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MA11-02 Maths Watch

Solving Linear Equations with Brackets

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In this lesson

In this video you'll learn about solving linear equations with brackets for GCSE Maths, with worked examples and the mistakes examiners report. By the end you'll be able to solve a linear equation that requires expanding one or more brackets first, by multiplying every term inside each bracket before balancing.

What it covers

  1. 0:42 What a bracket means + checkable question
  2. 2:25 Worked example
  3. 4:25 The minus-sign trap
  4. 6:51 Exam technique
  5. 8:42 Your turn now, with one equation and no help from me
  6. 11:06 What's next

Key words

About this video

GCSE Maths - Solving Linear Equations with Brackets | Linear Equations 2/9 (2026/27 exams)

In this video you'll learn about solving linear equations with brackets for GCSE Maths, with worked examples and the mistakes examiners report.

By the end you'll be able to solve a linear equation that requires expanding one or more brackets first, by multiplying every term inside each bracket before balancing.

For: AQA, Cambridge iGCSE, Edexcel, Edexcel iGCSE, Eduqas, OCR GCSE/iGCSE Maths · Foundation (Cambridge: Core)
Watch first: {{video:G-ALGBASE-3}}, {{video:G-ALGBASE-4}}, {{video:G-EXPFAC-1}}

Specifications: AQA 8300, Cambridge iGCSE 0580, Edexcel 1MA1, Edexcel iGCSE 4MA1, Eduqas C300QS, OCR J560

Video code: MA11-02 - search YouTube for "ScholaFly MA11-02" to come straight back to this video.

Videos in this chapter:
MA11-01 — Solving Linear Equations by Balancing
MA11-02 — Solving Linear Equations with Brackets
MA11-03 — Using and Rearranging a Formula (Subject Appears Once)
MA11-04 — Rearranging Harder Formulae (Subject Appears Twice, or Under a Power/Root)
MA11-05 — Writing an Expression or Formula from a Context
MA11-06 — Setting Up and Solving an Equation from a Context
MA11-07 — Recalling Circle, Pythagoras and Trig Formulae
MA11-08 — Choosing Between the Quadratic Formula, Sine Rule, Cosine Rule and Area Formula
MA11-09 — Using the Kinematics (SUVAT) Formulae

#GCSEMaths #Maths

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Read the transcript

Picture four friends at the cinema, all buying the same ticket, and all holding the same three pound voucher. The till takes three pounds off the first ticket, then charges the other three friends the full price. You would spot that in a second. The discount belongs to every ticket, not just the first one in the queue. A bracket in algebra works exactly the same way, and the same slip ruins an equation before the solving has even begun.

So what is a bracket actually asking you to do? Four lots of the bracket x minus three means four lots of the whole thing inside it, both parts of it. The four multiplies the x, and the four multiplies the three as well. Two terms inside the bracket, so two multiplications. Picture two arrows, one running from the outside number to each term inside. Draw them both, every single time.

Try one yourself. Expand three lots of the bracket two x plus five. Is it six x plus five, or six x plus fifteen?

Take your pick. I'll wait.

The answer is six x plus fifteen. Three multiplies the two x, and three multiplies the five as well.

Six x plus five is what a single arrow gives you. The second term just sat there and was never touched.

Every bracket in this video has a single number or term outside it, never another bracket. Two brackets multiplied together is a different method, taught in the video on expanding double brackets.

Here comes a full equation, brackets and all. Four lots of the bracket x minus three equals two x plus six.

Balancing works by moving whole terms across the equals sign. While that x is trapped inside a bracket it is not yet a term on its own, so the bracket comes off first.

Arrow one: four times x gives four x. Arrow two: four times minus three gives minus twelve.

So the left side becomes four x minus twelve, and the equation now reads four x minus twelve equals two x plus six.

From here it is ordinary balancing, and I will run it briskly. Take two x off both sides, and you are left with two x minus twelve equals six.

Add twelve to both sides for two x equals eighteen, then divide both sides by two, and x equals nine.

Check it back into the original. Four lots of nine minus three is four lots of six, which is twenty four, and two times nine plus six is twenty four as well.

If you want that balancing built up slowly, the video called Solving Linear Equations by Balancing does exactly that job.

Some equations arrive with a bracket on each side. Nothing new happens there, you simply expand both of them before any balancing starts.

Change one thing now. Put a minus sign in front of the bracket, and a new trap opens up.

Solve five minus two lots of the bracket x minus three equals x.

The number doing the multiplying here is not two. It is minus two, and that minus travels into the bracket with it.

So arrow two is minus two times minus three. Does that land on minus six, or on plus six?

Pick one. I'll wait.

It lands on plus six, because a negative multiplied by a negative gives a positive.

Put the two versions side by side. The wrong line reads five minus two x minus six. The right line reads five minus two x plus six.

Both signs inside the bracket changed on the way out. Two arrows, both signs, and your handle for this video is now complete.

Tidy the left side. Five plus six is eleven, so the equation now reads eleven minus two x equals x.

Add two x to both sides for eleven equals three x, then divide both sides by three, and x is eleven over three.

Leave that as the fraction eleven thirds, or write it as three and two thirds, because both of those are exact. Do not round it to three point six seven unless the question asks you for a decimal.

A fraction is not a sign that you went wrong somewhere. Real equations land on fractions all the time.

One wrong expansion, and every line after it inherits the mistake, however neat the rest of the working looks.

Here is one line from an examiner report, about a question where a bracket had to be expanded on the way to an answer.

Those who expanded the bracket sometimes introduced an error by only multiplying the first term in the bracket to achieve eight d minus five equals twenty eight but then correctly followed through with an answer d equals four point one two five.

Read the end of that again, because the balancing afterwards was correct. It was the expansion at the very start that went wrong, and a tidy finish cannot rescue that.

Here is a line from a second examiner report, on a question where a bracket had a minus sign in front of it.

The most common misconceptions were to expand minus one across the second bracket as minus five c minus one or to multiply the second bracket by four to achieve minus twenty c minus four.

Those are the same two slips you have just practised. A term left unmultiplied, and a sign that never flipped on the way out.

So before you balance anything, look back at the line you have just written and count. Two terms inside the bracket means two multiplications, every time.

Your turn now, with one equation and no help from me.

Solve three lots of the bracket two x minus one equals x plus seventeen.

Pause the video and solve it, expanding the bracket before you balance anything.

Both arrows first. Three times two x is six x, and three times minus one is minus three. So the line reads six x minus three equals x plus seventeen. Now balance. Take x off both sides for five x minus three equals seventeen, add three to both sides for five x equals twenty, and x equals four.

If you multiplied both terms inside and landed on four, this skill is yours.

Time to gather it up, and the whole of this video fits into four words. Two arrows, both signs. Two arrows means the number outside multiplies every term inside, never only the first one. Both signs means a minus in front of the bracket travels inside and changes what comes back out. Expand before you balance, because a term trapped inside a bracket is not yet a term you can move across. Then solve it just as you normally would, and when you have your answer, put it back into the original equation, brackets and all, to check that both sides really do agree.

Next in the chapter: Using and Rearranging a Formula, where the subject appears once.

For more, visit scholafly.com, or watch the next video.

Related terms

For: AQA GCSE 8300, Edexcel GCSE 1MA1, Eduqas GCSE C300, Edexcel IGCSE 4MA1, OCR GCSE J560, Cambridge IGCSE 0580

On the specification

BoardSpecStatement
AQA GCSE 8300A17Solve linear equations in one unknown algebraically
Edexcel GCSE 1MA1A17Solve linear equations in one unknown algebraically (including those with the unknown on both sides of the equation); find approximate solutions using a graph
Eduqas GCSE C300FA14Solve linear equations in one unknown algebraically (including those with the unknown on both sides of the equation); find approximate solutions using a graph
Eduqas GCSE C300HA17Solve linear equations in one unknown algebraically (including those with the unknown on both sides of the equation); find approximate solutions using a graph
Edexcel IGCSE 4MA1F2.4ASolve linear equations, with integer or fractional coefficients, in one unknown in which the unknown appears on either side or both sides of the equation
OCR GCSE J5606.03aSolve linear equations in one unknown algebraically.
OCR GCSE J5606.03dUse a graph to find the approximate solution of a linear equation.
Cambridge IGCSE 0580C2.5Construct simple expressions, equations and formulas.
Cambridge IGCSE 0580E2.5Construct expressions, equations and formulas.
For teachers

This GCSE Maths lesson teaches solving linear equations with brackets. By the end, students should be able to solve a linear equation that requires expanding one or more brackets first, by multiplying every term inside each bracket before balancing. It works through two worked examples and the mistakes examiners report.