When an equation contains brackets, the first step is always to expand the brackets before you balance the equation. Once the brackets are removed and like terms are collected, you solve exactly as you would any linear equation. This two-phase approach — expand then balance — is the core skill tested in Year 8 and Year 9 algebra.

What does it mean to expand brackets?

Expanding brackets means multiplying everything inside the brackets by the term directly outside. The rule is:

a(b + c) = ab + ac

This is the distributive law. It works whether a is a positive number, a negative number, or a letter (variable).

Examples of expanding brackets

  • 3(x + 4) = 3x + 12
  • 5(2x − 3) = 10x − 15
  • −2(x + 6) = −2x − 12 (multiply each term by −2, taking care with signs)
  • 4(3x − 1) = 12x − 4

Sign rule: When the term outside the brackets is negative, every sign inside the bracket flips.

Step-by-step method for solving equations with brackets

Step 1 — Expand the brackets

Multiply every term inside the brackets by the number (or letter) immediately outside.

Step 2 — Collect like terms

Simplify the equation by grouping x terms together and number terms together.

Step 3 — Balance the equation

Use inverse operations to isolate x:

  • Add or subtract to move number terms to one side.
  • Multiply or divide to find the value of x.

Step 4 — Check your answer

Substitute your answer back into the original equation, including the brackets, to verify it works.

Worked examples

Worked example 1 — single set of brackets

Solve 3(x + 4) = 21.

Step 1 — Expand: 3x + 12 = 21

Step 2 — Subtract 12 from both sides: 3x = 21 − 12 = 9

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

Step 4 — Check: 3(3 + 4) = 3 × 7 = 21 ✓

Worked example 2 — brackets with subtraction inside

Solve 5(2x − 3) = 25.

Step 1 — Expand: 10x − 15 = 25

Step 2 — Add 15 to both sides: 10x = 40

Step 3 — Divide by 10: x = 4

Check: 5(2 × 4 − 3) = 5(8 − 3) = 5 × 5 = 25 ✓

Worked example 3 — brackets on both sides

Solve 4(x + 2) = 2(x + 8).

Step 1 — Expand both sides: 4x + 8 = 2x + 16

Step 2 — Subtract 2x from both sides: 2x + 8 = 16

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

Step 4 — Divide by 2: x = 4

Check: Left: 4(4 + 2) = 4 × 6 = 24. Right: 2(4 + 8) = 2 × 12 = 24 ✓

Worked example 4 — negative term outside the bracket

Solve −3(x − 5) = 6.

Step 1 — Expand (multiply each term by −3): −3x + 15 = 6

Step 2 — Subtract 15 from both sides: −3x = −9

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

Check: −3(3 − 5) = −3 × (−2) = 6 ✓

Worked example 5 — brackets plus extra terms

Solve 2(x + 3) + x = 15.

Step 1 — Expand: 2x + 6 + x = 15

Step 2 — Collect like terms: 3x + 6 = 15

Step 3 — Subtract 6: 3x = 9

Step 4 — Divide by 3: x = 3

Check: 2(3 + 3) + 3 = 2 × 6 + 3 = 12 + 3 = 15 ✓

Common mistakes and how to avoid them

Mistake 1 — Only multiplying the first term inside the bracket

A very common error is writing 3(x + 4) = 3x + 4 instead of 3x + 12. The number outside must multiply every term inside the brackets.

Mistake 2 — Getting signs wrong with a negative outside

−2(x − 5) is NOT −2x − 10. The correct expansion is −2x + 10, because −2 × −5 = +10. Write out each multiplication explicitly until sign work becomes automatic.

Mistake 3 — Moving terms across the equals sign without changing the sign

When you move +12 from the left side to the right side, it becomes −12. This is "balancing" — what you do to one side you do to the other.

Where this skill appears in the KS3 curriculum

The Department for Education's KS3 mathematics programme of study requires pupils to "solve linear equations in one variable" and to "manipulate algebraic expressions, expanding products of two binomials." BBC Bitesize's KS3 algebra section covers expanding brackets and solving equations as linked topics. Equations with brackets appear in virtually every GCSE maths paper, so getting confident with the method at KS3 is time well invested.

Practice problems

Try these yourself, then check by substituting back:

  1. 2(x + 5) = 18
  2. 3(x − 2) = 12
  3. 5(3x + 1) = 40
  4. 4(x + 1) = 2(x + 7)
  5. −2(x − 3) = 10

Answers: 1. x = 4, 2. x = 6, 3. x = 13/5 = 2.6, 4. x = 5, 5. x = −2

Frequently asked questions

Why do I have to expand the brackets before balancing the equation?

You cannot add or subtract terms that are inside brackets with terms that are outside, because the brackets act like a grouping instruction — multiply first. Once you expand, every term is separate, and you can legally move them across the equals sign using inverse operations. Skipping the expansion step means you are treating 3(x + 4) as 3x + 4, which is wrong and will give you an incorrect value of x.

What if there are two sets of brackets on the same side?

Expand both sets first, then collect like terms before balancing. For example, 2(x + 3) + 4(x − 1) expands to 2x + 6 + 4x − 4, which simplifies to 6x + 2. Then solve as normal. The order in which you expand the two sets does not matter, as long as you expand both before collecting terms.

How do I know if my answer is correct?

Substitute your value of x back into the original equation — including the brackets, before you expanded them. Work out both sides separately. If they are equal, the answer is correct. If not, check each step of the expansion and the sign work, as these are the most common sources of error.

Does this method work when the unknown is x² or higher?

No. This step-by-step expand-and-balance method is for linear equations, where x appears to the power of 1 only. When the equation contains x², the approach is different (factorising or using the quadratic formula), and you will encounter that at GCSE. At KS3, all equations with brackets are linear.

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