1.4Variables on Both Sides
When variable terms appear on both sides of an equation, collect them on one side first — then solve as usual. Master this skill and you can handle any linear equation.
Why This Matters
When variables appear on both sides, you're modeling situations where two quantities are changing simultaneously — like comparing two pricing plans or two moving objects. This skill is essential in Algebra 2 and Physics.
Workbook
Lesson, vocabulary, worked examples, and practice problems.
Essential Question
When a variable appears on both sides of an equation, what is the first move — and why does it work?
Lesson Overview
Some equations have variable terms on both sides of the equals sign, such as 5x + 3 = 2x + 12. You cannot isolate the variable until all variable terms are on the same side. The strategy is to add or subtract a variable term from both sides to eliminate it from one side — then the equation reduces to a familiar multi-step form. Choose to move the smaller variable term so the coefficient stays positive. After collecting variables on one side and constants on the other, finish with the standard inverse-operation steps. Always check by substituting back.
Worked Examples
Solve: 5x + 3 = 2x + 12
Subtract 2x: 3x + 3 = 12
Subtract 3: 3x = 9
Divide by 3: x = 3
Check: 5(3)+3=18; 2(3)+12=18 ✓
Solve: 7x − 4 = 3x + 16
Subtract 3x: 4x − 4 = 16
Add 4: 4x = 20
Divide by 4: x = 5
Check: 7(5)−4=31; 3(5)+16=31 ✓
Solve: 2x + 9 = 5x − 6
Subtract 2x: 9 = 3x − 6
Add 6: 15 = 3x
Divide by 3: x = 5
Check: 2(5)+9=19; 5(5)−6=19 ✓
Solve: 3(x + 4) = 5x − 2
Distribute: 3x + 12 = 5x − 2
Subtract 3x: 12 = 2x − 2
Add 2: 14 = 2x
Divide by 2: x = 7
Check: 3(11)=33; 5(7)−2=33 ✓
Solve: 6x + 5 = 2(3x + 1) + 3 (Special Case)
Distribute: 6x + 5 = 6x + 2 + 3
Combine: 6x + 5 = 6x + 5
Subtract 6x: 5 = 5 (always true)
Infinitely many solutions.
Guided Practice
Guided Practice Video: Variables on Both Sides
Watch the guided practice walkthrough for variables on both sides, then complete the problems below.
Video by Sang Real Math
Watch on YouTube ↗Answers are in the Answer Key section.
Solve: 6x + 1 = 4x + 9
Hint: Subtract 4x from both sides. You'll get 2x + 1 = 9. Then solve the two-step equation.
Solve: 8x − 3 = 5x + 12
Hint: Subtract 5x from both sides to get 3x − 3 = 12. Then add 3 and divide by 3.
Solve: 3x + 10 = 7x − 6
Hint: Subtract 3x from both sides (keep coefficient positive on right): 10 = 4x − 6.
Solve: 2(x + 5) = 4x − 2
Hint: Distribute to get 2x + 10 = 4x − 2. Then subtract 2x from both sides.
Solve: 5(x − 3) = 3(x + 1)
Hint: Distribute both sides first. Then collect variable terms on one side.
Key Vocabulary
Variables on Both Sides
An equation where variable terms appear on both sides of the equals sign.
Example: 5x + 3 = 2x + 12
Collect Variable Terms
Use addition or subtraction to move all variable terms to one side of the equation.
Example: Subtract 2x from both sides of 5x + 3 = 2x + 12.
Identity
An equation true for every value of the variable. Has infinitely many solutions.
Example: 2x + 4 = 2(x + 2) → 0 = 0
Contradiction
An equation that is never true. Has no solution.
Example: 3x + 1 = 3x + 5 → 1 = 5
Practice Questions
Interactive Practice — 5 Questions
Solve: 5x + 3 = 2x + 12
What is the best first move for 3x + 7 = 9x − 5?
What type of solution does 4(x + 2) = 4x + 8 have?
Solve: 3(x + 4) = 5x − 2
Plan A: $25/month + $0.10/text. Plan B: $15/month + $0.20/text. At how many texts are they equal?
Independent Practice
Answers are in the Answer Key section.
Independent Practice
Solve: 7x + 2 = 4x + 14
Solve: 9x − 5 = 6x + 10
Solve: 3x + 8 = 8x − 7
Solve: 2(x + 6) = 5x − 3
Solve: 5(x + 2) = 5x + 10 (Identify the solution type)
Common Mistakes
Moving the variable term to the right side, making the coefficient negative and causing sign errors.
Move the smaller variable term to the side with the larger coefficient to keep it positive.
Forgetting to distribute before collecting variables.
Always clear parentheses first — otherwise you'll collect terms that aren't fully simplified.
Sign errors when moving negative terms — e.g., to move −3x from the right, add 3x to both sides.
To move −3x from the right, add 3x to both sides (not subtract).
Misidentifying special cases when the variable cancels.
If variables cancel and the statement is true (5 = 5) → infinite solutions. If false (3 = 7) → no solution.
Math Tips
Move the variable term with the smaller coefficient to keep the remaining coefficient positive.
After collecting variables on one side, the equation looks just like a two-step equation.
Always distribute and combine like terms on each side before collecting variable terms across sides.
If the variable completely cancels, check whether the remaining statement is true or false.