Edexcel International GCSE Mathematics A 4MA1: Expressions, formulae and linear equations – Revision Notes

These are condensed notes for the final weeks. For full explanations and longer worked examples, use the study guide for 2.3–2.4.

The notes cover sections 2.3 Expressions and formulae and 2.4 Linear equations of Topic 2, Equations, formulae and identities, in the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), for the January and June series examined on that specification. Both sections are on the Foundation (1F, 2F) and Higher (1H, 2H) papers. The extra Higher statement on changing the subject is marked Higher tier only. A calculator may be used on every paper.

When you have revised, try the practice questions for 2.3–2.4. Expanding and factorising, which you need here, are in the algebraic manipulation revision notes. The course hub and printable checklist cover the rest of 4MA1.

Key words (2.3A)

Term	Meaning	Example
Expression	Terms with no equals sign	5k − 2m
Equation	Two sides that are equal for a particular value of the unknown	4x − 1 = 11
Formula	A rule linking variables	P = 2l + 2w
Unknown	A letter standing for one particular number	x in 4x − 1 = 11 (x = 3)
Variable	A letter that can take many values	l and w in P = 2l + 2w
Subject	The letter on its own on one side of a formula	P in P = 2l + 2w

You simplify or evaluate an expression, solve an equation, and use or rearrange a formula.

Notation (2.3B)

• Number first, then letters: 7ab, not a7b.
• 1x is written x; −1y is written −y.
• n × n = n², n × n × n = n³, a × b = ab.
• Division is written as a fraction: (k + 1) ÷ 3 = (k + 1)/3.
• “Subtract 2 from p, then multiply by 5” is 5(p − 2). Brackets show the order.

Substitution (2.3C)

Method in steps

1. Write the expression out with each letter replaced by its value in brackets.
2. Powers first, then × and ÷, then + and −.
3. With fractions, work out the numerator and denominator separately before dividing.

Worked reminder. Evaluate 2a² − b when a = −3 and b = −0.5.

2(-3)^2 - (-0.5) = 2(9) + 0.5 = 18.5

Must-know: (−3)² = 9 but −3² = −9. On a calculator, type the brackets.

Formulae from words, diagrams and facts (2.3D–E)

Method in steps

1. Choose a letter for each quantity and state it with units (“let C be the cost in pounds”).
2. Translate each phrase: “per” means multiply, “plus a fixed charge” means add a constant.
3. Write the formula with the subject on the left.
4. Check it with an easy number.

Worked reminder (words). A plumber charges a call-out fee of £45 plus £30 per hour. The cost of h hours is C = 45 + 30h.

Worked reminder (diagram). A regular hexagon has sides of (x + 2) cm. Its perimeter is P = 6(x + 2) = 6x + 12.

Changing the subject (2.3F)

Undo the operations on the new subject in reverse order, doing the same to both sides.

Operation on the subject	Undo it by
+ a	− a
× a	÷ a
÷ a (subject in a numerator)	× a
squared	square root (positive root for a length)
square root	square
cubed	cube root

Worked reminder. Make v the subject of E = ½mv².

2E = mv^2
v^2 = 2E/m
v = √(2E/m)

Higher tier only: subject appears twice

Method in steps

1. Clear fractions (multiply by the denominator).
2. Expand any brackets.
3. Collect all terms with the subject on one side, all others on the other.
4. Factorise the subject out.
5. Divide by the bracket.

Worked reminder. Make y the subject of x = (y + 2)/(y − 5).

x(y - 5) = y + 2
xy - 5x = y + 2
xy - y = 5x + 2
y(x - 1) = 5x + 2
y = (5x + 2)/(x - 1)

With a power or root of the subject: isolate the power or root first, then apply the inverse (square to remove a square root, cube root to remove a cube).

Solving linear equations (2.4A)

Method in steps

1. Clear fractions: multiply every term by the lowest common multiple of the denominators.
2. Expand brackets, watching the sign in front of each bracket.
3. Collect the unknown on one side (the side where its coefficient is bigger keeps it positive).
4. Collect numbers on the other side.
5. Divide by the coefficient. Give an exact fraction or exact decimal.
6. Check by substituting into the original equation.

Worked reminder (minus before a bracket). Solve 6 − (x − 4) = 2x + 1.

6 - x + 4 = 2x + 1
10 - x = 2x + 1
9 = 3x
x = 3

Worked reminder (fraction on each side). Solve (x + 2)/4 = (x − 1)/3.

3(x + 2) = 4(x - 1)      (multiply both sides by 12)
3x + 6 = 4x - 4
x = 10

Fractional coefficients such as (3/4)x = 6 are solved the same way: multiply both sides by 4, then divide by 3, so x = 8.

The Higher tier example in the specification has three fractions in one equation. The method is identical: one LCM, multiply every term, then solve.

Setting up equations (2.4B)

Fact	Equation form
Angles in a triangle	sum = 180°
Angles on a straight line	sum = 180°
Angles in a quadrilateral	sum = 360°
Angles around a point	sum = 360°
Perimeter	sum of all sides
Consecutive integers	n, n + 1, n + 2
Consecutive odd or even numbers	n, n + 2, n + 4
“A is 5 more than B”	A = B + 5
“A is 3 times B”	A = 3B

Method: define the letter, write every quantity in terms of it, form the equation, solve, then answer the question asked (all the angles, both ages, the cost).

Must-know distinctions

• Unknown vs variable: an equation has a particular answer; a formula holds for many values.
• Expression vs equation: you cannot “solve” 3x + 2; you can only simplify or evaluate it.
• (−a)² vs −a²: for a = 4, (−4)² = 16 and −4² = −16.
• 4(x − 2) vs 4x − 2: clearing a fraction multiplies the whole of the other side.
• Subject once vs twice: once means undo in reverse order; twice (Higher tier only) means collect and factorise.
• Solving vs answering: finding x is often not the final answer; the question may want angles, ages or a total.

Quick self-test

1. Write p × q × p × 5 in its simplest form.
2. Evaluate 4m − n² when m = −1.5 and n = −3.
3. Evaluate (a + b)/c when a = 3/4, b = −1/4 and c = 1/8.
4. A bike costs £8 to hire plus £2.50 per hour. Write a formula for the cost £C for h hours.
5. Make t the subject of w = 6t − k.
6. Make r the subject of C = 2πr.
7. Solve 9 − 2x = 3x − 11.
8. Solve 2(3x − 1) = 7 − (x − 5).
9. Solve (x − 3)/2 + (x + 1)/3 = 4.
10. Angles on a straight line are 2y°, (3y + 20)° and (y − 8)°. Find y and the three angles.
11. (Higher tier only) Make x the subject of ax + 3 = bx − 5.
12. (Higher tier only) Make d the subject of k = 3√d + 2.

Answers

1. 5p²q
2. 4(−1.5) − (−3)² = −6 − 9 = −15
3. (3/4 − 1/4) ÷ 1/8 = 1/2 × 8 = 4
4. C = 8 + 2.5h
5. w + k = 6t, so t = (w + k)/6
6. r = C/(2π)
7. 9 + 11 = 3x + 2x, 20 = 5x, x = 4
8. 6x − 2 = 12 − x, 7x = 14, x = 2
9. Multiply by 6: 3(x − 3) + 2(x + 1) = 24, so 5x − 7 = 24 and x = 31/5 = 6.2
10. 6y + 12 = 180, 6y = 168, y = 28; angles 56°, 104°, 20° (sum 180°)
11. ax − bx = −8, x(a − b) = −8, so x = 8/(b − a) (equivalently −8/(a − b))
12. k − 2 = 3√d, √d = (k − 2)/3, so d = ((k − 2)/3)² = (k − 2)²/9

Where marks are usually lost

• Typing −3² instead of (−3)² when substituting a negative value.
• Multiplying only the first term when clearing a fraction, as in 4x − 2 for 4(x − 2).
• Expanding −(x − 4) as −x − 4 instead of −x + 4.
• Dividing only part of one side when changing the subject: x = 4y + 5/3 instead of x = (4y + 5)/3.
• Taking a square root before isolating the squared term.
• Leaving the subject on both sides of the final answer (Higher tier only).
• Rounding a solution such as x = 31/5 to 6 when an exact answer is expected.
• Stopping at the value of the letter when the question asks for the angles, the ages or the price.
• Writing a formula with no definition of the letters or units when the question asks you to state them.

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).