OCR GCSE Mathematics: Algebraic formulae, rearranging and kinematics (J560)

This guide follows the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026), and covers section 6.02, Algebraic formulae (6.02a to 6.02e). Most of it is for both tiers. Four formulae are marked Higher tier only where they appear. The specification says there is no formula sheet: every formula you must know is listed under 6.02d, and any other formula needed is given in the paper.

Related pages: revision notes, practice questions, exam preparation, the OCR GCSE Mathematics hub, the printable checklist and the free 10-minute diagnostics.

What this unit covers

Ref	What you must be able to do	Tier
6.02a	Formulate simple formulae and expressions from real-world contexts	Both tiers
6.02b	Substitute positive numbers into simple formulae; substitute positive or negative numbers into formulae with powers, roots and algebraic fractions	Both tiers
6.02c	Change the subject: subject once only; subject twice; power or reciprocal of the subject	Both tiers
6.02d	Recall and use circumference and area of a circle, Pythagoras’ theorem, SOH-CAH-TOA	Both tiers
6.02d	Recall and use the quadratic formula, sine rule, cosine rule, area = ½ab sin C	Higher tier only
6.02e	Use v = u + at, s = ut + ½at², v² = u² + 2as with constant acceleration	Both tiers

Paper 2 (Foundation) and Paper 5 (Higher) do not permit a calculator, so be ready to leave answers in terms of π or as a surd when a question asks.

6.02a Forming formulae

A formula is a rule written with letters. To build one, name each quantity with a letter, decide which parts change and which stay fixed, then connect them.

• A fixed charge is a number on its own.
• A charge per unit is a number multiplied by the letter.
• Add the two.

Worked example 1. A plumber charges £35 for the call-out plus £22 for each hour worked. Write a formula for the total cost £C for h hours, and find the cost of 3.5 hours.

C = 35 + 22h
h = 3.5:  C = 35 + 22 × 3.5 = 35 + 77 = 112

The cost is £112.

Worked example 2. A rectangle has length 2 cm less than three times its width w cm. Write a formula for the perimeter P, then find w when P = 44.

length = 3w − 2
P = 2(w + 3w − 2) = 2(4w − 2) = 8w − 4
44 = 8w − 4  →  8w = 48  →  w = 6

The width is 6 cm and the length is 16 cm. Check: 2 × (6 + 16) = 44.

6.02b Substituting into formulae

Replace each letter by its value, put every substituted number in brackets, then work in the usual order: brackets, powers and roots, multiplication and division, addition and subtraction.

Worked example 3 (simple). A turkey takes T = 20m + 40 minutes to cook, where m is its mass in kg. Find T when m = 2.5.

T = 20 × 2.5 + 40 = 50 + 40 = 90 minutes

Worked example 4 (negatives and powers). Find y when x = −2 in y = 2x³ − x² + 5.

y = 2(−2)³ − (−2)² + 5
  = 2(−8) − 4 + 5
  = −16 − 4 + 5
  = −15

The trap is the second term: (−2)² is +4, so −x² gives −4. Written without brackets, −2² would mean −4, a different value.

Worked example 5 (algebraic fraction). Find K when a = −3 and b = −5 in K = (a² − b)/(a + b).

numerator:   (−3)² − (−5) = 9 + 5 = 14
denominator: −3 + (−5) = −8
K = 14/(−8) = −7/4 = −1.75

Evaluate the top and bottom separately before dividing. If a formula contains a square root, work out everything under the root first. A root sign covers the whole expression beneath it.

Worked example 5b (root). Find T when L = 2.5 and g = 10 in T = 2π√(L/g).

L/g = 2.5/10 = 0.25
T = 2π × √0.25 = 2π × 0.5 = π

6.02c Changing the subject

The subject is the letter on its own, usually on the left. To change it, do the same operation to both sides, undoing operations in the reverse of the order in which they were applied to the new subject.

Subject appears once

Worked example 6. Make l the subject of P = 2l + 2w.

P − 2w = 2l
l = (P − 2w)/2     (or  l = P/2 − w)

Worked example 7. Make b the subject of A = ½(a + b)h.

2A = (a + b)h
2A/h = a + b
b = 2A/h − a

Power or root of the subject

Isolate the power or root first, then undo it. Undo a square with a square root, and undo a square root by squaring.

Worked example 8. Make r the subject of A = πr².

A/π = r²
r = √(A/π)

Worked example 9. Make y the subject of x = 3√(y/z).

x/3 = √(y/z)
x²/9 = y/z
y = x²z/9

Square the whole of the left side, including the 3, not just x.

Worked example 9b. Make t the subject of s = ½at², and of v = x/t.

s = ½at²  →  2s = at²  →  t² = 2s/a  →  t = √(2s/a)
v = x/t   →  vt = x    →  t = x/v

Worked example 10. Rearrange v² = u² + 2as to give a formula for u.

v² − 2as = u²
u = √(v² − 2as)

Subject appears twice

Remove any fraction, expand brackets, collect every term containing the subject on one side, factorise, then divide.

Worked example 11. Make t the subject of h = (3t − 5)/(t + 2).

h(t + 2) = 3t − 5
ht + 2h = 3t − 5
ht − 3t = −5 − 2h
t(h − 3) = −5 − 2h
t = (−5 − 2h)/(h − 3)  =  (2h + 5)/(3 − h)

Reciprocal of the subject

Worked example 12. Make v the subject of 1/f = 1/u + 1/v.

1/v = 1/f − 1/u = (u − f)/(fu)
v = fu/(u − f)

Do not flip each term separately. Combine into a single fraction first, then invert the whole thing.

6.02d Standard formulae

You must recall these, since none is supplied.

Formula	Use
C = 2πr = πd	Circumference of a circle
A = πr²	Area of a circle
a² + b² = c²	Pythagoras, c is the hypotenuse
sin θ = o/h, cos θ = a/h, tan θ = o/a	Right-angled triangles

Worked example 13. A pond has diameter 6.4 m. Find its circumference and area to 3 significant figures.

C = π × 6.4 = 20.1 m
r = 3.2,  A = π × 3.2² = 32.2 m²

Worked example 14. A circle has circumference 18π cm. Find its area in terms of π.

2πr = 18π  →  r = 9
A = π × 9² = 81π cm²

Worked example 15. A ladder 4.5 m long leans on a wall with its foot 1.2 m from the wall. How high up the wall does it reach?

height² + 1.2² = 4.5²
height² = 20.25 − 1.44 = 18.81
height = √18.81 = 4.34 m

To find a shorter side, subtract. To find the hypotenuse, add.

Worked example 16. In a right-angled triangle the side opposite angle θ is 7 cm and the hypotenuse is 12 cm. Find θ. Then find the side opposite a 38° angle when the adjacent side is 9.5 cm.

sin θ = 7/12  →  θ = sin⁻¹(7/12) = 35.7°
tan 38° = opposite/9.5  →  opposite = 9.5 × tan 38° = 7.42 cm

Higher tier only: quadratic formula

For ax² + bx + c = 0, Higher tier only:

x = (−b ± √(b² − 4ac))/(2a)

Worked example 17. Solve 4x² − 5x − 3 = 0, giving answers to 2 decimal places.

a = 4, b = −5, c = −3
b² − 4ac = 25 + 48 = 73
x = (5 ± √73)/8
x = 1.69  or  x = −0.44

Higher tier only: sine rule, cosine rule and area

For a triangle with sides a, b, c opposite angles A, B, C (Higher tier only):

Sine rule:    a/sin A = b/sin B = c/sin C
Cosine rule:  a² = b² + c² − 2bc cos A
Area:         ½ab sin C

Use the sine rule when you know a side and its opposite angle plus one more piece. Use the cosine rule when you know two sides and the angle between them, or all three sides.

Worked example 18 (sine rule). In triangle ABC, A = 48°, B = 71° and a = 8.5 cm. Find b.

b/sin 71° = 8.5/sin 48°
b = 8.5 × sin 71° / sin 48° = 10.8 cm

Worked example 19 (cosine rule, side). In triangle ABC, b = 6.4 cm, c = 8.3 cm and A = 57°. Find a.

a² = 6.4² + 8.3² − 2 × 6.4 × 8.3 × cos 57°
a² = 51.99
a = 7.21 cm

Worked example 20 (cosine rule, angle). A triangle has sides 5, 7 and 9. Find the largest angle C, opposite the side 9.

cos C = (5² + 7² − 9²)/(2 × 5 × 7) = (−7)/70 = −0.1
C = 95.7°

A negative cosine means an obtuse angle.

Worked example 21 (area). Two sides of a triangle are 6.4 cm and 9.1 cm with 54° between them.

Area = ½ × 6.4 × 9.1 × sin 54° = 23.6 cm²

6.02e Kinematics formulae

These apply when acceleration a is constant. Here u is initial velocity, v is final velocity, s is displacement from the position at t = 0, and t is time. The specification says you do not need to know the definition of each letter, so a question will tell you, but learn them anyway.

v = u + at
s = ut + ½at²
v² = u² + 2as

Choose the formula that contains the three quantities you are given and the one you want.

Worked example 22. A car moving at 12 m/s accelerates at 1.5 m/s² for 8 s. Find its final velocity and the distance covered.

v = 12 + 1.5 × 8 = 24 m/s
s = 12 × 8 + ½ × 1.5 × 8² = 96 + 48 = 144 m

Worked example 23. A ball is thrown upwards at 14 m/s. Taking upwards as positive, a = −9.8 m/s². How high does it rise before stopping?

At the top v = 0:  0 = 14² + 2(−9.8)s
196 = 19.6s  →  s = 10 m

Worked example 24. A train slows from 30 m/s to rest over 150 m. Find a.

0 = 30² + 2a(150)
a = −900/300 = −3 m/s²

The negative sign shows deceleration.

Worked example 25 (Higher tier only). An object starts at 2 m/s with a = 3 m/s². How long does it take to travel 20 m?

20 = 2t + 1.5t²  →  3t² + 4t − 40 = 0
t = (−4 ± √(16 + 480))/6 = (−4 ± 22.27)/6
t = 3.05 s   (reject the negative root)

Common errors

• Squaring a negative without brackets, so −3² is read as −9 instead of 9.
• Multiplying the fixed charge by the hours, giving 57h instead of 35 + 22h.
• Moving a term across an equals sign without changing its sign.
• Leaving the subject on both sides, as in l = P − 2w − l.
• Dividing only one term by a number, such as (P − 2w)/2 written as P − w.
• Using Pythagoras to find a short side by adding instead of subtracting.
• Using the sine rule when the given angle is not opposite a given side.
• Stopping at ± in the quadratic formula without calculating both values.

Next steps

Condense this unit using the revision notes, then test yourself with the practice questions.

Official syllabus

OCR GCSE (9-1) Mathematics J560, Version 2.1 (August 2026), for first assessment in 2017, published by OCR. Section 6.02, Algebraic formulae.