1. KS4 Mathematics
A3 Formulae

2. A3.1 Substituting into formulae
Contents
A3 Formulae
A3.1 Substituting into formulae
A3.2 Problems that lead to equations to solve
A3.3 Changing the subject of a formula
A3.4 Manipulating more difficult formulae
A3.5 Generating formulae

3. Formulae
A formula is a special type of equation that links two or more physical variables.
For example in the formula,
P = 2(l + w)
P represents the perimeter of a rectangle and l and w represent its length and width.
We can use this formula to work out the perimeter of any rectangle given its length and width.
Remind pupils that the letter symbols in formulae are called variables because their value can vary. Unlike equations where the unknown can be any real number the variables in formulae are often limited to values determined by the context. For example, if the variable in a formula represents a length it can only take positive values. If it represents a number of objects then it can only take whole number values.
We do this by substituting the values we are given into the formula.

4. Formulae
Because formulae deal mainly with real-life quantities such as length, mass, temperature or time, the given variables often have units attached.
Units shouldn’t be included in the formula itself.
The units that have to be used are usually defined in the formula. For example,
S = d t
Point out that the letters representing each variable often start with the same letter as the word for the variable they are representing.
This formula doesn’t mean much unless we say “S is the average speed in m/s, d is the distance travelled in metres, and t is the time taken in seconds”.

5. Formulae S = d t Use the formula to find the speed of a car that
travels 2 km in 1 minute and 40 seconds.
Write the distance and the time using the correct units before substituting them into the formula,
2 kilometres = 2000 metres
1 minute and 40 seconds = 100 seconds
Now substitute these numerical values into the formula,
Ask pupils how we could have found the average speed of the car in km/h rather than m/s.
S =
2000
100
We can write the units at the end.
= 20 m/s

6. Substituting into formulaeh w l
The surface area S of a cuboid is given by the formula
S = 2lw + 2lh + 2hw
In formulae involving length, area and volume the units are not usually given. However, we must make sure that all the units are the same before substituting them into the formula.
where l is the length, w is the width and h is the height.
What is the surface area of a cuboid with a length of 1.5 m, a width of 32 cm and a height of 250 mm?

7. Substituting into formulae
What is the surface area of a cuboid with a length of 1.5 m, a width of 32 cm and a height of 250 mm?
Before we can use the formula we must write all of the amounts using the same units.
l = 150 cm, w = 32 cm and h = 25 cm
Next, substitute the values into the formula without the units.
S = 2lw + 2lh + 2hw
Remind pupils that 1 m2 = 1 m × 1 m = 100 cm × 100 cm = cm2 and so establish that cm2 is equal to 1.87 m2. The units used in the answer must be consistent with the units used in the formula. That doesn’t mean they’re the same units (as can be seen here, where cm lead to cm2).
= (2 × 150 × 32) + (2 × 150 × 25) + (2 × 32 × 25) =
= cm2
Don’t forget to write the units in at the end.

8. Substituting into formulae
The distance d, in metres, that an object falls after being dropped is given by the formula,
d = 4.9t2
where t is the time in seconds.
Suppose a boy drops a rock from a 100 metre high cliff.
How far will the rock have fallen after:
a) 2 seconds
b) 3 seconds
c) 5 seconds?
When t = 2,
When t = 3,
When t = 5,
Explain that this formula is true for any object (whatever its mass) that is dropped on earth (disregarding air resistance). This is because gravity causes objects to fall with the same acceleration regardless of their mass.
Talk through each substitution.
There is something wrong with the answer to part c. Ask pupils if they know what it is.
The cliff is only 100 metres high so the rock cannot fall metres. We can therefore conclude that the rock would have hit the ground by then.
d =
4.9 × 22
d =
4.9 × 32
d =
4.9 × 52
= 4.9 × 4
= 4.9 × 9
= 4.9 × 25
= 19.6 metres
= 44.1 metres
= metres

9. Substituting into formulae

10. A3.2 Problems that lead to equations to solve
Contents
A3 Formulae
A3.1 Substituting into formulae
A3.2 Problems that lead to equations to solve
A3.3 Changing the subject of a formula
A3.4 Manipulating more difficult formulae
A3.5 Generating formulae

11. Problems that lead to equations to solve
Formulae are usually (but not always) arranged so that a single variable is written on the left-hand side of the equals sign. For example, in the formula
v = u + at
v is called the subject of the formula.
If we are given the values of u, a and t, we can find v by substituting these values into the formula.
Suppose instead that we are given to values of v, u and a, and asked to find t.
Explain that if the unknown is not in the subject of a formula we will be left with an equation to solve.
When these values are substituted we are left with an equation to solve in t.

12. Problems that lead to equations to solve
For example, suppose v = 20, u = 5 and a = 3. Find t.
Substituting these values into v = u + at gives us the equation,
20 = 5 + 3t
We can then solve this equation as usual.
swap sides,
5 + 3t = 20
Discuss how to solve this equation. We start by swapping the left-hand side with the right-hand side so that the unknown is on the left-hand side as required.
subtract 5 from both sides,
3t = 15
divide both sides by 3.
t = 5 seconds

13. Problems that lead to equations to solve
The formula used to find the area A of a trapezium with parallel sides a and b and height h is:
A = (a + b)h 1 2
We are told the area and the lengths of the parallel sides and we need to find the height.
Explain that we can substitute the values we are given into the formula to make an equation with one unknown, h.
An alternative would be to rewrite the formula with h as the subject. This is demonstrated in A3.3 Changing the subject of a formula.
What is the height of a trapezium with an area of 40 cm2 and parallel sides of length 7 cm and 9 cm?

14. Problems that lead to equations to solve
A = (a + b)h 1 2
Substituting A = 40, a = 7 and b = 9 into gives
40 = (7 + 9)h 1 2
40 = ×16h 1 2
Simplifying,
40 = 8h
swap both sides,
8h = 40
This slide shows how to solve the problem by substituting the given values to make an equation.
Check the solution by substituting h = 5, a = 7 and b = 9 into the original formula to make sure that it gives an area of 40 cm2.
divide by 8.
h = 5
So the height of the trapezium is 5 cm.

15. Problems that lead to equations to solve

16. A3.3 Changing the subject of a formula
Contents
A3 Formulae
A3.1 Substituting into formulae
A3.2 Problems that lead to equations to solve
A3.3 Changing the subject of a formula
A3.4 Manipulating more difficult formulae
A3.5 Generating formulae

17. The subject of a formula
Here is a formula you may know from physics:
V = IR
where V is voltage, I is current and R is resistance.
V is called the subject of the formula.
The subject of a formula always appears in front of the equals sign without any other numbers or operations.
Sometimes it is useful to rearrange a formula so that one of the other variables is the subject of the formula.
Suppose, for example, that we want to make I the subject of the formula V = IR.

18. Changing the subject of the formula
V is the subject of this formula
The formula:
V = IR
can be written as a function diagram: I
× R V
The inverse of this is: I
÷ R V
Ask pupils what do we do to I to get V and establish that we multiply it by R.
Reveal the first diagram showing the operation × R.
Ask pupils how we can find the inverse of this.
Reveal the second diagram corresponding to V ÷ R = I which gives us the formula I = V/R.
Give a numerical example. For example, ask pupils to give you the value of I when V=12 Volts and R=3 Ohms.
Ask pupils how we could make R the subject of the formula (R = V/I).
So:
I is now the subject of this formula
I = V R

19. Matchstick pattern Look at this pattern made from matchsticks:
Number, n 1 2 3 4
Number of
Matches, m 3 5 7 9
The formula for the number of matches, m, in pattern number n is given by the formula:
Go through each step on the slide and then ask,
If we are given m, in this case m = 47, how can we find n?
What have we done to n?
We’ve multiplied it by 2 and added 1.
What is the inverse of this? How do we ‘undo’ times 2 and add 1. Remember, we have to reverse the order of the operations as well as the operations themselves.
Establish that we need to subtract 1 and divide by 2. (47 – 1) ÷ 2 is 23.
We can check that this is correct by verifying that 2 x = 47.
m = 2n + 1
Which pattern number will contain 47 matches?

20. Changing the subject of the formula
m is the subject of this formula
The formula:
m = 2n + 1
can be written as a function diagram: n
× 2
+ 1 m
The inverse of this is: n
÷ 2
– 1 m
We can rearrange this formula using inverse operations.
Writing the formula as n = (m – 1)/2 allows us to find the pattern number given the number of matches.
n is the subject of this formula or
n =
m – 1 2

21. Changing the subject of the formula
To find out which pattern will contain 47 matches, substitute 47 into the rearranged formula.
n =
m – 1 2
n =
47 – 1 2
n = 46 2
Check the solution by substituting 23 into the original formula m = 2n + 1 to get 47.
n = 23
So, the 23rd pattern will contain 47 matches.

22. Changing the subject of the formula
We can also change the subject by performing the same operations on both sides of the equals sign. For example, to make C the subject of
F = 9C 5
subtract 32,
F – 32 = 9C 5
multiply by 5,
5(F – 32) = 9C
This formula converts degrees Celsius to degrees Fahrenheit. This slide demonstrates how to change the subject of the formula using inverse operations.
Ask pupils how we could write the formula using functions. We could start with the input C, multiply it by 9, divide it by 5 and add 32. The inverse of this is to start with F, subtract 32, multiply by 5 and divide by 9.
Remind pupils that we are trying to rearrange the formula so that the C appears to the left of the equals sign on its own.
divide by 9.
5(F – 32) 9
= C
5(F – 32) 9
C =

23. Change the subject of the formula 1
Ask a volunteer to come to the board and use the pen tool to change the subject of the given formula.
Ask them to justify each step in their working.

24. Find the equivalent formulae
Ask pupils to decide which of the formulae shown can be rearranged to give the formula shown.

25. A3.4 Manipulating more difficult formulae
Contents
A3 Formulae
A3.1 Substituting into formulae
A3.2 Problems that lead to equations to solve
A3.3 Changing the subject of a formula
A3.4 Manipulating more difficult formulae
A3.5 Generating formulae

26. Formulae where the subject appears twice
Sometimes the variable that we are making the subject of a formula appears twice. For example,
S = 2lw + 2lh + 2hw
where S is the surface area of a cuboid, l is its length, w is its width and h is its height.
Make w the subject of the formula.
To do this we must collect all terms containing a w on the same side of the equals sign.
We can then isolate w by factorizing.

27. Formulae where the subject appears twice
S = 2lw + 2lh + 2hw
Let’s start by swapping the left-hand side and the right-hand side so that the terms with w’s are on the left.
2lw + 2lh + 2hw = S
subtract 2lh from both sides,
2lw + 2hw = S – 2lh
factorize,
w(2l + 2h) = S – 2lh
Notice that we do not factorize 2lw + 2hw completely. We only take out the w to isolate it.
w =
S – 2lh
2l + 2h
divide by 2l + 2h.

28. Formulae involving fractions
When a formula involves fractions we usually remove these by multiplying before changing the subject.
For example, if two resistors with a resistance a and b ohms respectively, are arranged in parallel their total resistance R ohms can be found using the formula, 1 R = a + b aΩ bΩ
Although R is on the left-hand side of this formula it can be rearranged to be in in form R = …
Make R the subject of the formula

29. Formulae involving fractions1 R = a + b
multiply both sides by Rab, = +
Rab R a b
simplify,
ab = Rb + Ra
factorize,
ab = R(b + a)
divide both sides by a + b,
= R ab
a + b
Notice that we do not factorize 2lw + 2hw completely. We only take out the w to isolate it.
R = ab
a + b

30. Formulae involving powers and roots
The length c of the hypotenuse of a right-angled triangle is given by
c = √a2 + b2
where a and b are the lengths of the shorter sides.
Make a the subject of the formula
square both sides,
c2 = a2 + b2
subtract b2 from both sides,
c2 – b2 = a2
square root both sides,
√c2 – b2 = a
a = √c2 – b2

31. Formulae involving powers and roots
The time T needed for a pendulum to make a complete swing is
T = 2π l g
where l is the length of the pendulum and g is acceleration due to gravity.
Make l the subject of the formula
When the variable that we wish to make the subject appears under a square root sign, we should isolate it on one side of the equation and then square both sides.

32. Formulae involving powers and roots
divide both sides by 2π, T 2π = l g
square both sides, T2
4π2 = l g
multiply both sides by g,
T2g
4π2
= l
l =
T2g
4π2

33. Change the subject of the formula 2
Ask a volunteer to come to the board and use the pen tool to change the subject of the given formula.
Ask them to justify each step in their working.

34. A3 Formulae Contents A3.1 Substituting into formulae
A3.2 Problems that lead to equations to solve
A3.3 Changing the subject of a formula
A3.4 Manipulating more difficult formulae
A3.5 Generating formulae

35. Writing formulae Write a formula to work out,
The cost, c, of b boxes of crisps at £3 each.
c = 3b
The distance left, d, of a 500 km journey after travelling k km.
d = 500 – k
For each example substitute some actual values to help pupils determine the correct operations in each formula.
The cost per person, c, if a meal costing m pounds is shared between p people.
c = m p

36. Writing formulae
The number of seats in a theatre, n, with 25 seats in each row, r.
n = 25r
The age of a boy Andy, a, if he is 5 years older than his sister Betty, b.
a = b + 5
The average weight, w, of Alex who weighs a kg, Bob who weighs b kg and Claire who weighs c kg.
w =
a + b + c 3
1 of 20

37. Writing formulae
A window cleaner charges a £10 call out fee plus £7 for every window that he cleans. Write a formula to find the total cost C when n windows are cleaned.
C = 7n + 10
Using this formula, how much would it cost to clean all 105 windows of Formula Mansion?
We substitute the value into the formula,
Talk through this example using the formula.
C = 7 × =
= 745
It will cost £745.

38. Writing formulae
C = 7n + 10
At another house the window cleaner made £94.
How many windows did he have to clean?
We substitute this value into the formula to give an equation,
94 = 7n + 10
swap both sides,
7n + 10 = 94
Talk through this example using the formula.
subtract 10 from both sides,
7n = 84
n = 12
There are 12 windows in the house.

39. Using formulae to write mathematical rules
When conducting a mathematical investigation, it is usually necessary to use formula to write rules and generalizations.
For example, Sophie is investigating patterns of shaded squares on a numbered grid.
She starts by looking at arrangements of numbers in two by two squares on a 100 grid. For example,
She notices that the sum of the numbers in this square is always equal to four times the number in the top left-hand square plus 22 and writes this as a formula. 34 35 44 45
Explain to pupils that they need to be able to generate formulae using both words and symbols in the context of most of the investigational work that they will do.
Such formulae should be justified by the context of the problem as shown on the following slide.

40. Number grid patterns
Discuss how each number in the pattern can be written in terms of one of the numbers. Ask pupils to five formulae for various arrangements and discuss any patterns that pupils notice.