Rearranging Formulae GCSE Maths | Worked Examples and Practice

You can follow the example in class. Then the question changes the letters, adds a fraction, and suddenly you have no idea where to start.

Rearranging formulae gets easier when you stop trying to move everything at once. Each line has one job: undo an operation while keeping both sides equal.

Rearranging a formula means rewriting it so a different letter is on its own. For example, to make x the subject of y = 3x + 6, subtract 6 from both sides, then divide both sides by 3:

x = (y − 6) ÷ 3

This GCSE Maths guide covers brackets, fractions, powers and questions where the subject appears twice. You’ll also find 11 practice questions with worked answers.

What does “make a letter the subject” mean?

The subject of a formula is the letter isolated on one side of the equals sign.

In this formula:

A = lw

A is the subject. The formula tells you how to calculate the area when you know the length and width.

To make w the subject, divide both sides by l:

A ÷ l = w

You can write this more neatly as:

w = A ÷ l

The relationship stays the same. You have rewritten it to calculate a different quantity. Here, l must be non-zero.

A quick notation reminder: Letters written next to each other mean multiplication. So lw means l × w, and 3x means 3 × x.

Rearranging, solving and substituting: what is the difference?

Rearranging is different from solving

If you solve 3x + 6 = 21, you find a number:

x = 5

If you rearrange y = 3x + 6 to make x the subject, your answer still contains y:

x = (y − 6) ÷ 3

Unless the question gives you values to use, there is no need to invent any.

How to rearrange a formula

Start by finding the letter you need to isolate. Look at what is happening to it, then undo those operations.

Operation	How to undo it
Add a quantity	Subtract that quantity
Subtract a quantity	Add that quantity
Multiply by a non-zero quantity	Divide by that quantity
Divide by a non-zero quantity	Multiply by that quantity
Square	Take a square root, considering positive and negative solutions
Cube	Take a cube root

Apply each operation to both sides of the equation.

For a simple chain of operations, work backwards.

In y = 3x + 6, x is multiplied by 3, then 6 is added. To get back to x, subtract 6 first, then divide by 3.

Before each line, ask:

“What am I undoing?”

How to rearrange formulae using inverse operations

An inverse operation undoes another operation. Addition and subtraction undo each other. Multiplication and division do the same.

Operation	Inverse operation
Add (a)	Subtract (a)
Subtract (a)	Add (a)
Multiply by non-zero (a)	Divide by (a)
Divide by non-zero (a)	Multiply by (a)

The rule that matters throughout is:

Apply the same operation to both sides of the equation.

For a simple chain of operations, undo them in reverse order. In (y=4x+7), (x) is multiplied by 4, then 7 is added. To isolate (x), subtract 7 first, then divide by 4.

If you are unsure how to read an expression containing several operations, revisit our BIDMAS guide. Understanding the order in which an expression is built makes it easier to undo.

Example 1: a two-step rearrangement

Make x the subject of y = 6x − 5.

The expression 6x − 5 means multiply x by 6, then subtract 5.

Undo the subtraction by adding 5 to both sides:

y + 5 = 6x

Now divide both sides by 6:

x = (y + 5) ÷ 6

The whole expression y + 5 must be divided by 6. That is why the brackets matter.

Compare these expressions:

• (y + 5) ÷ 6 means add y and 5, then divide the result by 6.
• y + 5 ÷ 6 means divide 5 by 6, then add y.

They give different answers.

Try it: Make x the subject of y = 8x + 3. Check your working against question 1 in the practice answers.

Example 2: rearranging formulae with brackets

Make x the subject of y = 3(x + 4).

The whole bracket is multiplied by 3. Undo that first by dividing both sides by 3:

y ÷ 3 = x + 4

Subtract 4 from both sides:

x = (y ÷ 3) − 4

You could expand the bracket first:

y = 3x + 12

Then subtract 12:

y − 12 = 3x

Finally, divide by 3:

x = (y − 12) ÷ 3

Both answers are correct. Dividing first saves a line of working here.

Watch out: (y ÷ 3) − 4 and (y − 4) ÷ 3 are different expressions. Check exactly which part you are dividing.

Example 3: rearranging formulae with fractions

Make x the subject of y = (x − 2) ÷ 5.

This means the whole expression x − 2 is divided by 5.

Multiply both sides by 5:

5y = x − 2

Add 2 to both sides:

x = 5y + 2

It can help to think of x − 2 as a package. First undo the division around the package, then deal with what is inside it.

Example 4: the subject is in the denominator

Make t the subject of v = d ÷ t.

Here, t is the denominator, which is the bottom part of the fraction.

Multiply both sides by t:

vt = d

Now divide both sides by v:

t = d ÷ v

The original formula requires t to be non-zero. To write t = d ÷ v, v must also be non-zero.

A formula triangle may help you remember this particular relationship between speed, distance and time. Writing the algebra gives you a method you can also use when a question contains extra terms.

Example 5: rearranging an area formula

Make h the subject of the trapezium area formula:

A = ½ × (a + b) × h

Multiply both sides by 2:

2A = (a + b)h

Then divide both sides by the whole bracket:

h = 2A ÷ (a + b)

Keep a + b together. Both lengths form the expression multiplying h.

For a trapezium, a and b are the lengths of the parallel sides, so their sum is positive.

Example 6: rearranging formulae with squares and roots

Make r the subject of A = πr².

First divide both sides by π:

A ÷ π = r²

Then take the square root:

r = √(A ÷ π)

Here, r represents a radius, so we use the non-negative square root.

A common slip: Dividing by 2 does not undo squaring. Squaring means multiplying a quantity by itself, so you need a square root.

Do you always need a plus-or-minus sign?

The context matters.

For an unrestricted algebraic variable:

x² = k

gives:

x = ±√k, provided k is zero or positive.

For example, x² = 9 has two solutions: x = 3 and x = −3.

A radius cannot be negative, so the circle formula uses the non-negative root.

Example 7: the subject appears twice

Make x the subject of ax + b = cx + d.

This needs a slightly different approach. Collect the terms containing x, then factorise.

Subtract cx from both sides:

ax − cx + b = d

Subtract b from both sides:

ax − cx = d − b

Factorise the left-hand side:

x(a − c) = d − b

Now divide both sides by a − c:

x = (d − b) ÷ (a − c)

This assumes a and c are different, so you are not dividing by zero.

Factorising is the turning point. It rewrites the two terms containing x as one product, allowing you to isolate x.

Challenge example: the subject appears twice in a fraction

Make x the subject of y = (x + 2) ÷ (x − 3).

Multiply both sides by x − 3:

y(x − 3) = x + 2

Expand the bracket:

yx − 3y = x + 2

Subtract x from both sides and add 3y to both sides:

yx − x = 3y + 2

Factorise:

x(y − 1) = 3y + 2

Divide both sides by y − 1:

x = (3y + 2) ÷ (y − 1)

The original expression requires x ≠ 3. The rearranged expression requires y ≠ 1.

If this feels much harder than the earlier examples, practise collecting terms and factorising before trying another question of this type.

Common rearranging mistakes and how to fix them

Mistake	Why it causes a problem	What to do instead
Changing only one side	The two sides may no longer be equal	Apply the same operation to both sides
Dividing only one term in a sum	Part of the expression is left unchanged	Put brackets around the whole expression before dividing
Losing a bracket	You change which terms an operation applies to	Keep grouped terms together
Dividing by 2 to undo a square	Doubling and squaring are different operations	Use a square root
Dividing before collecting repeated subjects	The required letter can remain elsewhere in your answer	Collect its terms and factorise
Finishing with 3x or x²	The requested letter is not isolated	Check what still needs undoing

“Move it across and change the sign” can be tempting shorthand. But it hides the operation that makes the step valid.

Writing “subtract 5 from both sides” gives you something you can check.

How to check your answer

Use the original formula and a simple value.

Suppose you started with:

y = 6x − 5

and rearranged it to get:

x = (y + 5) ÷ 6

Choose x = 3. The original formula gives:

y = 6 × 3 − 5 = 13

Now put y = 13 into your rearranged formula:

x = (13 + 5) ÷ 6 = 3

You get back to your starting value.

A numerical check can reveal a missing bracket or sign error. One successful check does not prove the formula works for every permitted value, so your algebraic working still matters.

Also check that the letter you were asked to isolate does not appear elsewhere in your final answer.

Rearranging formulae practice questions

Write one operation per line. Start with questions 1–4, then move on when you can explain your steps.

Start here

1. Make x the subject: y = 8x + 3
2. Make a the subject: P = 2a + 2b
3. Make p the subject: q = 4(p − 6)
4. Make m the subject: n = (m + 7) ÷ 3

Build your confidence

5. Make w the subject: A = lw, where l ≠ 0.
6. Make r the subject: C = 2πr
7. Make h the subject: V = πr²h, where r > 0.
8. Make x the subject: y = 5 ÷ x, where y ≠ 0 and the original formula requires x ≠ 0.

Stretch yourself

9. Make r the subject: V = ⁴⁄₃ × π × r³
10. Make x the subject: y = px + qx, where p + q ≠ 0.
11. Make x the subject: ax + b = cx − d, where a ≠ c.

Practice answers with working

Answer 1

Start with:

y = 8x + 3

Subtract 3:

y − 3 = 8x

Divide by 8:

x = (y − 3) ÷ 8

Answer 2

Start with:

P = 2a + 2b

Subtract 2b:

P − 2b = 2a

Divide by 2:

a = (P − 2b) ÷ 2

You can also write this as:

a = (P ÷ 2) − b

Answer 3

Start with:

q = 4(p − 6)

Divide by 4:

q ÷ 4 = p − 6

Add 6:

p = (q ÷ 4) + 6

Answer 4

Start with:

n = (m + 7) ÷ 3

Multiply by 3:

3n = m + 7

Subtract 7:

m = 3n − 7

Answer 5

Start with:

A = lw

Divide by l:

w = A ÷ l

Answer 6

Start with:

C = 2πr

Divide by 2π:

r = C ÷ (2π)

The brackets show that you are dividing by the entire quantity 2π.

Answer 7

Start with:

V = πr²h

Divide by πr²:

h = V ÷ (πr²)

Answer 8

Start with:

y = 5 ÷ x

Multiply by x:

xy = 5

Divide by y:

x = 5 ÷ y

Answer 9

Start with:

V = ⁴⁄₃ × π × r³

Multiply by 3:

3V = 4πr³

Divide by 4π:

r³ = 3V ÷ (4π)

Take the cube root:

r = ∛(3V ÷ (4π))

The symbol ∛ means cube root.

Answer 10

Start with:

y = px + qx

Factorise:

y = x(p + q)

Divide by p + q:

x = y ÷ (p + q)

Answer 11

Start with:

ax + b = cx − d

Subtract cx and subtract b from both sides:

ax − cx = −d − b

Factorise:

x(a − c) = −d − b

Divide by a − c:

x = (−d − b) ÷ (a − c)

What to practise next

If you made a mistake, look for the first line where the working went wrong.

Was it a sign? A missing bracket? An unfamiliar factorising step? That tells you what to practise next.

If questions 1–4 felt difficult, spend more time on inverse operations. If questions 10 and 11 caused problems, revisit collecting like terms and factorising.

For exam questions, use AQA’s GCSE Maths assessment resources, choosing papers for your course and tier. If you study with another board, use its corresponding papers and mark schemes.

Using this skill in GCSE revision

Start with questions containing one or two inverse operations. Once those feel secure, introduce brackets, fractions and powers. Leave questions with the subject on both sides until you can confidently collect terms and factorise.

If an answer is wrong, find the first incorrect line. That tells you more than simply comparing final answers. You might understand the method but have copied a sign incorrectly, or you might need more practice dividing an entire expression.

For exam-board practice, use the questions and mark schemes available through AQA’s GCSE Maths assessment resources. Choose material that matches your course and tier.

Parents supporting revision can ask, “What are you undoing in this step?” This gives the student a chance to explain their thinking without someone taking over the calculation. Our guide to building maths confidence before GCSE offers more advice on keeping support calm and manageable.

FAQs

What is rearranging a formula?

It means rewriting a formula so a different letter is isolated as the subject. The relationship between the quantities stays the same.

How do I know which operation to do first?

In a simple chain, identify the last operation applied to the required letter and undo that first. In (y=5x+2), subtract 2 before dividing by 5. More complex expressions may require expanding brackets or collecting terms first.

Do signs always change when letters cross the equals sign?

No. A term’s sign changes in familiar addition or subtraction rearrangements because you have added or subtracted that term from both sides. Multiplication, division and powers need their own inverse operations.

What if the subject appears twice?

Collect the terms containing that letter on one side, then factorise. Once the letter appears as a single factor, divide by the expression multiplying it, provided that expression is non-zero.

Can two different-looking answers both be correct?

Yes. For example, (\frac{y-12}{3}) and (\frac{y}{3}-4) are equivalent. Keep your answer clear and follow any instruction to simplify.

Should I substitute numbers before rearranging?

If the question asks you to change the subject, rearrange using the letters. If it asks you to calculate a quantity from given values, substituting first can sometimes make the working shorter.

A quick recap

Rearranging a formula changes its subject without changing the relationship it describes. Apply the same operation to both sides of the equation. For simple chains, undo operations in reverse order. Keep brackets and fraction numerators together. If the subject appears twice, collect its terms and factorise. Finish by checking that the required letter is isolated and that each line follows from the one before it.

Need help making algebra make sense?

If you can follow a worked example but struggle to begin a question alone, bring that question to a tutor. A Battersea House GCSE Maths tutor can work through your approach, explain the sticking point and give you similar questions to try independently.

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