Solving Equations with Unknowns on Both Sides: KS3 Maths

When an equation has the unknown letter on both sides, the first step is always to collect all the letter terms on one side and all the number terms on the other. Once the letters are on one side only, you solve as normal.

Why unknowns on both sides need a different first step

In a simple equation like 3x = 12, the unknown is only on the left, so you divide straight away: x = 4.

When the equation is 5x + 3 = 2x + 12, the unknown x appears on both sides. If you divide immediately you get a fraction with x still in it — that is not solved. Instead, collect the x terms onto one side first.

The balancing method

The key principle is the balance model: whatever you do to one side of an equation, you must do to the other side. This keeps the equation true.

Worked example 1: basic unknowns on both sides

Solve 5x + 3 = 2x + 12.

Step 1 — Subtract 2x from both sides to remove x from the right:
5x − 2x + 3 = 12
3x + 3 = 12

Step 2 — Subtract 3 from both sides:
3x = 9

Step 3 — Divide both sides by 3:
x = 3

Check: LHS = 5(3) + 3 = 18. RHS = 2(3) + 12 = 18. LHS = RHS ✓

Answer: x = 3

Worked example 2: larger x coefficient on the right

Solve 4x − 1 = 7x − 10.

Here it is easier to move x to the right (where the coefficient is bigger), so the result is positive.

Step 1 — Subtract 4x from both sides:
−1 = 3x − 10

Step 2 — Add 10 to both sides:
9 = 3x

Step 3 — Divide both sides by 3:
x = 3

Check: LHS = 4(3) − 1 = 11. RHS = 7(3) − 10 = 11. LHS = RHS ✓

Answer: x = 3

Worked example 3: negative result

Solve 2x + 8 = 5x − 1.

Step 1 — Subtract 2x from both sides:
8 = 3x − 1

Step 2 — Add 1 to both sides:
9 = 3x

Step 3 — Divide both sides by 3:
x = 3

Now try an equation that produces a negative answer.

Solve 6x + 2 = 8x + 10.

Step 1 — Subtract 6x from both sides:
2 = 2x + 10

Step 2 — Subtract 10 from both sides:
−8 = 2x

Step 3 — Divide both sides by 2:
x = −4

Check: LHS = 6(−4) + 2 = −22. RHS = 8(−4) + 10 = −22. LHS = RHS ✓

Answer: x = −4

Equations with brackets on both sides

When brackets appear, always expand first, then collect like terms.

Worked example 4: single bracket on each side

Solve 3(x + 2) = 2(x + 5).

Step 1 — Expand:
3x + 6 = 2x + 10

Step 2 — Subtract 2x from both sides:
x + 6 = 10

Step 3 — Subtract 6 from both sides:
x = 4

Check: LHS = 3(4 + 2) = 18. RHS = 2(4 + 5) = 18. LHS = RHS ✓

Answer: x = 4

Worked example 5: expanding two brackets, negative coefficient

Solve 5(2x − 1) = 3(x + 8).

Step 1 — Expand:
10x − 5 = 3x + 24

Step 2 — Subtract 3x from both sides:
7x − 5 = 24

Step 3 — Add 5 to both sides:
7x = 29

Step 4 — Divide by 7:
x = 29/7 ≈ 4.14 (leave as a fraction unless asked to round)

Check: LHS = 5(2 × 29/7 − 1) = 5(58/7 − 7/7) = 5 × 51/7 = 255/7. RHS = 3(29/7 + 8) = 3(29/7 + 56/7) = 3 × 85/7 = 255/7. LHS = RHS ✓

Answer: x = 29/7

A summary table of the method

Step | What to do | Example using 5x + 3 = 2x + 12
1 | Collect x terms on one side | Subtract 2x: 3x + 3 = 12
2 | Collect number terms on other side | Subtract 3: 3x = 9
3 | Isolate x by dividing | Divide by 3: x = 3
4 | Check in the original equation | 5(3) + 3 = 18; 2(3) + 12 = 18 ✓

Common mistakes

Mistake 1 — Subtracting the wrong term.
Always subtract the smaller x coefficient from both sides so the remaining x term is positive. Negative x terms are still correct but are an extra source of sign errors.

Mistake 2 — Forgetting to move the number term.
After collecting x on one side, the number term on the same side must also be moved. Missing step 2 gives an equation like 3x + 3 = 9, and students often jump to x = 3 without dealing with the +3.

Mistake 3 — Sign error when subtracting a negative.
If one term is negative, subtracting it adds it: 4x − (−2x) = 6x. Write it out carefully rather than trying to do it mentally.

Mistake 4 — Not checking the answer.
Substituting back in is always worth doing. It takes 30 seconds and catches arithmetic errors before they cost marks.

How this fits the KS3 national curriculum

The Department for Education's KS3 Mathematics Programme of Study requires pupils to "solve linear equations in one variable, including those with unknowns on both sides of the equation, and where the solution is a negative number or a fraction." The AQA GCSE Mathematics specification covers this as a Year 8 and Year 9 algebra skill, linking it directly to GCSE preparation.

Frequently asked questions

Which side should I move the x term to?

Move x to the side where its coefficient is larger. This ensures the resulting x coefficient is positive, which makes the arithmetic simpler. Both sides give the same answer — choose whichever avoids a negative x coefficient in the working.

What if x cancels out completely?

If subtracting x terms from both sides leaves 0 = 0, the equation is true for all values of x (infinitely many solutions). If it leaves something like 0 = 5, the equation has no solution. Both cases are unusual at KS3 but may appear as extension questions.

Can I move the number terms first instead of the x terms?

Yes — the order (collect x first or collect numbers first) does not matter as long as you are consistent and do the same operation to both sides. Most teachers recommend collecting x terms first because it makes the next step more visible.

How is this used at GCSE?

At GCSE the same technique is applied to more complex equations — with fractions, larger brackets, and quadratics. Mastering unknowns on both sides at KS3 is the essential stepping stone to all equation work in Years 10 and 11.

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