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Substitution and Rearranging Formulae

Substituting into formulae, writing your own, and changing the subject of a formula.

  • 10 key terms
  • All boards

Learning Objectives

  1. 1Substitute positive and negative numbers into expressions and formulae.
  2. 2Write and use formulae to solve problems in context.
  3. 3Change the subject of a formula when the subject appears once.
  4. 4Rearrange harder formulae, including squares, roots and a subject that appears twice (Higher tier).

Formulae: rules with letters

A formula is a rule that links quantities, such as \(A = lw\) or \(v = u + at\). You will be asked to put numbers into formulae (substitution), to build your own formula from a situation, and to rearrange a formula so that a different letter is on its own. All three rely on the same idea: follow the order of operations carefully, and undo steps in reverse order.

Substitution

Replace each letter with its value and then work out the answer, using BIDMAS.

  • Use brackets for negatives

    If \(a = -3\), then \(2a^2 = 2 \times (-3)^2 = 2 \times 9 = 18\). Writing \(-3^2\) would give \(-9\), which is wrong.

  • Powers before multiplying

    \(3x^2\) means \(3 \times x^2\), so with \(x = 2\) it is \(3 \times 4 = 12\), whereas \((3x)^2 = 6^2 = 36\).

  • Formulae

    In \(v = u + at\) with \(u = 5\), \(a = 2\) and \(t = 4\), the value is \(v = 5 + 2 \times 4 = 13\).

  • Context

    A taxi costs \(C = 3 + 2m\) pounds for \(m\) miles, so a 10 mile journey costs \(3 + 20 = \pounds 23\).

Substituting with negatives

Work out the value of \(2a^2 - 3b\) when \(a = 3\) and \(b = -2\).

Show the solutionHide the solution
  1. 1 Write the substitution with brackets \(2 \times 3^2 - 3 \times (-2)\).
  2. 2 Powers first \(3^2 = 9\), so the expression is \(2 \times 9 - 3 \times (-2)\).
  3. 3 Multiply \(18 - (-6)\).
  4. 4 Subtract a negative \(18 + 6 = 24\).

Answer24

Substituting into a longer formula

Work out the value of \(s = ut + \tfrac{1}{2}at^2\) when \(u = 4\), \(a = 2\) and \(t = 3\).

Show the solutionHide the solution
  1. 1 Substitute \(s = 4 \times 3 + \tfrac{1}{2} \times 2 \times 3^2\).
  2. 2 Evaluate the power \(3^2 = 9\), so \(s = 12 + \tfrac{1}{2} \times 2 \times 9\).
  3. 3 Multiply \(\tfrac{1}{2} \times 2 \times 9 = 9\).
  4. 4 Add \(12 + 9 = 21\).

Answer21

Writing formulae

Turn a description into algebra by spotting what stays the same and what changes.

  • A fixed amount

    A fixed charge is a number on its own, such as a call-out fee of \(\pounds 35\).

  • A rate

    A cost per hour or per mile is multiplied by the variable, such as \(22h\).

  • Put them together

    Cost \(C = 35 + 22h\) for \(h\) hours of work.

  • Use it both ways

    Put in \(h = 4\) to find the cost. Put in a cost and solve to find the hours.

Changing the subject

Rearranging uses the same steps as solving an equation, except that the answer is a formula.

  • The subject

    The subject of a formula is the letter on its own on one side, such as \(y\) in \(y = 3x + 5\).

  • Do the same to both sides

    Undo the operations in the reverse of the order they were applied, treating the other letters like numbers.

  • Fractions

    If the subject is in a fraction, multiply both sides by the denominator first.

  • Check

    Substitute numbers into both versions of the formula. If they agree, your rearrangement is right.

Making t the subject

Make \(t\) the subject of \(v = u + at\).

Show the solutionHide the solution
  1. 1 Subtract \(u\) from both sides \(v - u = at\).
  2. 2 Divide both sides by \(a\) \(\dfrac{v - u}{a} = t\).
  3. 3 Write with the subject on the left \(t = \dfrac{v - u}{a}\).

Answer\(t = \dfrac{v - u}{a}\)

Making x the subject of a fraction

Make \(x\) the subject of \(y = \dfrac{2x - 1}{3}\).

Show the solutionHide the solution
  1. 1 Multiply both sides by 3 \(3y = 2x - 1\).
  2. 2 Add 1 to both sides \(3y + 1 = 2x\).
  3. 3 Divide both sides by 2 \(x = \dfrac{3y + 1}{2}\).
  4. 4 Check If \(x = 5\) then \(y = \dfrac{9}{3} = 3\), and \(\dfrac{3 \times 3 + 1}{2} = 5\).

Answer\(x = \dfrac{3y + 1}{2}\)

Harder rearranging (Higher tier)

Squares, roots and a subject that appears twice need extra steps.

  • Squares and roots

    To make \(r\) the subject of \(A = \pi r^2\), divide by \(\pi\) to get \(\dfrac{A}{\pi} = r^2\) and then take the square root: \(r = \sqrt{\dfrac{A}{\pi}}\).

  • Subject in two places

    Multiply out any fractions, get every term containing the subject on one side, factorise the subject out, and then divide.

  • Take care with brackets

    Subtract the whole of an expression using brackets, so that no signs go wrong.

A subject that appears twice

Make \(x\) the subject of \(y = \dfrac{x + 3}{x - 2}\).

Show the solutionHide the solution
  1. 1 Remove the fraction \(y(x - 2) = x + 3\).
  2. 2 Expand \(xy - 2y = x + 3\).
  3. 3 Collect the \(x\) terms on one side \(xy - x = 3 + 2y\).
  4. 4 Factorise \(x(y - 1) = 3 + 2y\).
  5. 5 Divide \(x = \dfrac{2y + 3}{y - 1}\).

Answer\(x = \dfrac{2y + 3}{y - 1}\)

Substituting and rearranging

Substituting

  • Numbers replace the letters
  • The answer is a single number
  • Follow BIDMAS and use brackets for negative values

Rearranging

  • The letters stay in place
  • The answer is a new formula
  • Undo operations in reverse order, using inverse operations

Using familiar formulae

Several formulae appear in both Maths and Science, and examiners expect you to recognise them.

  • Speed

    \(s = \dfrac{d}{t}\). A car that travels 150 km in 2.5 hours has speed \(150 \div 2.5 = 60\) km/h.

  • Density

    Density \(= \dfrac{\text{mass}}{\text{volume}}\). A block of mass 240 g and volume 30 cm\(^3\) has density 8 g/cm\(^3\).

  • Area of a trapezium

    \(A = \tfrac{1}{2}(a + b)h\), where \(a\) and \(b\) are the parallel sides and \(h\) is the height.

  • Perimeter of a rectangle

    \(P = 2(l + w)\), which can be rearranged to find a missing side.

Using a formula

The area of a trapezium is \(A = \tfrac{1}{2}(a + b)h\). Work out \(A\) when \(a = 7\), \(b = 11\) and \(h = 5\).

Show the solutionHide the solution
  1. 1 Substitute \(A = \tfrac{1}{2}(7 + 11) \times 5\).
  2. 2 Brackets first \(7 + 11 = 18\), so \(A = \tfrac{1}{2} \times 18 \times 5\).
  3. 3 Multiply \(\tfrac{1}{2} \times 18 = 9\), and \(9 \times 5 = 45\).

Answer45

Making a letter the subject when there are brackets

Make \(a\) the subject of \(P = 2(a + b)\).

Show the solutionHide the solution
  1. 1 Divide both sides by 2 \(\dfrac{P}{2} = a + b\).
  2. 2 Subtract \(b\) from both sides \(\dfrac{P}{2} - b = a\).
  3. 3 Write with the subject first \(a = \dfrac{P}{2} - b\), which is the same as \(a = \dfrac{P - 2b}{2}\).

Answer\(a = \dfrac{P}{2} - b\)

The subject is in a fraction term

Make \(x\) the subject of \(y = \dfrac{3x}{4} + 2\).

Show the solutionHide the solution
  1. 1 Subtract 2 from both sides \(y - 2 = \dfrac{3x}{4}\).
  2. 2 Multiply both sides by 4 \(4(y - 2) = 3x\).
  3. 3 Divide both sides by 3 \(x = \dfrac{4(y - 2)}{3}\).

Answer\(x = \dfrac{4(y - 2)}{3}\)

Squares and roots in formulae (Higher tier)

When the subject is squared, take a square root as the last step.

  • Isolate the squared term

    To make \(u\) the subject of \(v^2 = u^2 + 2as\), subtract \(2as\) from both sides to get \(u^2 = v^2 - 2as\).

  • Take the square root of the whole side

    \(u = \sqrt{v^2 - 2as}\). The root covers everything on the right-hand side.

  • Positive and negative roots

    A square root can be positive or negative, but in context, such as a speed, you normally use the positive one.

  • Check with numbers

    Put \(v = 5\), \(a = 2\) and \(s = 3\) into the original, which gives \(u^2 = 25 - 12 = 13\), and then check your rearranged formula gives \(\sqrt{13}\).

Substituting to find an unknown

Sometimes you are given the answer from a formula and must find the input.

  • Substitute what you know

    If \(C = 12 + 8d\) and \(C = 84\), then \(84 = 12 + 8d\).

  • Solve the equation

    \(72 = 8d\), so \(d = 9\).

  • Or rearrange first

    \(d = \dfrac{C - 12}{8}\), then substitute \(C = 84\) to get \(d = 9\).

  • Choose the quicker method

    Substituting first is usually faster for a single value, and rearranging is better if you need many values.

Test yourself

  1. 1

    Work out \(3x^2\) when \(x = 4\).

    Show answerHide answer

    48.

  2. 2

    Work out \(5 - 2a\) when \(a = -3\).

    Show answerHide answer

    11.

  3. 3

    Make \(x\) the subject of \(y = x - 6\).

    Show answerHide answer

    \(x = y + 6\).

  4. 4

    Make \(x\) the subject of \(y = 2x\).

    Show answerHide answer

    \(x = \dfrac{y}{2}\).

  5. 5

    Make \(t\) the subject of \(d = st\).

    Show answerHide answer

    \(t = \dfrac{d}{s}\).

Exam technique: substitution and rearranging

Neat layout makes these questions straightforward.

  • Write the formula first

    Then substitute, so that the examiner can see where each number came from.

  • Brackets for negatives

    Always put a negative value in brackets when you substitute it.

  • One step per line

    Rearranging is solving an equation with letters, so give each step its own line.

  • Check the subject

    The subject must be on its own, and it must not appear on both sides of the final formula.

Summary and exam focus

  • Put negative numbers in brackets when substituting, and remember that \(2a^2\) means \(2 \times a^2\).
  • To write a formula, separate the fixed amount from the rate and combine them.
  • To change the subject, reverse the function machine: use inverse operations in reverse order.
  • Multiply by the denominator first when the subject sits in a fraction.
  • On the Higher tier, collect the subject terms together and factorise when it appears twice.

Exam focus

Make \(a\) the subject of the formula \(v^2 = u^2 + 2as\). (3 marks) (3 marks)

Treat \(u^2\) and \(2s\) as if they were numbers, so you subtract \(u^2\) from both sides first and then divide both sides by \(2s\). Show each step on its own line, and write the final answer with \(a\) alone on the left.

Key terms

The words this lesson expects you to use. Each one is linked from the first place it appears above.

Output
The result of putting an input through a formula or a function machine.
Substitute
Replace a letter in an expression or formula with a number.
Formula
A rule that links quantities using letters.
Subject of a formula
The letter on its own on one side of the equals sign.
Rearrange
Change a formula so that a different letter is the subject.
Variable
A letter that can take different values.
Inverse operation
The operation that undoes another, such as division undoing multiplication.
Function machine
A diagram showing the operations applied to an input, in order, to give an output.
Constant
A quantity that has a fixed value and does not change.
Evaluate
Work out the numerical value of an expression.

Questions and answers

17 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Work out 2 marks Easier

Work out the value of \(5p - 2q\) when \(p = 3\) and \(q = -4\).

Mark scheme — 2 marks available

  • \(15\) or \(-2 \times (-4) = +8\) — M1
  • 23 — A1

Model answer

\(5 \times 3 - 2 \times (-4) = 15 + 8 = 23\).

2. Exam question Work out 2 marks Core

Work out the value of \(3x^2\) when \(x = -4\).

Mark scheme — 2 marks available

  • \((-4)^2 = 16\) — M1
  • 48 — A1

Model answer

The power comes first: \((-4)^2 = 16\). Then \(3 \times 16 = 48\).

3. Exam question Work out 4 marks Core

A formula for the distance \(s\) is \(s = \tfrac{1}{2}(u + v)t\). (a) Work out \(s\) when \(u = 3\), \(v = 11\) and \(t = 6\). [2 marks] (b) Make \(v\) the subject of the formula. [2 marks]

Mark scheme — 4 marks available

  • (a) \(\tfrac{1}{2} \times 14 \times 6\) or \(7 \times 6\) — M1
  • (a) 42 — A1
  • (b) \(\dfrac{2s}{t} = u + v\) — M1
  • (b) \(v = \dfrac{2s}{t} - u\) — A1

Model answer

(a) \(s = \tfrac{1}{2} \times (3 + 11) \times 6 = \tfrac{1}{2} \times 14 \times 6 = 42\). (b) Multiply both sides by 2: \(2s = (u + v)t\). Divide by \(t\): \(\dfrac{2s}{t} = u + v\). Subtract \(u\): \(v = \dfrac{2s}{t} - u\).

4. Exam question Make 2 marks Core

Make \(x\) the subject of \(y = 4x + 7\).

Mark scheme — 2 marks available

  • \(y - 7 = 4x\) — M1
  • \(x = \dfrac{y - 7}{4}\) — A1

Model answer

Subtract 7 from both sides to get \(y - 7 = 4x\), then divide by 4 to get \(x = \dfrac{y - 7}{4}\).

5. Exam question Make 2 marks Stretch

Make \(r\) the subject of the formula \(A = \pi r^2\).

Mark scheme — 2 marks available

  • \(\dfrac{A}{\pi} = r^2\) — M1
  • \(r = \sqrt{\dfrac{A}{\pi}}\) — A1

Model answer

Divide both sides by \(\pi\) to get \(\dfrac{A}{\pi} = r^2\). Take the square root of both sides to get \(r = \sqrt{\dfrac{A}{\pi}}\).

6. Exam question Make 3 marks Core

Make \(a\) the subject of the formula \(v^2 = u^2 + 2as\).

Mark scheme — 3 marks available

  • \(v^2 - u^2 = 2as\) — M1
  • Divides both sides by \(2s\) — M1
  • \(a = \dfrac{v^2 - u^2}{2s}\) — A1

Model answer

Subtract \(u^2\) from both sides to get \(v^2 - u^2 = 2as\). Divide both sides by \(2s\) to get \(a = \dfrac{v^2 - u^2}{2s}\).

7. Exam question Work out 4 marks Core

The cost, \(C\) pounds, of hiring a bike for \(d\) days is given by \(C = 12 + 8d\). (a) Work out the cost of hiring the bike for 5 days. [1 mark] (b) Paul paid \(\pounds 84\). For how many days did he hire the bike? [2 marks] (c) Make \(d\) the subject of the formula. [1 mark]

Mark scheme — 4 marks available

  • (a) \(\pounds 52\) — B1
  • (b) \(8d = 72\) or \(84 - 12 = 72\) — M1
  • (b) 9 days — A1
  • (c) \(d = \dfrac{C - 12}{8}\) — B1

Model answer

(a) \(12 + 8 \times 5 = 12 + 40 = \pounds 52\). (b) \(12 + 8d = 84\), so \(8d = 72\) and \(d = 9\) days. (c) \(C - 12 = 8d\), so \(d = \dfrac{C - 12}{8}\).

8. Multiple choice 1 mark Core

Make \(a\) the subject of \(P = 2(a + b)\).

  1. A \(a = P - 2b\)
  2. B \(a = 2P - b\)
  3. C \(a = \dfrac{P}{2} - b\) Correct
  4. D \(a = \dfrac{P}{2} + b\)

Why: Divide both sides by 2 to get \(\dfrac{P}{2} = a + b\), then subtract \(b\).

9. Multiple choice 1 mark Core

Use \(A = \tfrac{1}{2}(a + b)h\) to work out \(A\) when \(a = 7\), \(b = 11\) and \(h = 5\).

  1. A 45 Correct
  2. B 18
  3. C 35
  4. D 90

Why: \(\tfrac{1}{2} \times 18 \times 5 = 45\).

10. Multiple choice 1 mark Easier

Work out the value of \(2a + b\) when \(a = 4\) and \(b = -3\).

  1. A \(-5\)
  2. B 5 Correct
  3. C 11
  4. D 1

Why: \(2 \times 4 + (-3) = 8 - 3 = 5\).

11. Multiple choice 1 mark Easier

Work out the value of \(x^2\) when \(x = -5\).

  1. A 25 Correct
  2. B \(-10\)
  3. C \(-25\)
  4. D 10

Why: \((-5) \times (-5) = 25\), because a negative multiplied by a negative is positive.

12. Multiple choice 1 mark Core

Work out the value of \(3x^2\) when \(x = -2\).

  1. A \(-36\)
  2. B 36
  3. C 12 Correct
  4. D \(-12\)

Why: The power is done first: \((-2)^2 = 4\), then \(3 \times 4 = 12\).

13. Multiple choice 1 mark Easier

Make \(x\) the subject of \(y = x + 7\).

  1. A \(x = 7 - y\)
  2. B \(x = y + 7\)
  3. C \(x = 7y\)
  4. D \(x = y - 7\) Correct

Why: Subtract 7 from both sides to get \(x = y - 7\).

14. Multiple choice 1 mark Core

Make \(x\) the subject of \(y = 4x - 1\).

  1. A \(x = \dfrac{y + 1}{4}\) Correct
  2. B \(x = 4(y + 1)\)
  3. C \(x = \dfrac{y - 1}{4}\)
  4. D \(x = \dfrac{y}{4} + 4\)

Why: Add 1 to both sides to get \(y + 1 = 4x\), then divide by 4.

15. Multiple choice 1 mark Core

Make \(t\) the subject of \(v = u + at\).

  1. A \(t = v - u - a\)
  2. B \(t = \dfrac{v - u}{a}\) Correct
  3. C \(t = \dfrac{v + u}{a}\)
  4. D \(t = a(v - u)\)

Why: Subtract \(u\) from both sides to get \(v - u = at\), then divide by \(a\).

16. Multiple choice 1 mark Core

What is the value of \((2x)^2\) when \(x = 3\)?

  1. A 12
  2. B 36 Correct
  3. C 6
  4. D 18

Why: \(2x = 6\), and \(6^2 = 36\). Note that \(2x^2\) would be \(2 \times 9 = 18\).

17. Multiple choice 1 mark Stretch

Make \(r\) the subject of \(A = \pi r^2\).

  1. A \(r = \dfrac{A^2}{\pi}\)
  2. B \(r = \dfrac{\sqrt{A}}{\pi}\)
  3. C \(r = \dfrac{A}{\pi}\)
  4. D \(r = \sqrt{\dfrac{A}{\pi}}\) Correct

Why: Divide by \(\pi\) to get \(r^2 = \dfrac{A}{\pi}\), then take the square root of both sides.