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🔟Elementary Algebra Unit 2 Review

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2.1 Solve Equations Using the Subtraction and Addition Properties of Equality

2.1 Solve Equations Using the Subtraction and Addition Properties of Equality

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
🔟Elementary Algebra
Unit & Topic Study Guides

Solving Equations Using the Subtraction and Addition Properties of Equality

Solving equations means finding the value of a variable that makes a statement true. The addition and subtraction properties of equality are your first tools for doing this: they let you add or subtract the same number from both sides of an equation to isolate the variable. Once you're comfortable with these properties, every other equation-solving technique builds on the same logic.

Verification of Linear Equation Solutions

A solution to an equation is any value that, when plugged in for the variable, makes both sides equal. Before you trust an answer, you should always verify it by substituting it back into the original equation.

Steps to verify a solution:

  1. Take the value you think is the solution and substitute it for the variable in the original equation.
  2. Simplify the left side by performing all operations.
  3. Simplify the right side by performing all operations.
  4. Compare the two sides.
    • If they're equal, the value is a solution.
    • If they're not equal, it's not a solution.

Example: Is x=3x = 3 a solution to 2x+1=72x + 1 = 7?

  • Left side: 2(3)+1=6+1=72(3) + 1 = 6 + 1 = 7
  • Right side: 77
  • 7=77 = 7 ✓, so yes, x=3x = 3 is a solution.

Example: Is x=4x = 4 a solution to 2x+1=72x + 1 = 7?

  • Left side: 2(4)+1=8+1=92(4) + 1 = 8 + 1 = 9
  • Right side: 77
  • 979 \neq 7, so no, x=4x = 4 is not a solution.

Properties for Equation Solving

Think of an equation like a balanced scale. Whatever you do to one side, you must do to the other side to keep it balanced. That's the core idea behind both properties.

Addition Property of Equality: If a=ba = b, then a+c=b+ca + c = b + c. You can add the same value to both sides without breaking the equality.

Subtraction Property of Equality: If a=ba = b, then ac=bca - c = b - c. You can subtract the same value from both sides without breaking the equality.

Your goal is to isolate the variable on one side of the equation. You do this by "undoing" whatever operation is attached to the variable.

Example using subtraction: Solve x+3=7x + 3 = 7

  1. The variable xx has 3 added to it, so subtract 3 from both sides: x+33=73x + 3 - 3 = 7 - 3

  2. Simplify: x=4x = 4

Example using addition: Solve x5=2x - 5 = 2

  1. The variable xx has 5 subtracted from it, so add 5 to both sides: x5+5=2+5x - 5 + 5 = 2 + 5

  2. Simplify: x=7x = 7

The value left on the side without the variable is your solution. You can always verify by plugging it back into the original equation.

Verification of linear equation solutions, OpenAlgebra.com: Free Algebra Study Guide & Video Tutorials: Solving Linear Equations: Part II

Equations with Simplification Steps

Sometimes an equation needs to be simplified before you can isolate the variable. This means combining like terms or distributing first.

Steps:

  1. Simplify each side of the equation separately by combining like terms.
  2. If there are parentheses, use the Distributive Property to remove them: a(b+c)=ab+aca(b + c) = ab + ac.
  3. Once each side is simplified, use the addition or subtraction property to isolate the variable.
  4. Simplify to find the solution.

Example: Solve 3x+2x4=113x + 2x - 4 = 11

  1. Combine like terms on the left: 5x4=115x - 4 = 11

  2. Add 4 to both sides: 5x=155x = 15

  3. (Dividing both sides by 5 uses a different property you'll learn soon): x=3x = 3

Example with distribution: Solve 2(x+3)=102(x + 3) = 10

  1. Distribute the 2: 2x+6=102x + 6 = 10
  2. Subtract 6 from both sides: 2x=42x = 4
  3. Divide both sides by 2: x=2x = 2

Note: Division is technically a separate property of equality. For now, the key skill is recognizing when to add or subtract to move constants away from the variable term.

Word Problems to Equations

Translating words into algebra is a skill that takes practice. The trick is identifying what's unknown, assigning it a variable, and recognizing which operation the words describe.

Common keyword translations:

WordsOperationExample
"sum," "more than," "increased by"Additionx+5=12x + 5 = 12
"difference," "less than," "decreased by"Subtractionx3=7x - 3 = 7
"product," "times," "multiplied by"Multiplication2x=182x = 18
"quotient," "divided by," "ratio"Divisionx4=6\frac{x}{4} = 6

Steps:

  1. Read the problem and identify the unknown quantity. Assign a variable to it (e.g., let xx = the unknown number).
  2. Use the keywords to translate the sentence into an equation.
  3. Solve the equation using the appropriate property of equality.

Example: "5 more than an unknown number is 12."

  • Let xx = the unknown number.
  • "5 more than" means add 5: x+5=12x + 5 = 12
  • Subtract 5 from both sides: x=7x = 7

Watch out for "less than" phrasing. "3 less than a number" translates to x3x - 3, not 3x3 - x. The number comes first, then you subtract.

Verification of linear equation solutions, OpenAlgebra.com: Free Algebra Study Guide & Video Tutorials: Solving Linear Equations: Part I

Real-World Applications of Linear Equations

Real-world problems follow the same process as word problems, but you also need to check that your answer makes sense in context.

Steps:

  1. Read the problem carefully. Identify what you know and what you need to find.
  2. Assign a variable to the unknown quantity.
  3. Write an equation using the given information.
  4. Solve the equation.
  5. Interpret the solution in context and check that it's reasonable.

Example: You earn $15 per hour plus a $50 bonus. Your total pay was $290. How many hours did you work?

  • Let xx = number of hours worked.
  • Equation: 15x+50=29015x + 50 = 290
  • Subtract 50 from both sides: 15x=24015x = 240
  • Divide both sides by 15: x=16x = 16
  • Interpretation: You worked 16 hours. That's reasonable (not negative, not impossibly large).

Always do that final reasonableness check. If you got a negative number for hours worked or a fractional number of people, something went wrong.

Fundamental Concepts in Equation Solving

These terms come up repeatedly, so make sure you're comfortable with them:

  • Algebraic expression: A combination of variables, numbers, and operations with no equals sign (e.g., 2x+32x + 3). An expression is not an equation.
  • Equation: A statement that two expressions are equal (e.g., 2x+3=72x + 3 = 7). It always has an equals sign.
  • Solving an equation: Finding the value(s) of the variable that make the equation true.
  • Isolating the variable: Rearranging the equation so the variable is alone on one side. This is the central strategy for every equation you'll solve in this course.
  • Number sense: Your intuition about whether an answer is reasonable. Use it to catch mistakes before they cost you points.