6 equals 11." This translates to kβˆ’6=11k - 6 = 11. Adding 6 to both sides gives k=17k = 17.

Explanation

This skill is like being a language translator, but for math! You look for clue words in a sentence to convert it into an equation. Words like 'is', 'gives', or 'results in' usually mean 'equals'.

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 1Current

    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 4

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

New Concept

This lesson introduces the core principle of balancing equations. You'll use the Addition and Subtraction Properties of Equality to isolate a variable and find its value, a foundational skill for solving more complex algebraic problems.

What’s next

Soon, you'll walk through interactive examples and solve practice problems using these properties to find the value of the unknown variable.

Section 2

Verify a 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.

To determine whether a number is a solution to an equation.
Step 1. Substitute the number in for the variable in the equation.
Step 2. Simplify the expressions on both sides of the equation.
Step 3. Determine whether the resulting equation is true. If it is true, the number is a solution. If it is not true, the number is not a solution.

Examples

  • Is y=4y=4 a solution to 5yβˆ’3=175y - 3 = 17? Substitute y=4y=4: 5(4)βˆ’3=20βˆ’3=175(4) - 3 = 20 - 3 = 17. Since 17=1717=17, yes, it is a solution.
  • Is x=βˆ’3x=-3 a solution to 2x+8=x+42x + 8 = x+4? Substitute x=βˆ’3x=-3: 2(βˆ’3)+8=22(-3) + 8 = 2 and βˆ’3+4=1-3+4=1. Since 2β‰ 12 \neq 1, it is not a solution.
  • Is a=12a = \frac{1}{2} a solution to 8aβˆ’1=38a - 1 = 3? Substitute a=12a=\frac{1}{2}: 8(12)βˆ’1=4βˆ’1=38(\frac{1}{2}) - 1 = 4 - 1 = 3. Since 3=33=3, yes, it is a solution.

Section 3

Subtraction Property of Equality

Property

For any numbers aa, bb, and cc,

Ifa=b,thenaβˆ’c=bβˆ’c\begin{array}{ll} \text{If} & a = b, \\ \text{then} & a - c = b - c \end{array}

When you subtract the same quantity from both sides of an equation, you still have equality.

Examples

  • To solve x+10=25x + 10 = 25, subtract 10 from both sides: x+10βˆ’10=25βˆ’10x + 10 - 10 = 25 - 10, which simplifies to x=15x = 15.
  • To solve y+4.2=8.7y + 4.2 = 8.7, subtract 4.2 from both sides: y+4.2βˆ’4.2=8.7βˆ’4.2y + 4.2 - 4.2 = 8.7 - 4.2, which simplifies to y=4.5y = 4.5.
  • To solve z+18=5z + 18 = 5, subtract 18 from both sides: z+18βˆ’18=5βˆ’18z + 18 - 18 = 5 - 18, which simplifies to z=βˆ’13z = -13.

Explanation

This property is like removing the same weight from both sides of a balanced scale. The scale stays balanced. We use this to 'undo' addition and isolate the variable to find its value.

Section 4

Addition Property of Equality

Property

For any numbers aa, bb, and cc,

Ifa=b,thena+c=b+c\begin{array}{ll} \text{If} & a = b, \\ \text{then} & a + c = b + c \end{array}

When you add the same quantity to both sides of an equation, you still have equality.

Examples

  • To solve pβˆ’8=14p - 8 = 14, add 8 to both sides: pβˆ’8+8=14+8p - 8 + 8 = 14 + 8, which simplifies to p=22p = 22.
  • To solve qβˆ’15=35q - \frac{1}{5} = \frac{3}{5}, add 15\frac{1}{5} to both sides: qβˆ’15+15=35+15q - \frac{1}{5} + \frac{1}{5} = \frac{3}{5} + \frac{1}{5}, which simplifies to q=45q = \frac{4}{5}.
  • To solve rβˆ’5.5=βˆ’10r - 5.5 = -10, add 5.5 to both sides: rβˆ’5.5+5.5=βˆ’10+5.5r - 5.5 + 5.5 = -10 + 5.5, which simplifies to r=βˆ’4.5r = -4.5.

Explanation

This property helps you 'undo' subtraction in an equation. By adding the same number to both sides, you keep the equation balanced and get the variable all by itself.

Section 5

Equations requiring simplification

Property

Usually, we will need to simplify one or both sides of an equation before using the Subtraction or Addition Properties of Equality. You should always simplify as much as possible before you try to isolate the variable. Simplify one side of the equation at a time.

Examples

  • Solve 10xβˆ’3βˆ’9x=1210x - 3 - 9x = 12. First, combine like terms on the left: (10xβˆ’9x)βˆ’3=12(10x - 9x) - 3 = 12, which gives xβˆ’3=12x - 3 = 12. Then add 3 to both sides to get x=15x = 15.
  • Solve 4(z+2)βˆ’3z=114(z + 2) - 3z = 11. First, distribute the 4: 4z+8βˆ’3z=114z + 8 - 3z = 11. Combine like terms: z+8=11z + 8 = 11. Then subtract 8 from both sides to get z=3z = 3.
  • Solve 2y+5+3y=4y+92y + 5 + 3y = 4y + 9. Combine terms on the left: 5y+5=4y+95y + 5 = 4y + 9. Subtract 4y4y from both sides: y+5=9y + 5 = 9. Then subtract 5 to get y=4y = 4.

Explanation

It's like tidying up before solving a puzzle. Combine all like terms on the left side, then do the same on the right. This makes the equation much simpler and easier to solve correctly.

Section 6

Translate to an equation and solve

Property

To translate an English sentence to an algebraic equation:
Step 1. Locate the β€œequals” word(s). Translate to an equals sign (=).
Step 2. Translate the words to the left of the β€œequals” word(s) into an algebraic expression.
Step 3. Translate the words to the right of the β€œequals” word(s) into an algebraic expression.

Examples

  • Translate and solve: "Twenty more than a number nn is 35." This translates to n+20=35n + 20 = 35. Subtracting 20 from both sides gives n=15n = 15.
  • Translate and solve: "The difference of 5y5y and 4y4y is -8." This translates to 5yβˆ’4y=βˆ’85y - 4y = -8. Simplifying the left side gives y=βˆ’8y = -8.
  • Translate and solve: "A number kk decreased by 6 equals 11." This translates to kβˆ’6=11k - 6 = 11. Adding 6 to both sides gives k=17k = 17.

Explanation

This skill is like being a language translator, but for math! You look for clue words in a sentence to convert it into an equation. Words like 'is', 'gives', or 'results in' usually mean 'equals'.

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 1Current

    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 4

    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