March 2025

Equations with Variables on Both Sides

When x appears on both sides of the equals sign (e.g., 3x + 2 = 2x + 8), we need to get all x terms on one side and all numbers on the other. The strategy: move variable terms first, then constants.

Step-by-step strategy

  1. Move variable terms — Add or subtract to get all x terms on one side.
  2. Move constants — Add or subtract to get numbers on the other side.
  3. Divide — Solve for x.

Example: 3x + 2 = 2x + 8

Step 1: Subtract 2x from both sides to move x terms to the left.

3x − 2x + 2 = 2x − 2x + 8
x + 2 = 8

Step 2: Subtract 2 from both sides.

x + 2 − 2 = 8 − 2
x = 6

Answer: x = 6. Check: 3(6) + 2 = 20, 2(6) + 8 = 20 ✓

Example: 5x − 4 = 2x + 11

Subtract 2x: 3x − 4 = 11. Add 4: 3x = 15. Divide by 3: x = 5.

When to move which side?

It doesn't matter — move the smaller x coefficient to avoid negatives if you prefer. For 2x + 7 = 5x − 2, subtracting 2x gives 7 = 3x − 2, then 9 = 3x, so x = 3.

Equations with parentheses

Distribute first, then solve. For 2(x + 3) = x + 10:

2x + 6 = x + 10
2x − x = 10 − 6
x = 4

Common mistakes

Subtracting the wrong term: Always move the smaller variable term to the other side to keep x positive (when possible). In 5x + 2 = 3x + 10, subtract 3x (not 5x) from both sides.

Distributing first: If there are parentheses, distribute before collecting terms. 2(x + 3) = x + 10 becomes 2x + 6 = x + 10.

More examples

Example: 7x − 4 = 3x + 12

7x − 4 = 3x + 12
7x − 3x = 12 + 4
4x = 16
x = 4

Example: 2(x + 5) = x + 16

2x + 10 = x + 16
2x − x = 16 − 10
x = 6

Practice problems

1. Solve: 6x + 1 = 4x + 9

Show answer
2x = 8, x = 4

2. Solve: 3(x − 2) = 2x + 1

Show answer
3x − 6 = 2x + 1, x = 7

3. Solve: 5x + 3 = 8x − 12

Show answer
−3x = −15, x = 5

Practice builds confidence. Arithmia helps you build the algebraic thinking that makes these problems easier.

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