Equations with the Unknown on Both Sides

When the unknown is on both sides

So far you may have solved equations like 3x + 2 = 14, where the unknown sits on only one side. But many equations have the unknown on both sides, for example:

5x − 3 = 2x + 9

The goal is still the same: find the value of x that makes both sides equal. The new skill is gathering all the x-terms onto one side before you finish.

If you are still building confidence with the basics, review solving linear equations first.

The balance rule still applies

An equation is a balance: the = sign means both sides are worth the same. To keep it balanced, whatever you do to one side you do to the other. This is what lets you safely subtract an x-term, or a number, from both sides at once.

A clear strategy

1. (If needed) expand any brackets and collect like terms on each side.
2. Choose a side for the unknowns. Remove the smaller x-term from both sides so the remaining coefficient stays positive.
3. Move the numbers to the other side using inverse operations.
4. Divide to leave a single x.
5. Check by substituting back into the original equation.

Worked example 1

Solve 5x − 3 = 2x + 9.

The smaller x-term is 2x, so subtract 2x from both sides:

5x − 3 = 2x + 9
5x − 2x − 3 = 9        (subtract 2x from both sides)
3x − 3 = 9
3x = 9 + 3             (add 3 to both sides)
3x = 12
x = 12 ÷ 3             (divide both sides by 3)
x = 4

Check: left side 5 × 4 − 3 = 17; right side 2 × 4 + 9 = 17. Equal, so x = 4. ✓

Worked example 2: unknown term on the right is bigger

Solve 2x + 8 = 6x.

Here 6x is the larger x-term, so subtract the smaller 2x from both sides. This leaves the x on the right, which is fine:

2x + 8 = 6x
8 = 6x − 2x            (subtract 2x from both sides)
8 = 4x
8 ÷ 4 = x             (divide both sides by 4)
2 = x

So x = 2. Check: left 2 × 2 + 8 = 12; right 6 × 2 = 12. ✓

Worked example 3: a subtracted unknown

Solve 9 − x = 2x + 3.

The x on the left is negative (−x). Add x to both sides to make it positive:

9 − x = 2x + 3
9 = 2x + x + 3        (add x to both sides)
9 = 3x + 3
9 − 3 = 3x            (subtract 3 from both sides)
6 = 3x
2 = x

Check: left 9 − 2 = 7; right 2 × 2 + 3 = 7. ✓

Worked example 4: tidy up first, then solve

Solve 4x + 2 + x = 3x + 14.

Collect like terms on the left first: 4x + x = 5x, so the left becomes 5x + 2.

5x + 2 = 3x + 14
5x − 3x + 2 = 14      (subtract 3x from both sides)
2x + 2 = 14
2x = 12               (subtract 2 from both sides)
x = 6                 (divide by 2)

Check: left 4 × 6 + 2 + 6 = 32; right 3 × 6 + 14 = 32. ✓

Worked example 5: brackets on both sides

Solve 3(x + 4) = 5(x − 2).

Expand both brackets first:

3(x + 4) = 5(x − 2)
3x + 12 = 5x − 10      (multiply out both sides)
12 = 5x − 3x − 10      (subtract 3x from both sides)
12 = 2x − 10
12 + 10 = 2x           (add 10 to both sides)
22 = 2x
11 = x

Check: left 3(11 + 4) = 3 × 15 = 45; right 5(11 − 2) = 5 × 9 = 45. ✓

If expanding brackets is new, see expanding brackets.

Worked example 6: a non-whole answer

Solve 4x + 1 = x + 8.

4x + 1 = x + 8
4x − x + 1 = 8        (subtract x)
3x + 1 = 8
3x = 7                (subtract 1)
x = 7 ÷ 3 = 2⅓ ≈ 2.33

Not every solution is a whole number — a fraction or decimal is perfectly valid.

Activity: solve and check

1. 4x = x + 9
2. 6x − 1 = 2x + 11
3. 10 − 2x = x + 1
4. 5(x − 1) = 2x + 7
5. 3x + 4 = 7x − 8

Answers:

1. x = 3 (subtract x → 3x = 9)
2. x = 3 (subtract 2x → 4x − 1 = 11 → 4x = 12)
3. x = 3 (add 2x → 10 = 3x + 1 → 9 = 3x)
4. x = 4 (expand → 5x − 5 = 2x + 7 → 3x = 12)
5. x = 3 (subtract 3x → 4 = 4x − 8 → 12 = 4x)

Where this leads

These moves are the heart of algebra. The very same balancing steps appear when you work with solving equations with brackets and when you rearrange formulae to make a different letter the subject.