Equations with Variables on Both Sides

TipLearning Objectives

By the end of this lesson, you’ll be able to:

  • Solve equations where the variable appears on both sides.
  • Decide which side to collect variable terms on.
  • Recognize when an equation has no solution or infinitely many solutions.

Key Ideas

When variables appear on both sides:

  1. Collect variable terms on one side.
  2. Collect constants on the other side.
  3. Solve the resulting simpler equation.

Some equations lead to special cases:

  • True statement (e.g., \(0 = 0\)) → infinitely many solutions
  • False statement (e.g., \(5 = 2\)) → no solution

Common Problem Types

1. Simple Variable Move

Solve:
\[ 5x + 3 = 2x + 15 \]

Subtract \(2x\): \(3x + 3 = 15\)
Subtract 3: \(3x = 12\)
Divide:
\[ x = 4 \]

2. Variables and Distribution

Solve:
\[ 2(x + 4) = 3x - 2 \]

Distribute: \(2x + 8 = 3x - 2\)
Subtract \(2x\): \(8 = x - 2\)
Add 2:
\[ x = 10 \]

3. No Solution / Infinite Solutions

  • \(3x + 2 = 3x + 2\)
    → always true
    → infinitely many solutions

  • \(4x + 1 = 4x - 5\)
    → \(1 = -5\) (false)
    → no solution

Strategies

  • Move the variable term from the smaller coefficient side to the larger one to avoid negatives (optional but helpful).
  • Always simplify both sides before moving terms.
  • After moving variables, the equation should look like a standard one- or two-step equation.

Worked Examples

Example 1

Solve:
\[ 4x - 7 = 2x + 5 \]

Step-by-step Solution:
1. Subtract \(2x\) → \(2x - 7 = 5\)
2. Add 7 → \(2x = 12\)
3. Divide → \(x = 6\)

Answer: \(6\)

Example 2

Solve:
\[ 3(x - 1) = 2(x + 4) - 3 \]

Step-by-step Solution:
1. Distribute → \(3x - 3 = 2x + 8 - 3\)
2. Combine → \(3x - 3 = 2x + 5\)
3. Subtract \(2x\) → \(x - 3 = 5\)
4. Add 3 → \(x = 8\)

Answer: \(8\)

WarningCommon Mistakes
  • Moving constants before simplifying each side.
  • Moving variable terms in the wrong direction (causing sign errors).
  • Forgetting to distribute before combining or moving terms.
  • Misinterpreting \(0 = 0\) (infinite solutions) or \(5 = 2\) (no solution).

Practice Problems

  1. \(6x + 4 = 2x + 20\)
  2. \(5x - 3 = 3x + 7\)
  3. \(2(x + 1) = x + 8\)
  4. \(4x + 6 = 4x - 10\)
  5. \(3(x - 2) - 1 = 2(x + 1)\)

1. \(6x + 4 = 2x + 20\)
Subtract \(2x\) → \(4x + 4 = 20\)
Subtract 4 → \(4x = 16\)
Divide → \(x = 4\)

2. \(5x - 3 = 3x + 7\)
Subtract \(3x\) → \(2x - 3 = 7\)
Add 3 → \(2x = 10\)
Divide → \(x = 5\)

3. \(2(x + 1) = x + 8\)
Distribute → \(2x + 2 = x + 8\)
Subtract \(x\) → \(x + 2 = 8\)
Subtract 2 → \(x = 6\)

4. \(4x + 6 = 4x - 10\)
Subtract \(4x\) → \(6 = -10\)
False → no solution

5. \(3(x - 2) - 1 = 2(x + 1)\)
Distribute → \(3x - 6 - 1 = 2x + 2\)
Combine → \(3x - 7 = 2x + 2\)
Subtract \(2x\) → \(x - 7 = 2\)
Add 7 → \(x = 9\)

Summary

  • Move variable terms to one side, constants to the other.
  • Simplify first, then solve normally.
  • True statements → infinitely many solutions; false statements → no solution.
  • Organize work carefully to avoid sign errors.
  • Move variables first, then isolate the constant.
  • Choose the side with the larger coefficient to avoid negatives.
  • Always simplify before deciding if an equation has 0, 1, or infinitely many solutions.
  • If variables cancel, check whether the leftover statement is true or false.