Is rearranging formulae on the Foundation paper too?

Yes, in part. Spec ref A5 is examined at both tiers, so a Foundation paper can ask you to make x the subject of something like y = 5x + 2. What Higher adds is the harder half of the spec point: formulae where the new subject appears twice and has to be factorised out, and formulae where a power or a root of the subject is involved.

How many marks is changing the subject worth on Edexcel Higher?

A one- or two-step rearrangement is normally 2 marks. The Higher-only version, where the subject appears on both sides or in the top and bottom of a fraction, is 3-4 marks, because there are separate marks for clearing the fraction, for collecting the terms and for the factorising. Across a full series of three papers expect somewhere around 2-6 marks in total.

What do I do when the letter I want appears twice?

Clear any fractions first, expand any brackets, then move every term containing that letter to one side and everything else to the other. Factorise the letter out, so the side reads (letter) times a bracket, and divide both sides by the whole bracket. For 5y - 3 = k(y + 2) that gives y(5 - k) = 2k + 3 and then y = (2k + 3)/(5 - k).

Does my answer have to look exactly like the mark scheme?

No. Edexcel mark schemes accept any correct equivalent form, so (y - 6)/(2) and 0.5y - 3 both score. Two things do matter: the subject must be on its own on one side, and a fraction written along a diagonal line can cost you if it is ambiguous about what is being divided, so write it as a proper stacked fraction or bracket the numerator.

Rearranging Formulae - Edexcel GCSE Higher Maths

Typically 2-4 marks per paper, and 4 in one go when the subject appears twice · Spec 1MA1

Rearranging a formula means rewriting it so that a different letter stands alone on one side of the equals sign. That letter is the subject. Edexcel files this under spec reference A5 on 1MA1, and it is one of the few algebra topics that runs across both tiers: a Foundation candidate is asked to make $x$ the subject of $y = 4x - 3$ , and Higher adds the cases that inverse operations alone will not finish, namely when the new subject appears twice and when it is trapped under a root or inside a square.

The wording barely moves from series to series. 'Make $t$ the subject of the formula' and 'Rearrange to make $x$ the subject' are the two stems you will meet. A one- or two-step rearrangement is worth 2 marks and sits early in the paper; the version with the subject on both sides, or spread across the numerator and denominator of a fraction, is worth 3-4 marks and lands in the last third of a Higher paper. It can appear on any of the three papers, and the calculator on Papers 2 and 3 does nothing for you here, because the answer is an expression rather than a number. The exception is the two-part question that rearranges a formula and then substitutes numbers into the rearranged version, where the second part is ordinary arithmetic.

The marks go missing in the same few places. Multiplying through by a denominator and leaving one term untouched is the commonest single slip, and once that line is written there is nothing left for the mark scheme to reward. On the Higher-only questions the method mark is for collecting every term containing the new subject on one side and factorising it out, so write the factorised line on its own: $ab + 7b = c$ becomes $b(a + 7) = c$ , and only then do you divide by the whole bracket. Check any rearrangement with numbers before you move on. If $y = 4x - 3$ has given you $x = \frac{y + 3}{4}$ , then feeding $y = 9$ back in should return $x = 3$ .

Worked example

Make

x

the subject of the formula

y = \frac{4x + 3}{x - 2}

[4 marks]

Clear the fraction by multiplying both sides by $x - 2$ , keeping the whole left-hand side in a bracket: $y(x - 2) = 4x + 3$ .
M1: multiplying both sides by the denominator, with $y$ bracketed against $x - 2$
Expand the bracket so the $x$ terms are visible: $xy - 2y = 4x + 3$ .
M1: correct expansion; $yx - 2$ here loses this mark and every mark after it
Collect the two terms containing $x$ on the left and everything else on the right: $xy - 4x = 2y + 3$ .
M1: all terms in $x$ on one side, all terms without $x$ on the other
Factorise the left-hand side and divide by the whole bracket: $x(y - 4) = 2y + 3$ , so $x = \frac{2y + 3}{y - 4}$ .
A1: correct expression; $\frac{2y + 3}{y} - 4$ is the version examiners see from dividing by $y$ alone

Practice questions

These are original questions written in Edexcel 1MA1 Higher style, running from a single inverse operation up to the factorising rearrangements that only Higher candidates are set. Type an expression the way you would write it on one line: a slash for the fraction bar, so $\frac{y + 3}{4}$ goes in as (y+3)/4, and a power for a root, so $\sqrt{y + 5}$ goes in as (y+5)^0.5 or (y+5)^(1/2). Put brackets round anything that sits under the line, so $\frac{100I}{PT}$ is 100I/(PT) and not 100I/PT. Any equivalent form is accepted, so 0.25y + 0.75 marks the same as (y+3)/4.

No account needed: the questions below are always free to check, plus 8 extra questions today. Your score so far updates as you go.

Question 1 Warm-up [1 mark]

Make $x$ the subject of the formula $y = x + 7$ .

Your expression (equivalent forms accepted, e.g. 2(x+3) or 2x+6)
Question 2 Exam pace [2 marks]

Make $x$ the subject of the formula $y = 9 - 2x$ .

Your expression (equivalent forms accepted, e.g. 2(x+3) or 2x+6)
Question 3 Exam pace [3 marks]

A taxi fare $C$ pounds for a journey of $m$ miles is $C = 3 + 1.4m$ . A journey costs £17. Work out the number of miles travelled.

Your answer (fractions like 1/2 are fine)
Question 4 Exam pace [1 mark]

The area of a circle of radius $r$ is $A = \pi r^2$ . Which rearrangement makes $r$ the subject?

Answer options for question 4
$r = \sqrt{\frac{A}{\pi}}$ $r = \frac{A}{2\pi}$ $r = \frac{\sqrt{A}}{\pi}$ $r = \frac{A}{\pi^2}$
Question 5 Stretch [4 marks]

Make $t$ the subject of the formula $p = \frac{t + 4}{t - 1}$ .

Your expression (equivalent forms accepted, e.g. 2(x+3) or 2x+6)
Question 6 Stretch [3 marks]

Make $m$ the subject of the formula $n = \frac{m + 3}{m}$ .

Your expression (equivalent forms accepted, e.g. 2(x+3) or 2x+6)

Common mistakes examiners see

Multiplying through by a denominator and missing a term. From $T = \frac{p}{5} + q$ a candidate writes $5T = p + q$ , leaving the $q$ unmultiplied, and hands in $p = 5T - q$ instead of $5T - 5q$ .

Multiply every term on both sides, then read the new line back and count the terms: three before, three after. If a term looks unchanged, it has been missed.
Dividing selectively instead of factorising when the subject appears twice. From $ab + 7b = c$ the working goes $b + 7b = \frac{c}{a}$ , which is the mark scheme's standard wrong answer for this style of question.

Once every term in the new subject is on one side, take that letter outside a bracket in a line of its own: $b(a + 7) = c$ . Divide by the entire bracket, giving $b = \frac{c}{a + 7}$ , never by one piece of it.
Rooting or squaring term by term. From $c^2 = a^2 + b^2$ a candidate writes $c = a + b$ , and from $y = 3x^2$ they write $x = \frac{\sqrt{y}}{3}$ , having rooted the $y$ but not the 3.

A root or a square applies to the whole of one side at once. Get the squared letter on its own first, so $x^2 = \frac{y}{3}$ , and only then take the square root of everything on the other side: $x = \sqrt{\frac{y}{3}}$ .
Losing the bracket when clearing a two-term denominator. Multiplying $y = \frac{3t + 2}{t - 1}$ by $t - 1$ and writing $yt - 1 = 3t + 2$ rather than $y(t - 1) = 3t + 2$ .

Write the bracket before you expand anything. The mark for this line is for a correct multiplication of the whole side, and a missing bracket kills it and the three marks that follow.
Sign slips when the new subject has a negative coefficient. From $y = 8 - 3x$ , reaching $3x = 8 - y$ correctly and then writing $x = \frac{y - 8}{3}$ because the $y$ looked like it should come first.

Copy the numerator across exactly as you wrote it: $x = \frac{8 - y}{3}$ . Test it with a value, since $x = 1$ gives $y = 5$ , and $\frac{8 - 5}{3}$ must come back as 1.

Frequently asked questions

Is rearranging formulae on the Foundation paper too?: Yes, in part. Spec ref A5 is examined at both tiers, so a Foundation paper can ask you to make $x$ the subject of something like $y = 5x + 2$ . What Higher adds is the harder half of the spec point: formulae where the new subject appears twice and has to be factorised out, and formulae where a power or a root of the subject is involved.
How many marks is changing the subject worth on Edexcel Higher?: A one- or two-step rearrangement is normally 2 marks. The Higher-only version, where the subject appears on both sides or in the top and bottom of a fraction, is 3-4 marks, because there are separate marks for clearing the fraction, for collecting the terms and for the factorising. Across a full series of three papers expect somewhere around 2-6 marks in total.
What do I do when the letter I want appears twice?: Clear any fractions first, expand any brackets, then move every term containing that letter to one side and everything else to the other. Factorise the letter out, so the side reads (letter) times a bracket, and divide both sides by the whole bracket. For $5y - 3 = k(y + 2)$ that gives $y(5 - k) = 2k + 3$ and then $y = \frac{2k + 3}{5 - k}$ .
Does my answer have to look exactly like the mark scheme?: No. Edexcel mark schemes accept any correct equivalent form, so $\frac{y - 6}{2}$ and $0.5y - 3$ both score. Two things do matter: the subject must be on its own on one side, and a fraction written along a diagonal line can cost you if it is ambiguous about what is being divided, so write it as a proper stacked fraction or bracket the numerator.

Rearranging Formulae - Edexcel GCSE Higher Maths

Worked example

Practice questions

Common mistakes examiners see

Frequently asked questions

Related topics