Edexcel International GCSE Mathematics A 4MA1: Expressions, formulae and linear equations – Study Guide

This study guide covers sections 2.3 Expressions and formulae and 2.4 Linear equations from Topic 2, Equations, formulae and identities, of the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), for the January and June series examined on that specification. 4MA1 is tiered: Foundation papers 1F and 2F and Higher papers 1H and 2H, each a 2-hour, 100-mark paper, and a calculator may be used on all of them. Almost everything here is on both tiers. The one extra Higher statement, changing the subject when it appears twice or appears with a power or root, is marked Higher tier only.

Index notation, expanding and factorising (sections 2.1–2.2) are covered in the use of symbols and algebraic manipulation study guide, with its own revision notes and practice questions. You need expanding brackets from there before you start the equations below. When you finish this guide, use the revision notes for 2.3–2.4 and then the practice questions for 2.3–2.4. The Edexcel IGCSE Mathematics course hub links the rest of the course, and the printable checklist lists every statement.

What this unit covers

Section	What you must be able to do	Tier
2.3A	Understand that a letter may stand for an unknown number or a variable	Both
2.3B	Use correct notation for expressions and formulae	Both
2.3C	Substitute positive and negative integers, decimals and fractions into expressions and formulae	Both
2.3D	Turn a formula given in words or in a diagram into letters and symbols, and use it	Both
2.3E	Derive a formula or expression	Both
2.3F	Change the subject of a formula where the subject appears once	Both
2.3A (Higher)	Change the subject where the subject appears twice or a power of the subject occurs	Higher tier only
2.4A	Solve linear equations with integer or fractional coefficients, unknown on one or both sides	Both
2.4B	Set up simple linear equations from given data	Both

For 2.4 the Higher content says “See Foundation Tier”: the skill is the same, but the Higher example in the specification has several fractions in one equation, so expect harder versions on 1H and 2H.

2.3A–B Letters, unknowns and notation

A letter can play two different roles.

• In an equation such as 3x + 4 = 19, x is an unknown: one particular number makes the equation true (here x = 5).
• In a formula such as A = lw, the letters are variables: they can take many values, and the formula links them.
• An expression such as 4a − 7 has no equals sign. You can simplify or evaluate it, but you cannot solve it.

Correct notation is what examiners expect to see in a final answer:

• Write 5n, not n5 or 5 × n. Number first, then letters, usually in alphabetical order: 3ab, not b3a.
• Write x, not 1x; write −y, not −1y.
• Write a × a as a², and p × p × p as p³.
• Write division as a fraction: m ÷ 4 becomes m/4.
• Use brackets to show what is grouped: “add 3 to n, then double” is 2(n + 3), not 2n + 3.

2.3C Substitution

Replace each letter with its value in brackets, then follow the order of operations (brackets, powers, ×÷, +−). Brackets matter most when a value is negative.

Worked example 1. Evaluate 3p² − 2pq when p = −2 and q = 1.5.

3p^2 - 2pq
= 3(-2)^2 - 2(-2)(1.5)
= 3(4) - (-6)
= 12 + 6
= 18

The power comes first: (−2)² = 4, so 3p² = 12, not (3 × −2)² = 36.

Worked example 2. The formula s = ut + ½at² gives distance travelled. Find s when u = 12, t = 4 and a = −2.5.

s = (12)(4) + 0.5(-2.5)(4)^2
  = 48 + 0.5(-2.5)(16)
  = 48 - 20
  = 28

Worked example 3 (fractions). Find the value of (2x + 1)/(x − 1) when x = 2/3.

Numerator:   2(2/3) + 1 = 4/3 + 3/3 = 7/3
Denominator: 2/3 - 1 = -1/3
Value: (7/3) / (-1/3) = (7/3) x (-3) = -7

Work out the numerator and denominator separately, write both down, then divide.

2.3D Formulae from words and diagrams

A formula in words becomes a formula in symbols when you choose a letter for each quantity and say what it stands for, including units.

Worked example 4. A recipe says: “Allow 40 minutes per kilogram, plus an extra 25 minutes.” Write a formula for the cooking time T minutes for a joint of mass w kg, and find T when w = 2.5.

T = 40w + 25
w = 2.5:  T = 40(2.5) + 25 = 100 + 25 = 125 minutes

Working backwards, T = 165 gives 40w + 25 = 165, so w = 3.5 kg: a linear equation from section 2.4.

For a diagram, read off each length or quantity, label anything not given, and combine them. Perimeter means add every outside edge; area of a compound shape means split it into rectangles or triangles first.

2.3E Deriving a formula or expression

Deriving means building the formula yourself from the facts.

Worked example 5. The length of a rectangle is 3 cm more than twice its width, w cm. Find formulae for its perimeter P and its area A.

Length = 2w + 3
P = 2(w + 2w + 3) = 2(3w + 3) = 6w + 6
A = w(2w + 3) = 2w^2 + 3w

If you are told P = 45, then 6w + 6 = 45, 6w = 39 and w = 6.5 cm.

Always simplify a derived formula fully, and check it with a quick number: if w = 1, the rectangle is 1 by 5, so P = 12 and 6(1) + 6 = 12.

2.3F Changing the subject (subject appears once)

The subject is the letter on its own on one side. To change it, undo what has been done to the new subject, in reverse order, doing the same to both sides.

Worked example 6. Make x the subject of y = (3x − 5)/4.

y = (3x - 5)/4
4y = 3x - 5          (multiply both sides by 4)
4y + 5 = 3x          (add 5)
x = (4y + 5)/3       (divide by 3)

Worked example 7. Make h the subject of A = ½(a + b)h, then find h when A = 42, a = 5 and b = 9.

2A = (a + b)h
h = 2A/(a + b)
h = 2(42)/(5 + 9) = 84/14 = 6

Worked example 8. Make r the subject of S = 4πr².

S/(4π) = r^2
r = √(S/(4π))

The Foundation content includes this kind of case (the specification’s Foundation example is A = πr²). In a real context r is a length, so take the positive root.

Higher tier only: subject twice, powers and roots

Subject appears twice

Collect every term containing the new subject on one side, everything else on the other, then factorise the subject out and divide.

Worked example 9. Make x the subject of p = (2x + 3)/(x − 1).

p(x - 1) = 2x + 3        (multiply by the denominator)
px - p = 2x + 3          (expand)
px - 2x = p + 3          (x-terms left, others right)
x(p - 2) = p + 3         (factorise)
x = (p + 3)/(p - 2)

Check with a number: p = 5 gives x = 8/3, and (2(8/3) + 3)/(8/3 − 1) = (25/3)/(5/3) = 5. Correct.

A power or root of the subject

Undo a power with the matching root and a root with the matching power, at the right stage.

Worked example 10. Make r the subject of V = (2/3)πr³, the volume of a hemisphere. Find r when V = 144π.

3V = 2πr^3
r^3 = 3V/(2π)
r = ∛(3V/(2π))
V = 144π:  r = ∛(432π/(2π)) = ∛216 = 6

Worked example 11. Make x the subject of y = √((x + 3)/2).

y^2 = (x + 3)/2      (square both sides first)
2y^2 = x + 3
x = 2y^2 - 3

2.4A Solving linear equations

A linear equation has the unknown only to the power 1. Keep it balanced: whatever you do to one side, do to the other. A good order is: clear fractions, expand brackets, collect the unknown on one side, then divide.

Worked example 12 (brackets, unknown on both sides). Solve 4(2x − 3) = 5x + 9.

8x - 12 = 5x + 9
3x - 12 = 9
3x = 21
x = 7

Worked example 13 (negative answer). Solve 5 − 2(x + 4) = 3x + 7.

5 - 2x - 8 = 3x + 7      (-2 multiplies both terms)
-3 - 2x = 3x + 7
-10 = 5x
x = -2

Collecting the x-terms on the side where they stay positive (here the right) cuts sign errors.

Worked example 14 (fractional coefficient). Solve (2/5)x − 3 = 1.

(2/5)x = 4
x = 4 x 5/2 = 10

Worked example 15 (unknown in a numerator). Solve (3x + 1)/4 = x − 2.

3x + 1 = 4(x - 2)        (multiply BOTH sides by 4)
3x + 1 = 4x - 8
9 = x
x = 9

Worked example 16 (two fractions, Higher style). Solve (x + 4)/3 − (2x − 1)/5 = 1.

LCM of 3 and 5 is 15. Multiply every term by 15:
5(x + 4) - 3(2x - 1) = 15
5x + 20 - 6x + 3 = 15    (-3 x -1 = +3)
-x + 23 = 15
x = 8

Check by substituting: (12)/3 − (15)/5 = 4 − 3 = 1. Correct.

Answers need not be whole numbers. Give an exact fraction or a decimal that terminates (x = 5.5 or 11/2); don’t round unless asked.

2.4B Setting up equations from given data

Let a letter stand for the unknown, write every other quantity in terms of it, form one equation from a fact you know, then solve and answer the actual question.

Facts you will use most often: angles in a triangle add to 180°, on a straight line to 180°, in a quadrilateral to 360°; perimeter is the sum of the sides; consecutive whole numbers are n, n + 1, n + 2; consecutive odd or even numbers are n, n + 2, n + 4.

Worked example 17. Three consecutive odd numbers add up to 87. Find them.

n + (n + 2) + (n + 4) = 87
3n + 6 = 87
3n = 81
n = 27        Numbers: 27, 29, 31

Worked example 18. Sana is 4 years older than Omar. In 6 years’ time the sum of their ages will be 50. How old is each now?

Omar now: x       Sana now: x + 4
In 6 years: (x + 6) + (x + 10) = 50
2x + 16 = 50
x = 17           Omar is 17, Sana is 21
Check: 23 + 27 = 50

The final sentence matters: the question asked for both ages, so x = 17 alone is incomplete.

Using your calculator

A calculator is allowed on every 4MA1 paper and helps with substitution involving decimals, fractions and π. Rearranging and solving must still be shown line by line, because most marks are for method.

Common errors

• Writing −2² for (−2)². With a negative value, (−2)² = 4 but −2² = −4.
• Multiplying only the first term when clearing a fraction: (3x + 1)/4 = x − 2 needs 4(x − 2), not 4x − 2.
• Losing the sign when subtracting a fraction: −3(2x − 1) = −6x + 3.
• Dividing only one term when changing the subject: from 4y + 5 = 3x, x = 4y + 5/3 is wrong; the whole of 4y + 5 is divided by 3.
• Taking the root too early: in y = √((x + 3)/2), square first, then rearrange.
• Higher: leaving the subject on both sides, as in x = (p + 3 + 2x)/p. The subject must not appear on the right.
• Solving for the letter and not answering the question (giving x = 17 when the question asked for both ages).

Next steps

Test recall with the revision notes, then try the practice questions. To find gaps elsewhere, use the free 10-minute diagnostics.

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification, Issue 2, November 2017, Pearson Education Limited (first assessment June 2018). Topic 2, Equations, formulae and identities: section 2.3 Expressions and formulae and section 2.4 Linear equations (Foundation and Higher tier content).