Algebra

Formulae and changing the subject

19 past-paper questions, easiest first. Try each one, then reveal the answer.

Foundation & Higher9 of 19 from Foundation papers

Revision guide by The Full Marks

Formulae and changing the subject: what you need to know

Changing the subject means rearranging a formula so a different letter is on its own. It uses the same inverse operations as solving an equation, and it comes up on both tiers, often with a substitution part first.

Foundation students rearrange formulae where the subject appears once, including some with squares or square roots; formulae where the subject appears twice are Higher.

Key facts

The subject is the letter on its own on one side, e.g. $v$ in $v = u + at$ .
Whatever you do to one side, do to the whole of the other side.
Undo operations in reverse order: last thing done to the letter is undone first.
The inverse of squaring is square rooting ( $\pm$ ), and the inverse of $\sqrt{\ }$ is squaring.
Substituting a negative number: use brackets. If $x = -3$ , then $x^2 = (-3)^2 = 9$ and $4x^2 = 4 \times 9 = 36$ .

Changing the subject

If the new subject is under a square root, get the root on its own, then square both whole sides.
Clear fractions by multiplying both sides by the denominator.
Expand any brackets if the new subject is inside them.
Get every term with the new subject on one side and everything else on the other.
Higher: if the subject appears twice, factorise it out, then divide by the bracket.

Worked examples

Example 1 Foundation · grade 4

Make $t$ the subject of $v = u + 5t$

Show the solution

Subtract $u$ from both sides: $v - u = 5t$
Divide both sides by 5: $t = \frac{v - u}{5}$

Example 2 Foundation · grade 6

Make $h$ the subject of $P = \sqrt{\frac{h - 4}{5}}$

Show the solution

Square both sides: $P^2 = \frac{h - 4}{5}$
Multiply both sides by 5: $5P^2 = h - 4$
Add 4 to both sides: $h = 5P^2 + 4$

Example 3 Higher · grade 7

Make $x$ the subject of $y = \frac{2x + 3}{x - 4}$

Show the solution

Multiply by $(x - 4)$ : $y(x - 4) = 2x + 3$
Expand: $xy - 4y = 2x + 3$
Collect $x$ terms: $xy - 2x = 4y + 3$
Factorise: $x(y - 2) = 4y + 3$
$x = \frac{4y + 3}{y - 2}$

Common mistakes

Dividing only one term: from $y = 3w - 10$ writing $w = \frac{y}{3} + 10$ . The answer is $w = \frac{y + 10}{3}$ .
Leaving the subject on both sides. If $x$ appears twice, collect and factorise.
Squaring only part of a side: $\sqrt{a + b}$ squared is $a + b$ , but $(y - 2)^2$ is not $y^2 - 4$ .

Exam tips

Write each step on a new line; method marks are given for correct steps even if the final line is wrong.
Your final answer must start with the new subject, e.g. $t = \ldots$
Some questions show someone else's working and ask what went wrong: name the exact step and say what should have been done instead.

Questions people ask

How do you change the subject of a formula?

Use inverse operations, one step at a time, doing the same to both sides until the letter you want is on its own.

What do you do if the subject appears twice?

Collect both terms on one side, factorise the letter out as a common factor, then divide by the bracket. This is a Higher skill.

Exam questions: Formulae and changing the subject (19, easiest first)

Formulae and changing the subject2 marksGrade 4No calculatorFoundation

November 2024 · Paper 1 Foundation · Q25

Rearrange a linear formula to make a different letter the subject.

Show answer

Answer	$w = \frac{y + 10}{3}$	2 marks

Worked solution Checked against the mark scheme

$y = 3w - 10$

Add 10: $y + 10 = 3w$

Divide by 3: $w = \frac{y + 10}{3}$

Watch out: Writing w = y/3 + 10 (dividing only one term).

Formulae and changing the subject4 marksGrade 4CalculatorFoundation

June 2024 · Paper 3 Foundation · Q29

Substitute a negative value into a formula and rearrange a formula to change its subject.

Show answer

(a)	$w = 15$	2 marks
(b)	$h = 3p + 5$	2 marks

Worked solution Checked against the mark scheme

(a) $w = 40 - (-5)^2 = 40 - 25 = 15$

(b) $p = \frac{h - 5}{3}$

Multiply by 3: $3p = h - 5$

Add 5: $h = 3p + 5$

Watch out: In (a), working out $-5^2$ as $-25$ to get $65$ .

Formulae and changing the subject2 marksGrade 4CalculatorOn both tiers

November 2022 · Paper 3 Foundation · Q21

Rearrange a linear formula to make a the subject.

Show answer

Answer	$a = \frac{p + 9}{3}$	2 marks

Worked solution Checked against the mark scheme

$p = 3a - 9$

Add 9: $p + 9 = 3a$

Divide by 3: $a = \frac{p + 9}{3}$

Watch out: Writing $a = p + 9 \div 3$ without brackets.

Formulae and changing the subject4 marksGrade 4CalculatorOn both tiers

June 2022 · Paper 3 Foundation · Q23

Substitute a negative number into a formula with a square, then rearrange a linear formula.

Show answer

(a)	$T = 25$	2 marks
(b)	$p = \frac{d - 4}{3}$	2 marks

Worked solution Checked against the mark scheme

(a) $(-3)^2 = 9$ , so $T = 4 \times 9 - 11 = 36 - 11 = 25$

(b) $d = 3p + 4$

Subtract 4: $d - 4 = 3p$

Divide by 3: $p = \frac{d - 4}{3}$

Watch out: In (a) squaring $4 \times -3$ to get 144, or using $-9$ for $(-3)^2$ .

Formulae and changing the subject2 marksGrade 4CalculatorFoundation

June 2019 · Paper 3 Foundation · Q19

Rearrange a linear formula to make x the subject.

Show answer

Answer	$x = \frac{y-4}{2}$	2 marks

Worked solution Checked against the mark scheme

$y = 2x + 4$

Subtract 4: $y - 4 = 2x$

Divide by 2: $x = \frac{y - 4}{2}$

Watch out: Dividing only y by 2, giving y/2 − 4.

Formulae and changing the subject2 marksGrade 4CalculatorFoundation

Specimen · Paper 2 Foundation · Q20

Show answer

Answer	$t = \frac{w - 11}{3}$	2 marks

Worked solution Checked against the mark scheme

$w = 3t + 11$

Subtract 11: $w - 11 = 3t$

Divide by 3: $t = \frac{w - 11}{3}$

Watch out: Dividing only $w$ by 3, giving $t = \frac{w}{3} - 11$ .

Formulae and changing the subject2 marksGrade 4CalculatorHigher

Specimen · Paper 2 Higher · Q1

Rearrange a simple linear formula to make t the subject.

Show answer

Answer	$t = \frac{w - 11}{3}$	2 marks

Worked solution Checked against the mark scheme

Take 11 from both sides: $3t = w - 11$

Divide both sides by 3: $t = \dfrac{w - 11}{3}$

Watch out: Writing $t = \frac{w}{3} - 11$ , dividing only part of the right-hand side by 3.

Formulae and changing the subject3 marksGrade 5CalculatorHigher

November 2023 · Paper 3 Higher · Q10

$Rearrange a formula with a fractional term to make m the subject.$

Show answer

Answer	$m = \dfrac{5(k - p)}{2}$	3 marks

Worked solution Checked against the mark scheme

$k = p + \dfrac{2m}{5}$

Subtract $p$ : $k - p = \dfrac{2m}{5}$

Multiply by 5: $5(k - p) = 2m$

Divide by 2: $m = \dfrac{5(k - p)}{2}$

Watch out: Multiplying only k by 5 and not p, giving 5k − p = 2m.

Formulae and changing the subject3 marksGrade 5No calculatorFoundation

November 2020 · Paper 1 Foundation · Q30

Rearrange a linear formula to change the subject, and simplify a negative power raised to a negative power.

Show answer

(a)	$q = \frac{p - 7}{6}$	2 marks
(b)	$m^6$	1 mark

Worked solution Checked against the mark scheme

(a) $p = 6q + 7$ . Subtract 7: $p - 7 = 6q$ . Divide by 6: $q = \frac{p - 7}{6}$ .

(b) Multiply the powers: $(-2) \times (-3) = 6$ , so $(m^{-2})^{-3} = m^6$ .

Watch out: In (b) adding the powers to get $m^{-5}$ .

Formulae and changing the subject1 markGrade 5No calculatorHigher

November 2019 · Paper 1 Higher · Q12

$Spot the error in someone's first step when rearranging a formula with a fraction.$

Show answer

Answer	He only doubled $\frac{q}{2}$ and $T$ ; the $5$ should also be doubled, giving $2T = q + 10$	1 mark

Worked solution Checked against the mark scheme

Multiplying both sides by 2 means every term is doubled:

$2T = q + 10$

Spencer left the 5 as 5. The correct result is $q = 2T - 10$ .

Watch out: Saying only that he should have subtracted 5 first, without saying what was wrong.

Formulae and changing the subject2 marksGrade 5CalculatorHigher

November 2019 · Paper 2 Higher · Q10

Rearrange a formula involving a square root to make k the subject.

Show answer

Answer	$k = 2m - y^2$	2 marks

Worked solution Checked against the mark scheme

$y = \sqrt{2m - k}$

Square both sides: $y^2 = 2m - k$

Add $k$ and subtract $y^2$ : $k = 2m - y^2$

Watch out: Squaring only one term, e.g. writing $y^2 = 2m - k^2$ .

Formulae and changing the subject4 marksGrade 5No calculatorOn both tiers

November 2018 · Paper 1 Foundation · Q21

Substitute into a kinematics formula to find v, then rearrange it to make s the subject.

Show answer

(a)	6 (or $-6$ )	2 marks
(b)	$s = \frac{v^2 - u^2}{2a}$	2 marks

Worked solution Checked against the mark scheme

(a) $v^2 = 12^2 + 2 \times (-3) \times 18 = 144 - 108 = 36$

$v = \sqrt{36} = 6$

(b) $v^2 - u^2 = 2as$

$s = \frac{v^2 - u^2}{2a}$

Watch out: In (a), forgetting the negative sign on $a$ and getting $v^2 = 252$ .

Formulae and changing the subject3 marksGrade 5CalculatorFoundation

June 2018 · Paper 3 Foundation · Q28

$Rearrange a formula containing a square root of a fraction to make g the subject.$

Show answer

Answer	$g = 2T^2 - 6$	3 marks

Worked solution Checked against the mark scheme

Square both sides: $T^2 = \frac{g + 6}{2}$ .

Multiply by 2: $2T^2 = g + 6$ .

Subtract 6: $g = 2T^2 - 6$ .

Watch out: Subtracting 6 before multiplying by 2, giving 2(T² − 6).

Formulae and changing the subject4 marksGrade 7CalculatorHigher

November 2024 · Paper 3 Higher · Q15

Rearrange a formula where the new subject appears in both the numerator and denominator.

Show answer

Answer	$p = \dfrac{5t + 6}{4 + 2t}$	4 marks

Worked solution Checked against the mark scheme

$t(5 - 2p) = 2(2p - 3)$

$5t - 2pt = 4p - 6$

$5t + 6 = 4p + 2pt$

$5t + 6 = p(4 + 2t)$

$p = \frac{5t + 6}{4 + 2t}$

Watch out: Not factorising, so p is left on both sides.

Formulae and changing the subject4 marksGrade 7No calculatorHigher

June 2023 · Paper 1 Higher · Q17

$Rearrange a formula with the subject appearing in a fraction's numerator and denominator.$

Show answer

Answer	$x = \frac{3y + 28}{8 - 5y}$	4 marks

Worked solution Checked against the mark scheme

Multiply both sides by $5x + 3$ : $5xy + 3y = 8x - 28$

Collect $x$ terms on one side: $3y + 28 = 8x - 5xy$

Factorise: $3y + 28 = x(8 - 5y)$

Divide: $x = \frac{3y + 28}{8 - 5y}$

Watch out: Not factorising out x, leaving x on both sides.

Formulae and changing the subject4 marksGrade 7No calculatorHigher

November 2021 · Paper 1 Higher · Q11

Equate the areas of two rectangles with algebraic sides and make y the subject.

Show answer

Answer	$y = \frac{30w}{6 - 21w}$	4 marks

Worked solution Checked against the mark scheme

$6(2w + y) = 7w(3y + 6)$

$12w + 6y = 21wy + 42w$

$6y - 21wy = 30w$

$y(6 - 21w) = 30w$

$y = \frac{30w}{6 - 21w}$ (or $\frac{10w}{2 - 7w}$ )

Watch out: Not factorising out $y$ , leaving $y$ on both sides of the answer.

Formulae and changing the subject4 marksGrade 7No calculatorHigher

November 2020 · Paper 1 Higher · Q17

$Rearrange a formula with the subject in both the numerator and denominator of a fraction.$

Show answer

Answer	$f = \frac{3 + 4d}{d + 3}$	4 marks

Worked solution Checked against the mark scheme

Multiply both sides by $(f - 4)$ : $d(f - 4) = 3(1 - f)$

Expand: $df - 4d = 3 - 3f$

Collect the $f$ terms: $df + 3f = 3 + 4d$

Factorise: $f(d + 3) = 3 + 4d$

Divide: $f = \frac{3 + 4d}{d + 3}$

Watch out: Getting a sign wrong when expanding $3(1 - f)$ .

Formulae and changing the subject3 marksGrade 7CalculatorHigher

June 2019 · Paper 2 Higher · Q15

$Rearrange an algebraic-fraction formula to make m the subject.$

Show answer

Answer	$m = \frac{f + 4}{f - 3}$	3 marks

Worked solution Checked against the mark scheme

$f(m - 1) = 3m + 4$

$fm - f = 3m + 4$

$fm - 3m = f + 4$

$m(f - 3) = f + 4$

$m = \frac{f + 4}{f - 3}$

Watch out: Forgetting to factorise out m after collecting terms.

Formulae and changing the subject3 marksGrade 7No calculatorHigher

Specimen · Paper 1 Higher · Q13

Rearrange a formula involving a square root and a cube to make k the subject.

Show answer

Answer	$k = \sqrt[3]{4m^2 - 1}$	3 marks

Worked solution Checked against the mark scheme

Square both sides: $m^2 = \frac{k^3 + 1}{4}$

Multiply by 4: $4m^2 = k^3 + 1$

Subtract 1: $k^3 = 4m^2 - 1$

Cube root: $k = \sqrt[3]{4m^2 - 1}$

Watch out: Taking the square root instead of squaring to remove the root sign.

Also tested in these questions

Their main topic is different, but part of each one tests this topic.

November 2023 · Paper 2 Higher · Q1: Expand and simplify a sum of two brackets, factorise a quadratic expression fully, and rearrange a linear formula.
November 2023 · Paper 2 Foundation · Q22: Expand and simplify two brackets, factorise a quadratic expression fully and rearrange a linear formula.
November 2022 · Paper 1 Higher · Q24: Find the ratio of radius to height when a sphere and cone have equal volumes, then when a sphere's surface area equals a cone's total surface area.
June 2022 · Paper 2 Higher · Q9: Use the pressure formula to find the missing width of the face of a box resting on the floor.
November 2021 · Paper 2 Higher · Q21: Rearrange a pendulum formula and use upper and lower bounds of length and time to find bounds for g.
Specimen · Paper 1 Higher · Q2: Use the pressure formula to find a contact area from a given force and pressure, stating the units.
Specimen · Paper 1 Foundation · Q21: Rearrange the pressure formula to find a contact area from a given force and pressure.