Chapter 8 · Higher tier

GCSE Maths Substitution and Formulae (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

Use the words expression, equation, formula, identity, term and subject correctly
Write expressions and formulae from words and from diagrams
Substitute positive and negative numbers into expressions, working in the correct order (BIDMAS)
Use scientific and geometric formulae such as speed, density, force, kinetic energy, area and volume
Change the subject of a formula using inverse operations (one step, two steps, brackets, fractions, squares and roots)
Higher: change the subject of a formula when the subject appears twice, by collecting terms and factorising

You should already know

Negative numbers and BIDMAS (Chapter 1)
Squares, cubes and square roots (Chapter 2)
Fractions (Chapter 3)
Collecting like terms and algebra notation (Chapter 7)

Where this leads

Solving linear equations (Chapter 9)
Straight-line graphs, y = mx + c (Chapter 12)
Speed, density and other compound measures (Chapter 16)
Areas and volumes (Chapters 18 and 19)

Edexcel GCSE (9–1) Mathematics 1MA1, Higher tier. Content references: A2, A5, A21.

Printable worksheets. Every lesson in this chapter has a free worksheet and an answer sheet to download and print.

8.1 Letters, expressions and formulae: Worksheet (PDF) · Answers (PDF) · lesson page
8.2 Substituting into expressions: Worksheet (PDF) · Answers (PDF) · lesson page
8.3 Substituting negative numbers: Worksheet (PDF) · Answers (PDF) · lesson page
8.4 Substituting into formulae: Worksheet (PDF) · Answers (PDF) · lesson page
8.5 Writing formulae: Worksheet (PDF) · Answers (PDF) · lesson page
8.6 Changing the subject: one step and two steps: Worksheet (PDF) · Answers (PDF) · lesson page
8.7 Changing the subject: brackets, fractions, powers and roots: Worksheet (PDF) · Answers (PDF) · lesson page
8.8 Higher: the subject appears twice: Worksheet (PDF) · Answers (PDF) · lesson page
8.9 Using formulae: problems and reasoning: Worksheet (PDF) · Answers (PDF) · lesson page
Check-up and exam-style questions: Worksheet (PDF) · Answers (PDF) · lesson page

Exercise P Do you remember?

Non-calculator

FWork out each of these.

(a)−3 + 7

(b)4 − (−5)

(c)−2 × 6

(d)(−3)²

Done

FWork out each of these, using the order of operations (BIDMAS).

(a)2 + 3 × 4²

(b)(2 + 3) × 4

Done

FSimplify by collecting like terms.

(a)3a + 2a

(b)5x − 2x + x

Done

FSolve each equation.

(a)x + 7 = 12

(b)3x = 21

(c)x4 = 5

Done

FWrite an expression for “3 more than twice n”.

Done

8.1

Letters, expressions and formulae

Edexcel 1MA1: A3, A21
Grade 1-4

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Letters, expressions and formulaeComing soon

Algebra uses letters to stand for numbers. A letter whose value can change is called a variable. If a bag holds n sweets, then n might be 20 today and 25 tomorrow. Writing a rule once with a letter saves us writing it again for every number.

Four words are used all the time in algebra. Exam questions expect you to know them without being told.

Word	Meaning	Example
Expression	Numbers and letters joined by +, −, ×, ÷. There is no equals sign.	3n + 5
Equation	Two expressions joined by an equals sign. It is true for only some values of the letter.	3n + 5 = 20 (true when n = 5)
Formula	A rule linking real quantities, written with an equals sign. You use it again and again with different numbers.	A = lw (area = length × width)
Identity	Two expressions that are equal for every value of the letter. We write ≡.	2(n + 3) ≡ 2n + 6

A term is one part of an expression, separated from the next by a + or − sign. In 3n + 5 the terms are 3n and 5.

Writing algebra in short form

To keep algebra tidy we do not write the multiplication sign. The rules are:

Algebra shorthand

3 × n is written 3n: the number goes first and there is no × sign.
n × n is written n² (say “n squared”), and n × n × n is n³.
a × b is written ab, with the letters in alphabetical order.
n ÷ 4 is written as a fraction: n4.
2 × (n + 1) is written 2(n + 1).

Turning words into algebra

Imagine Sam has n sweets and Tia has 5 more than Sam. Draw it as a bar. Tia has Sam’s amount n and then 5 extra, so Tia has n + 5 sweets. Drawing a bar like this is a good way to see what the words mean before you write any algebra.

Sam has n sweets. Tia has the same again plus 5 more, so n + 5.

Some words always point to the same operation. Use this table to translate, but always read the whole sentence.

Words in the question	Operation	Example in algebra
more than, plus, sum, total, increased by	add (+)	7 more than n is n + 7
less than, minus, fewer, decreased by, take away	subtract (−)	7 less than n is n − 7
times, multiplied by, product, twice, double, triple	multiply (×)	double n is 2n
divided by, shared equally between, half of	divide (÷)	half of n is n2

Finally, an expression can be pictured as a function machine: you put a number in, the machine does the operations in order, and a number comes out. The expression 3n + 2 is the machine “multiply by 3, then add 2”.

A function machine for 3n + 2. Put in n = 4: 4 × 3 = 12, then 12 + 2 = 14.

Watch out

“Less than” swaps the order. “5 less than n” means start from n and take 5 away: n − 5. It is not 5 − n. Test it: if n = 8, then 5 less than 8 is 3, and 8 − 5 = 3. Also remember 2n (double n) is not the same as n² (n × n): with n = 5 they are 10 and 25.

Worked example 8.1.1

Write an expression for “7 less than n”.

Key words. A variable is a letter that stands for a number that can change. An expression is a group of terms made from numbers and letters, with no equals sign.

Method. Take the sentence one piece at a time, write each piece in algebra, then join the pieces.

“7 less than” “Less than” means take away. The 7 is taken away from the number.
− 7
“n” The letter n stands for the number we start with. It comes first because we start from it.
n
Put the pieces together. Join the pieces in the same order as the words, using the operations the words describe.
n − 7

Check. Test with a number. Let n = 10. In words: 7 less than 10 = 3. The expression gives 10 − 7 = 3. They agree. ✓

Watch out. Read the words carefully. “5 less than n” is n − 5 (start from n), not 5 − n.

Worked example 8.1.2

A pen costs p pence. A notebook costs 90 pence. Write an expression for the total cost, in pence, of 4 pens and one notebook.

Key words. A variable is a letter that stands for a number that can change. An expression is a group of terms made from numbers and letters, with no equals sign.

Method. Take the sentence one piece at a time, write each piece in algebra, then join the pieces.

“A pen costs p pence” We let the letter p stand for the price of one pen, in pence.
p
“4 pens” 4 pens cost 4 lots of p, which is 4 × p. In algebra shorthand that is 4p.
4p
“and a notebook costing 90 pence” “And” tells us to add the notebook. The notebook is a fixed 90 pence.
+ 90
Put the pieces together. Join the pieces in the same order as the words, using the operations the words describe.
4p + 90

Check. Test with a number. Let p = 30. In words: 4 pens at 30p each is 120p, plus 90p = 210. The expression gives 4 × 30 + 90 = 120 + 90 = 210. They agree. ✓

Watch out. Read the words carefully. “5 less than n” is n − 5 (start from n), not 5 − n.

Worked example 8.1.3

Write down whether each is an expression, an equation, a formula or an identity: (a) 5n − 2 (b) 5n − 2 = 18 (c) 2(n + 3) ≡ 2n + 6.

Method. For each one, look for an equals sign, then decide which word fits.

(a) 5n − 2 There is no equals sign.
It is an expression. It is a phrase made of the terms 5n and −2.
(b) 5n − 2 = 18 There is an equals sign, but it is only true for one value of n. Try n = 1: 5 × 1 − 2 = 3, not 18. Try n = 4: 5 × 4 − 2 = 18, which works.
It is an equation. Solving it gives n = 4.
(c) 2(n + 3) ≡ 2n + 6 It uses ≡. Expanding the bracket: 2 × n = 2n and 2 × 3 = 6, so the left side is exactly 2n + 6 for every n.
It is an identity.

Check. For (c) try n = 10: left = 2(10 + 3) = 26 and right = 2 × 10 + 6 = 26. They agree. ✓

Watch out. An equals sign alone does not make something an equation. Check whether it is true for all values (identity) or only some (equation).

Worked example 8.1.4

n is a whole number. Write an expression for the sum of three consecutive whole numbers, starting with n, and simplify it. Explain why the answer is always a multiple of 3.

Method. Consecutive numbers go up by 1 each time. Add them, collect like terms, then look for a factor of 3.

Write the three numbers. Consecutive whole numbers follow each other: each is 1 more than the one before.
n, n + 1, n + 2
Add them. The sum means add all three.
n + (n + 1) + (n + 2)
Collect like terms. Add the n terms together (n + n + n = 3n) and add the numbers together (1 + 2 = 3).
3n + 3
Factorise. Both terms have a common factor of 3.
3n + 3 = 3(n + 1)
Explain. Anything written as 3 × (a whole number) is a multiple of 3, and n + 1 is a whole number.
So the sum is always a multiple of 3.

Check. Try n = 4: 4 + 5 + 6 = 15 = 3 × 5. Try n = 10: 10 + 11 + 12 = 33 = 3 × 11. ✓

Watch out. Do not stop at 3n + 3. To explain “multiple of 3” you must show the factor of 3, by writing 3(n + 1).

8.2

Substituting into expressions

Edexcel 1MA1: A2
Grade 2-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Substituting into expressionsComing soon

To substitute means to replace each letter with the number it stands for. Take the expression 3x + 2 when x = 5. Wherever we see x we write 5 instead, giving 3 × 5 + 2, and then we work it out: 15 + 2 = 17.

Notice that we put the × sign back in. In algebra 3x means 3 × x. If you just replaced x by 5 you would write 35, which is thirty-five and is wrong. The number 3 is next to the letter, not next to the number 5.

The order of working: BIDMAS

After substituting you have an ordinary calculation, so use the usual order of operations. Many people remember it as BIDMAS: Brackets, Indices, Division and Multiplication, Addition and Subtraction. Do the top row first.

BIDMAS. Division and multiplication share a row, and so do addition and subtraction. When two operations are on the same row, work from left to right.

Powers

The small number in x² is a power (an index). It tells us how many times to multiply: x² means x × x, not 2 × x. When x = 4, x² is 4 × 4 = 16 (a square of side 4 has 16 little squares), but 2x is just 2 × 4 = 8. In an expression like 2x², only the x is squared. Square first, then multiply by 2.

4² = 4 × 4 = 16

2 × 4 = 8

A square 4 by 4 contains 16 squares, which is 4². Two rows of 4 contain only 8, which is 2 × 4.

Brackets, fractions and roots

Work out the inside of a bracket first. The line of a fraction acts like brackets too: work out the top and the bottom separately, then divide the top by the bottom. A square root sign works the same way: work out what is under the sign first, then take the root.

Key ideas

Substitute: replace every letter with its number. Put the × sign back where a number touches a letter.
Put brackets round negative numbers when you substitute (see 8.3).
Then follow BIDMAS: brackets, powers, then multiply/divide, then add/subtract (left to right).
x² means x × x. In 5x² only the x is squared.
For a fraction, work out the top and the bottom first, then divide.

Watch out

3x with x = 4 is 3 × 4 = 12, not 34. And 2x² with x = 3 is 2 × 9 = 18, not (2 × 3)² = 36. Square first, then multiply.

Worked example 8.2.1

Work out the value of 3x + 2 when x = 5.

Key words. To substitute means to replace each letter with the number it stands for. An expression is a mixture of letters, numbers and operations (like 3x + 2). A formula is a rule that links quantities and has an equals sign (like A = lw). BIDMAS is the order for working out: Brackets, Indices, Division/Multiplication, Addition/Subtraction.

Method. Replace each letter by its value, then work out the answer using the order BIDMAS.

Write down the expression. Start by copying the expression exactly.
3x + 2
Substitute: x = 5. Substitute means replace each letter with its number. A number written next to a letter means multiply, so we show the × sign.
3x + 2 = 3 × 5 + 2
Do the multiplications and divisions. In BIDMAS, multiplication and division come before addition and subtraction. If there is more than one, work from left to right.
3 × 5 = 15
= 15 + 2
Do the additions and subtractions. Addition and subtraction come last in BIDMAS. If there is more than one, work from left to right.
15 + 2 = 17
= 17

Check. Put the numbers straight into a calculator: 3 × 5 + 2 gives 17. ✓

Watch out. A number next to a letter means multiply: 3x is 3 × x. If x = 5, then 3x = 15, not 35.

Worked example 8.2.2

Work out the value of 2a² − b when a = 3 and b = 4.

Method. Replace each letter by its value, then work out the answer using the order BIDMAS.

Write down the expression. Start by copying the expression exactly.
2a² − b
Substitute: a = 3, b = 4. Substitute means replace each letter with its number. A number written next to a letter means multiply, so we show the × sign.
2a² − b = 2 × 3² − 4
Work out the powers (indices). BIDMAS tells us the order: Brackets, Indices (powers and roots), Division and Multiplication, Addition and Subtraction. A power such as 5² means 5 × 5, and it is done before multiplying, dividing, adding or subtracting.
3² = 3 × 3 = 9
= 2 × 9 − 4
Do the multiplications and divisions. In BIDMAS, multiplication and division come before addition and subtraction. If there is more than one, work from left to right.
2 × 9 = 18
= 18 − 4
Do the additions and subtractions. Addition and subtraction come last in BIDMAS. If there is more than one, work from left to right.
18 − 4 = 14
= 14

Check. Put the numbers straight into a calculator: 2 × 3² − 4 gives 14. ✓

Watch out. In 5x² only the x is squared, not the 5. Square first, then multiply by 5.

Worked example 8.2.3

Work out the value of 2a + bc when a = 4, b = 7 and c = 5.

Method. Replace each letter by its value, then work out the answer using the order BIDMAS.

Write down the expression. Start by copying the expression exactly.
2a + bc
Substitute: a = 4, b = 7, c = 5. Substitute means replace each letter with its number. A number written next to a letter means multiply, so we show the × sign.
2a + bc = 2 × 4 + 75
Top of the fraction first: do the multiplications and divisions. Inside the top of the fraction we still use BIDMAS, so multiplying and dividing come before adding and subtracting.
2 × 4 = 8
= 8 + 75
Top of the fraction first: do the additions and subtractions. Inside the top of the fraction we still use BIDMAS. Only adding and subtracting are left, so work from left to right.
8 + 7 = 15
= 155
Divide the top by the bottom. The fraction bar means divide. Both the top and the bottom are now single numbers, so we can divide.
155 means 15 ÷ 5 = 3
= 3

Check. Put the numbers straight into a calculator: (2 × 4 + 7)/5 gives 3. ✓

Watch out. A number next to a letter means multiply: 3x is 3 × x. If x = 5, then 3x = 15, not 35.

Worked example 8.2.4

Work out the value of √x² + y² when x = 6 and y = 8.

Method. Replace each letter by its value, then work out the answer using the order BIDMAS.

Write down the expression. Start by copying the expression exactly.
√x² + y²
Substitute: x = 6, y = 8. Substitute means replace each letter with its number.
√x² + y² = √6² + 8²
Under the root sign first: work out the powers (indices). Inside the root sign we still use BIDMAS, so powers come first.
6² = 6 × 6 = 36
8² = 8 × 8 = 64
= √36 + 64
Under the root sign first: do the additions and subtractions. Inside the root sign we still use BIDMAS. Only adding and subtracting are left, so work from left to right.
36 + 64 = 100
= √100
Work out the root. A square root asks: which number multiplied by itself gives this? The root sign works like a bracket, so the number under it is worked out first.
√100 = 10 because 10 × 10 = 100
= 10

Check. Put the numbers straight into a calculator: √6² + 8² gives 10. ✓

Watch out. In 5x² only the x is squared, not the 5. Square first, then multiply by 5.

Worked example 8.2.5

Work out the value of 3x² − y2(x − y) when x = 4 and y = 2.

Method. Replace each letter by its value, then work out the answer using the order BIDMAS.

Write down the expression. Start by copying the expression exactly.
3x² − y2(x − y)
Substitute: x = 4, y = 2. Substitute means replace each letter with its number. A number written next to a letter means multiply, so we show the × sign.
3x² − y2(x − y) = 3 × 4² − 22 × (4 − 2)
Brackets first: do the additions and subtractions. Inside the brackets we still use BIDMAS. Only adding and subtracting are left, so work from left to right.
4 − 2 = 2
= 3 × 4² − 22 × 2
Top of the fraction first: work out the powers (indices). Inside the top of the fraction we still use BIDMAS, so powers come first.
4² = 4 × 4 = 16
2 × 2 = 4
= 3 × 16 − 24
Top of the fraction first: do the multiplications and divisions. Inside the top of the fraction we still use BIDMAS, so multiplying and dividing come before adding and subtracting.
3 × 16 = 48
= 48 − 24
Top of the fraction first: do the additions and subtractions. Inside the top of the fraction we still use BIDMAS. Only adding and subtracting are left, so work from left to right.
48 − 2 = 46
= 464
Divide the top by the bottom. The fraction bar means divide. Both the top and the bottom are now single numbers, so we can divide.
464 means 46 ÷ 4 = 11.5
= 11.5

Check. Put the numbers straight into a calculator: (3 × 4² − 2)/(2 × (4 − 2)) gives 11.5. ✓

Watch out. In 5x² only the x is squared, not the 5. Square first, then multiply by 5.

8.3

Substituting negative numbers

Edexcel 1MA1: A2
Grade 3-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Substituting negative numbersComing soon

Letters can stand for negative numbers as well as positive ones. Substituting a negative number is where most mistakes happen, so the rules need to be secure. Remember from Chapter 1 how negative numbers behave.

Top: 2 + (−3) = −1. Adding a negative moves left. Bottom: 2 − (−3) = 5. Subtracting a negative moves right.

These are the rules you need.

Calculation	Rule	Example
Add a negative	same as subtracting	5 + (−3) = 5 − 3 = 2
Subtract a negative	same as adding	5 − (−3) = 5 + 3 = 8
Positive × negative (or negative × positive)	answer is negative	4 × (−3) = −12
Negative × negative	answer is positive	(−4) × (−3) = 12
Dividing	follows the same signs rules as multiplying	−12 ÷ 4 = −3 and −12 ÷ (−4) = 3

When you substitute a negative number, always put it in brackets. If x = −4, then 3x becomes 3 × (−4), and 5 − x becomes 5 − (−4). The brackets stop the two signs running together, and they remind you which rule to use.

Squaring a negative number

A power only applies to the thing directly in front of it. Compare these three, all with x = −3:

Expression	Substitute	Working	Value
x²	(−3)²	(−3) × (−3)	9
−x²	−(−3)²	square first: 9, then the minus sign in front makes it negative	−9
(−x)²	(−(−3))²	−(−3) = 3, then 3²	9

Squaring a negative number always gives a positive answer, because a negative times a negative is positive. Cubing a negative number gives a negative answer: (−2)³ = (−2) × (−2) × (−2) = 4 × (−2) = −8.

On a calculator, use the (−) key for a negative number, and type brackets: (−3)². If you type −3² without brackets, the calculator gives −9.

Key ideas

Put every negative number in brackets when you substitute it.
Subtracting a negative = adding. Adding a negative = subtracting.
Same signs multiply to give a positive; different signs give a negative.
(−3)² = 9 but −3² = −9. The brackets decide what is squared.
Squaring a negative gives a positive. Cubing a negative gives a negative.

Watch out

With x = −5, the expression −x² is −25, not 25. The power is done before the minus sign in front of it. Also, 5 − (−2) is 5 + 2 = 7, not 3.

Worked example 8.3.1

Work out the value of 3x + 7 when x = −4.

Method. Replace each letter by its value, then work out the answer using the order BIDMAS.

Write down the expression. Start by copying the expression exactly.
3x + 7
Substitute: x = −4. Substitute means replace each letter with its number. A number written next to a letter means multiply, so we show the × sign. Put brackets round any negative number so that its sign is safe.
3x + 7 = 3 × (−4) + 7
Do the multiplications and divisions. In BIDMAS, multiplication and division come before addition and subtraction. If there is more than one, work from left to right.
3 × (−4) = −12 (a negative times a positive gives a negative)
= −12 + 7
Do the additions and subtractions. Addition and subtraction come last in BIDMAS. If there is more than one, work from left to right.
−12 + 7 = −5
= −5

Check. Put the numbers straight into a calculator, with brackets round negative numbers: 3 × (−4) + 7 gives −5. ✓

Watch out. Always put brackets round a negative number when you substitute it. Without them the minus sign is easy to lose, for example 5 − −3 looks wrong but 5 − (−3) is clear.

Worked example 8.3.2

Work out the value of a − 2b when a = −3 and b = −4.

Method. Replace each letter by its value, then work out the answer using the order BIDMAS.

Write down the expression. Start by copying the expression exactly.
a − 2b
Substitute: a = −3, b = −4. Substitute means replace each letter with its number. A number written next to a letter means multiply, so we show the × sign. Put brackets round any negative number so that its sign is safe.
a − 2b = −3 − 2 × (−4)
Do the multiplications and divisions. In BIDMAS, multiplication and division come before addition and subtraction. If there is more than one, work from left to right.
2 × (−4) = −8 (a negative times a positive gives a negative)
= −3 − (−8)
Do the additions and subtractions. Addition and subtraction come last in BIDMAS. If there is more than one, work from left to right.
−3 − (−8) = −3 + 8 = 5 (subtracting a negative number is the same as adding)
= 5

Check. Put the numbers straight into a calculator, with brackets round negative numbers: −3 − 2 × (−4) gives 5. ✓

Watch out. Always put brackets round a negative number when you substitute it. Without them the minus sign is easy to lose, for example 5 − −3 looks wrong but 5 − (−3) is clear.

Worked example 8.3.3

Work out the value of x² − 3x when x = −4.

Method. Replace each letter by its value, then work out the answer using the order BIDMAS.

Write down the expression. Start by copying the expression exactly.
x² − 3x
Substitute: x = −4. Substitute means replace each letter with its number. A number written next to a letter means multiply, so we show the × sign. Put brackets round any negative number so that its sign is safe.
x² − 3x = (−4)² − 3 × (−4)
Work out the powers (indices). BIDMAS tells us the order: Brackets, Indices (powers and roots), Division and Multiplication, Addition and Subtraction. A power such as 5² means 5 × 5, and it is done before multiplying, dividing, adding or subtracting.
(−4)² = (−4) × (−4) = 16 (a negative times a negative is positive)
= 16 − 3 × (−4)
Do the multiplications and divisions. In BIDMAS, multiplication and division come before addition and subtraction. If there is more than one, work from left to right.
3 × (−4) = −12 (a negative times a positive gives a negative)
= 16 − (−12)
Do the additions and subtractions. Addition and subtraction come last in BIDMAS. If there is more than one, work from left to right.
16 − (−12) = 16 + 12 = 28 (subtracting a negative number is the same as adding)
= 28

Check. Put the numbers straight into a calculator, with brackets round negative numbers: (−4)² − 3 × (−4) gives 28. ✓

Watch out. Put a negative number in brackets before you square it: (−3)² = 9, but −3² = −9. The brackets decide what gets squared.

Worked example 8.3.4

Work out the value of 2x² − 3xy + y² when x = −2 and y = 3.

Method. Replace each letter by its value, then work out the answer using the order BIDMAS.

Write down the expression. Start by copying the expression exactly.
2x² − 3xy + y²
Substitute: x = −2, y = 3. Substitute means replace each letter with its number. A number written next to a letter means multiply, so we show the × sign. Put brackets round any negative number so that its sign is safe.
2x² − 3xy + y² = 2 × (−2)² − 3 × (−2) × 3 + 3²
Work out the powers (indices). BIDMAS tells us the order: Brackets, Indices (powers and roots), Division and Multiplication, Addition and Subtraction. A power such as 5² means 5 × 5, and it is done before multiplying, dividing, adding or subtracting.
(−2)² = (−2) × (−2) = 4 (a negative times a negative is positive)
3² = 3 × 3 = 9
= 2 × 4 − 3 × (−2) × 3 + 9
Do the multiplications and divisions. In BIDMAS, multiplication and division come before addition and subtraction. If there is more than one, work from left to right.
2 × 4 = 8
3 × (−2) = −6 (a negative times a positive gives a negative)
= 8 − (−6) × 3 + 9
Do the multiplications and divisions. In BIDMAS, multiplication and division come before addition and subtraction. If there is more than one, work from left to right.
−6 × 3 = −18 (a negative times a positive gives a negative)
= 8 − (−18) + 9
Do the additions and subtractions. Addition and subtraction come last in BIDMAS. If there is more than one, work from left to right.
8 − (−18) = 8 + 18 = 26 (subtracting a negative number is the same as adding)
= 26 + 9
Do the additions and subtractions. Addition and subtraction come last in BIDMAS. If there is more than one, work from left to right.
26 + 9 = 35
= 35

Check. Put the numbers straight into a calculator, with brackets round negative numbers: 2 × (−2)² − 3 × (−2) × 3 + 3² gives 35. ✓

Watch out. Put a negative number in brackets before you square it: (−3)² = 9, but −3² = −9. The brackets decide what gets squared.

8.4

Substituting into formulae

Edexcel 1MA1: A2, A5
Grade 2-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Substituting into formulaeComing soon

A formula is a rule written with letters. Each letter stands for a quantity (a length, a time, a mass…). To use a formula we substitute the numbers we know and work out the one we do not. It is exactly what you did in 8.2 and 8.3, but now the letters mean something real, so the answer needs units.

In A = lw the letter on its own on the left, A, is called the subject of the formula. We use the formula to work out the subject. (In 8.6 we will learn to change which letter is the subject.)

Here are formulae you should know. Most appear in the GCSE exam without being given, so learn them.

Quantity	Formula	What the letters mean (usual units)
Area of a rectangle	A = lw	l length, w width (cm², m²)
Perimeter of a rectangle	P = 2l + 2w	distance round the outside (cm, m)
Area of a triangle	A = 12bh	b base, h perpendicular height
Area of a trapezium	A = 12(a + b)h	a and b the parallel sides, h the distance between them
Volume of a cuboid	V = lwh	length, width, height (cm³, m³)
Circumference of a circle	C = πd	d diameter
Area of a circle	A = πr²	r radius
Speed	s = dt	distance d and time t (km/h, m/s)
Density	D = mV	mass m and volume V (g/cm³)
Pressure	P = FA	force F and area A (N/m²)
Force	F = ma	mass m (kg), acceleration a (m/s²), force in newtons (N)
Final speed	v = u + at	u starting speed, a acceleration, t time
Kinetic energy	E = 12mv²	mass m (kg), speed v (m/s), energy in joules (J)
Electric potential difference	V = IR	current I (amps), resistance R (ohms), volts (V)

A trapezium (A = ½(a + b)h) and a cuboid (V = lwh). The height h of a trapezium is measured at right angles to the parallel sides.

Using a formula: four steps

Write the formula.
Substitute each known value (with its × sign, and brackets round negatives).
Work out the answer in the correct order (BIDMAS).
Write the answer with its units.

Units must match. If the speed formula is used with distance in km and time in hours, the answer is in km/h. If the time is given in minutes, change it to hours first (30 minutes is 0.5 hours). On a calculator, use brackets for fractions: the top and the bottom each go in their own brackets, for example (3 + 5) ÷ (2 × 4), and use the π button rather than typing 3.14. Do not round until the very end.

Key ideas

A formula links quantities. The letter on its own on the left is the subject.
Write the formula, substitute, work out with BIDMAS, then write the answer with units.
Keep units consistent: km with hours gives km/h, metres with seconds gives m/s.
On a calculator use brackets for the top and bottom of a fraction, and use the π button.
Round only at the end, and say how you rounded (decimal places or significant figures).

Watch out

In E = ½mv² the speed is squared, not the mass and not the whole product. With m = 4 and v = 10: square the 10 first (100), then ½ × 4 × 100 = 200. Squaring the whole product (½ × 4 × 10)² = 400 would be wrong.

Worked example 8.4.1

A coach travels 150 km in 2.5 hours. Use s = dt to work out its average speed.

Method. Write the formula, replace each letter by its value, then work out the right-hand side using the order BIDMAS.

Write down the formula. This is the rule we are going to use.
s = dt
Substitute: d = 150, t = 2.5. Substitute means replace each letter with its number.
s = 1502.5
Divide the top by the bottom. The fraction bar means divide. Both the top and the bottom are now single numbers, so we can divide.
1502.5 means 150 ÷ 2.5 = 60
s = 60
Write the answer with its units. The answer should say what was being worked out.
s = 60 km/h

Check. Put the numbers straight into a calculator: 150/2.5 gives 60. ✓

Watch out. Work in the order of BIDMAS: multiply and divide before you add and subtract.

Worked example 8.4.2

The area of a trapezium is A = ½(a + b)h. Work out A when a = 7 cm, b = 11 cm and h = 6 cm.

Method. Write the formula, replace each letter by its value, then work out the right-hand side using the order BIDMAS.

Write down the formula. This is the rule we are going to use.
A = 12(a + b)h
Substitute: a = 7, b = 11, h = 6. Substitute means replace each letter with its number. A number written next to a letter means multiply, so we show the × sign.
A = 12 × (7 + 11) × 6
Brackets first: do the additions and subtractions. Inside the brackets we still use BIDMAS. Only adding and subtracting are left, so work from left to right.
7 + 11 = 18
A = 12 × 18 × 6
Do the multiplications and divisions. In BIDMAS, multiplication and division come before addition and subtraction. If there is more than one, work from left to right.
12 × 18 = 9
A = 9 × 6
Do the multiplications and divisions. In BIDMAS, multiplication and division come before addition and subtraction. If there is more than one, work from left to right.
9 × 6 = 54
A = 54
Write the answer with its units. The answer should say what was being worked out.
A = 54 cm²

Check. Put the numbers straight into a calculator: 1/2 × (7 + 11) × 6 gives 54. ✓

Watch out. A number next to a letter means multiply: 3x is 3 × x. If x = 5, then 3x = 15, not 35.

Worked example 8.4.3

The kinetic energy of a moving object is E = ½mv² joules. Work out E when the mass is m = 4 kg and the speed is v = 10 m/s.

Method. Write the formula, replace each letter by its value, then work out the right-hand side using the order BIDMAS.

Write down the formula. This is the rule we are going to use.
E = 12mv²
Substitute: m = 4, v = 10. Substitute means replace each letter with its number. A number written next to a letter means multiply, so we show the × sign.
E = 12 × 4 × 10²
Work out the powers (indices). BIDMAS tells us the order: Brackets, Indices (powers and roots), Division and Multiplication, Addition and Subtraction. A power such as 5² means 5 × 5, and it is done before multiplying, dividing, adding or subtracting.
10² = 10 × 10 = 100
E = 12 × 4 × 100
Do the multiplications and divisions. In BIDMAS, multiplication and division come before addition and subtraction. If there is more than one, work from left to right.
12 × 4 = 2
E = 2 × 100
Do the multiplications and divisions. In BIDMAS, multiplication and division come before addition and subtraction. If there is more than one, work from left to right.
2 × 100 = 200
E = 200
Write the answer with its units. The answer should say what was being worked out.
E = 200 J

Check. Put the numbers straight into a calculator: 1/2 × 4 × 10² gives 200. ✓

Watch out. In 5x² only the x is squared, not the 5. Square first, then multiply by 5.

Worked example 8.4.4

Use A = πr² to work out the area of a circle of radius 6.3 cm. Use a calculator and give your answer to 1 decimal place.

Method. Write the formula, substitute the numbers, use a calculator, then round as asked.

Write down the formula. This is the rule we are going to use.
A = πr²
Substitute: r = 6.3. Replace each letter with its number. Keep π as it is: the calculator has a π button that is more accurate than 3.14.
A = π × 6.3²
Work it out on the calculator. Type it in carefully. Use brackets for the top and bottom of a fraction, and use the power key for squares. The calculator follows BIDMAS for you, but only if you type it correctly.
A = π × 6.3²
Calculator display: 124.68981242…
Round to 1 decimal place. Look at the digit just after the last one we keep. Here it is 8, which is 5 or more, so we round up.
124.68981242… = 124.7 (1 decimal place)
Write the answer with its units. Say what the number means.
A = 124.7 cm²

Check. Estimate with rounded numbers (r = 6, π = 3): 3 × 6² = 3 × 36 = 108. This is close to 124.7, so the answer is sensible. ✓

Watch out. Do not round until the very end. Rounding in the middle of a calculation makes the final answer less accurate.

Worked example 8.4.5

The distance travelled by a braking car is s = ut + ½at². Work out s when u = 12, t = 5 and a = −2.

Method. Write the formula, replace each letter by its value, then work out the right-hand side using the order BIDMAS.

Write down the formula. This is the rule we are going to use.
s = ut + 12at²
Substitute: u = 12, t = 5, a = −2. Substitute means replace each letter with its number. A number written next to a letter means multiply, so we show the × sign. Put brackets round any negative number so that its sign is safe.
s = 12 × 5 + 12 × (−2) × 5²
Work out the powers (indices). BIDMAS tells us the order: Brackets, Indices (powers and roots), Division and Multiplication, Addition and Subtraction. A power such as 5² means 5 × 5, and it is done before multiplying, dividing, adding or subtracting.
5² = 5 × 5 = 25
s = 12 × 5 + 12 × (−2) × 25
Do the multiplications and divisions. In BIDMAS, multiplication and division come before addition and subtraction. If there is more than one, work from left to right.
12 × 5 = 60
12 × (−2) = −1 (a negative times a positive gives a negative)
s = 60 + (−1) × 25
Do the multiplications and divisions. In BIDMAS, multiplication and division come before addition and subtraction. If there is more than one, work from left to right.
−1 × 25 = −25 (a negative times a positive gives a negative)
s = 60 + (−25)
Do the additions and subtractions. Addition and subtraction come last in BIDMAS. If there is more than one, work from left to right.
60 + (−25) = 60 − 25 = 35 (adding a negative number is the same as subtracting)
s = 35
Write the answer with its units. The answer should say what was being worked out.
s = 35 m

Check. Put the numbers straight into a calculator, with brackets round negative numbers: 12 × 5 + 1/2 × (−2) × 5² gives 35. ✓

Watch out. Always put brackets round a negative number when you substitute it. Without them the minus sign is easy to lose, for example 5 − −3 looks wrong but 5 − (−3) is clear.

8.5

Writing formulae

Edexcel 1MA1: A21, A5
Grade 2-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Writing formulaeComing soon

So far the formulae were given to you. Now you will write your own. Writing a formula means describing a rule in algebra, so that you can use it for any number.

Follow these steps:

Choose letters and say what each stands for, including units (for example C = cost in pounds, n = number of tickets).
Split the rule into a fixed part (paid once, never changes) and a changing part (a rate multiplied by the amount).
Write the formula with the subject on its own on the left.
Test it with a small easy number to see if it gives the right answer.

Many real rules have the shape total = fixed charge + rate × amount. A taxi that charges £3 to start and £2 per mile has cost C = 3 + 2m. A phone contract of £8 a month plus 4p for each minute has cost C = 8 + 0.04m pounds (we must write 4p as £0.04 so the units match).

Formulae from diagrams

When a shape has sides given by expressions, label every side, then add them for the perimeter or multiply for the area. Remember that a rectangle has two lengths and two widths, so its perimeter is 2 × length + 2 × width, and any bracket you write needs the 2 to multiply both terms inside it.

A pattern of matchstick triangles. 1 triangle uses 3 sticks, 2 triangles use 5, 3 triangles use 7. Each new triangle adds 2 sticks, and the very first stick adds 1 more: m = 2n + 1.

Key ideas

Choose letters and say what they stand for (with units).
Typical shape: total = fixed part + (rate × amount).
For shapes: label the sides, then add (perimeter) or multiply (area). Use brackets when a sum is multiplied.
Always test your formula with a simple number.
Make sure the units match (pounds with pounds, minutes with minutes).

Watch out

Do not mix units. If the rate is in pence per minute and the total is in pounds, change the pence to pounds first. And write the fixed charge on its own, not multiplied: £3 start plus £2 per mile is 3 + 2m, not 5m.

Worked example 8.5.1

A phone contract costs £8 a month plus 4p for every minute of calls. Write a formula for the monthly cost, C pounds, when m minutes of calls are made.

Key words. A variable is a letter that stands for a number that can change. An expression is a group of terms made from numbers and letters, with no equals sign.

Method. Take the sentence one piece at a time, write each piece in algebra, then join the pieces.

“£8 a month” A fixed amount, paid once. It does not depend on the minutes.
8
“4p for each minute” 4p = £0.04. With m minutes the cost is 0.04 × m.
0.04m
“plus” The fixed charge and the minute charge are added.
+
Put the pieces together. Join the pieces in the same order as the words, using the operations the words describe.
C = 8 + 0.04m

Check. Test with a number. Let m = 100. In words: £8 plus 100 minutes at £0.04 = £8 + £4 = 12. The expression gives 8 + 0.04 × 100 = 8 + 4 = 12. They agree. ✓

Watch out. Read the words carefully. “5 less than n” is n − 5 (start from n), not 5 − n.

A rectangle with length x + 3 and width 2x − 1.

Worked example 8.5.2

The diagram shows a rectangle. Its length is x + 3 and its width is 2x − 1 (in cm). Write a simplified expression for the perimeter, P.

Method. The perimeter of a rectangle is two lengths plus two widths. Write it, expand the brackets, then collect like terms.

Write the perimeter. There are 2 lengths and 2 widths.
P = 2(x + 3) + 2(2x − 1)
Expand both brackets. Multiply every term inside each bracket by the 2.
2 × x = 2x and 2 × 3 = 6
2 × 2x = 4x and 2 × (−1) = −2
P = 2x + 6 + 4x − 2
Collect like terms. 2x + 4x = 6x and 6 − 2 = 4.
P = 6x + 4 cm

Check. Try x = 5: length = 8, width = 9, perimeter = 2 × 8 + 2 × 9 = 34. And 6 × 5 + 4 = 34. ✓

Watch out. Do not forget to multiply the second term in each bracket: 2(2x − 1) = 4x − 2, not 4x − 1.

Worked example 8.5.3

Matchsticks make a row of triangles (see the diagram). Write a formula for the number of matchsticks, m, needed for n triangles, and check it for 3 triangles.

Key words. A variable is a letter that stands for a number that can change. An expression is a group of terms made from numbers and letters, with no equals sign.

Method. Take the sentence one piece at a time, write each piece in algebra, then join the pieces.

“each extra triangle adds 2 sticks” Look at the pattern: 3, 5, 7. It goes up by 2 each time, so n triangles give 2 × n = 2n sticks.
2n
“the first triangle needs 3 sticks, which is one more than 2” For n = 1, 2n = 2 but we need 3, so we add 1.
+ 1
Put the pieces together. Join the pieces in the same order as the words, using the operations the words describe.
m = 2n + 1

Check. Test with a number. Let n = 3. In words: 3 triangles use 7 sticks = 7. The expression gives 2 × 3 + 1 = 6 + 1 = 7. They agree. ✓

Watch out. Read the words carefully. “5 less than n” is n − 5 (start from n), not 5 − n.

Worked example 8.5.4

A rectangle has length (x + 4) cm and width (x − 1) cm. Write and simplify an expression for its area, A.

Method. The area of a rectangle is length × width, so multiply the two brackets. Expand carefully and collect like terms.

Write the area. Area = length × width.
A = (x + 4)(x − 1)
Multiply every term in the first bracket by every term in the second. There are 4 products (First, Outer, Inner, Last).
First: x × x = x²
Outer: x × (−1) = −x
Inner: 4 × x = 4x
Last: 4 × (−1) = −4
Add the four products.
A = x² − x + 4x − 4
Collect like terms. −x + 4x = 3x.
A = x² + 3x − 4 cm²

Check. Try x = 3: length 7, width 2, area 14. And 3² + 3 × 3 − 4 = 9 + 9 − 4 = 14. ✓

Watch out. Check your signs when multiplying by a negative number: 4 × (−1) = −4.

8.6

Changing the subject: one step and two steps

Edexcel 1MA1: A5
Grade 3-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Changing the subject: one step and two stepsComing soon

In a formula such as F = ma, the letter on its own on the left, F, is the subject. The formula is set up to work out F. But suppose we know the force and the mass and want the acceleration. We would like a formula with a on its own. Changing the subject (also called rearranging the formula) means rewriting the formula so that a different letter is on its own. Here F = ma becomes a = Fm.

Think of a function machine

The formula y = 3x + 2 says: take x, multiply by 3, then add 2, and the result is y. To find x from y we run the machine backwards, doing the inverse (opposite) of each step, in reverse order: first subtract 2, then divide by 3.

Top: the formula y = 3x + 2 as a machine. Bottom: the machine run backwards gives x = (y − 2) ÷ 3.

Each step undoes the one before it. This table lists the inverse (opposite) operations:

Operation	Inverse (undoes it)
add a number	subtract the same number
subtract a number	add the same number
multiply by a number	divide by the same number
divide by a number	multiply by the same number
square	square root
square root	square

A formula is like a pair of scales in balance. If you do something to one side, you must do exactly the same to the other side, or it will no longer balance. So at each step we write what we do to both sides.

Key ideas

The subject is the letter on its own on one side. To change the subject, get the letter you want on its own.
See what has been done to the letter, then undo it in reverse order using inverse operations.
Do the same to both sides every time, and to the whole of each side.
Write the answer with the new subject on the left. Check it with numbers.

Watch out

Do the same to the whole side. In y = 3x + 2, after subtracting 2 we have y − 2 = 3x. Dividing by 3 means the whole left side is divided by 3: x = (y − 2) ÷ 3. It is wrong to write x = y − 2 ÷ 3, which only divides the 2.

Worked example 8.6.1

Make l the subject of the formula A = lw.

Key words. The subject of a formula is the letter on its own on one side, such as v in v = u + at. To change the subject means to rearrange the formula so that a different letter is on its own. An inverse operation undoes another: adding and subtracting are inverses, multiplying and dividing are inverses, squaring and square rooting are inverses.

Method. Change the subject by undoing the operations one at a time, in reverse order, doing the same to both sides each time, until l is on its own.

Find the letter we want, and see what is done to it. Treat the formula like a function machine. To get l on its own we must undo everything that has been done to it, in the reverse order.
Formula: A = lw
Reading from the subject outwards, what happens to l: it is multiplied by w.
To undo it we reverse the order: divide by w.
Divide both sides by w. l is being multiplied by w (letters or numbers written next to each other are multiplied). The inverse of multiplying is dividing, so divide both sides by w.
Aw = lww
Aw = l
Write it with the subject on the left. The subject is the letter on its own. By convention we write it on the left of the equals sign. Swapping the two sides does not change the meaning.
l = Aw

Check. Pick numbers for the other letters: A = 10, w = 5. The new formula gives
l = 105 = 2
Now put l = 2 back into the original formula. Left side: 10. Right side: 2 × 5 = 10. Both sides are equal, so the rearranging is correct. ✓

Watch out. Do the same thing to both sides, and treat every term on a side as a whole. For example, if you divide by 2, the whole side is divided by 2, not just one term.

Worked example 8.6.2

Make x the subject of the formula y = 3x + 2.

Method. Change the subject by undoing the operations one at a time, in reverse order, doing the same to both sides each time, until x is on its own.

Find the letter we want, and see what is done to it. Treat the formula like a function machine. To get x on its own we must undo everything that has been done to it, in the reverse order.
Formula: y = 3x + 2
Reading from the subject outwards, what happens to x: it is multiplied by 3; then 2 is added.
To undo it we reverse the order: subtract 2, then divide by 3.
Subtract 2 from both sides. 2 is being added to the part containing x. The inverse (opposite) of adding is subtracting, so subtract 2 from both sides. Doing the same to both sides keeps the formula true.
y − 2 = 3x + 2 − 2
y − 2 = 3x
Divide both sides by 3. x is being multiplied by 3 (letters or numbers written next to each other are multiplied). The inverse of multiplying is dividing, so divide both sides by 3.
y − 23 = 3x3
y − 23 = x
Write it with the subject on the left. The subject is the letter on its own. By convention we write it on the left of the equals sign. Swapping the two sides does not change the meaning.
x = y − 23

Check. Pick numbers for the other letters: y = 8. The new formula gives
x = 8 − 23 = 63 = 2
Now put x = 2 back into the original formula. Left side: 8. Right side: 3 × 2 + 2 = 6 + 2 = 8. Both sides are equal, so the rearranging is correct. ✓

Watch out. Do the same thing to both sides, and treat every term on a side as a whole. For example, if you divide by 2, the whole side is divided by 2, not just one term.

Worked example 8.6.3

Make l the subject of the formula P = 2l + 2w.

Method. Change the subject by undoing the operations one at a time, in reverse order, doing the same to both sides each time, until l is on its own.

Find the letter we want, and see what is done to it. Treat the formula like a function machine. To get l on its own we must undo everything that has been done to it, in the reverse order.
Formula: P = 2l + 2w
Reading from the subject outwards, what happens to l: it is multiplied by 2; then 2w is added.
To undo it we reverse the order: subtract 2w, then divide by 2.
Subtract 2w from both sides. 2w is being added to the part containing l. The inverse (opposite) of adding is subtracting, so subtract 2w from both sides. Doing the same to both sides keeps the formula true.
P − 2w = 2l + 2w − 2w
P − 2w = 2l
Divide both sides by 2. l is being multiplied by 2 (letters or numbers written next to each other are multiplied). The inverse of multiplying is dividing, so divide both sides by 2.
P − 2w2 = 2l2
P − 2w2 = l
Write it with the subject on the left. The subject is the letter on its own. By convention we write it on the left of the equals sign. Swapping the two sides does not change the meaning.
l = P − 2w2

Check. Pick numbers for the other letters: P = 6, w = 2. The new formula gives
l = 6 − 2 × 22 = 6 − 42 = 22 = 1
Now put l = 1 back into the original formula. Left side: 6. Right side: 2 × 1 + 2 × 2 = 2 + 4 = 6. Both sides are equal, so the rearranging is correct. ✓

Watch out. Do the same thing to both sides, and treat every term on a side as a whole. For example, if you divide by 2, the whole side is divided by 2, not just one term.

Worked example 8.6.4

Make x the subject of the formula y = 10 − 3x.

Method. Change the subject by undoing the operations one at a time, in reverse order, doing the same to both sides each time, until x is on its own.

Find the letter we want, and see what is done to it. Treat the formula like a function machine. To get x on its own we must undo everything that has been done to it, in the reverse order.
Formula: y = 10 − 3x
Reading from the subject outwards, what happens to x: it is multiplied by 3; then it is subtracted from 10.
To undo it we reverse the order: add 3x, then subtract y, then divide by 3.
Add 3x to both sides. The part containing x is being subtracted from 10. To get rid of the minus in front of it, add 3x to both sides so that it becomes positive.
y + 3x = 10 − 3x + 3x
y + 3x = 10
Subtract y from both sides. Now x is in a positive part, but y is still added to it. The inverse of adding is subtracting.
10 − y = y + 3x − y
10 − y = 3x
Divide both sides by 3. x is being multiplied by 3 (letters or numbers written next to each other are multiplied). The inverse of multiplying is dividing, so divide both sides by 3.
10 − y3 = 3x3
10 − y3 = x
Write it with the subject on the left. The subject is the letter on its own. By convention we write it on the left of the equals sign. Swapping the two sides does not change the meaning.
x = 10 − y3

Check. Pick numbers for the other letters: y = 4. The new formula gives
x = 10 − 43 = 63 = 2
Now put x = 2 back into the original formula. Left side: 4. Right side: 10 − 3 × 2 = 10 − 6 = 4. Both sides are equal, so the rearranging is correct. ✓

Watch out. Do the same thing to both sides, and treat every term on a side as a whole. For example, if you divide by 2, the whole side is divided by 2, not just one term.

8.7

Changing the subject: brackets, fractions, powers and roots

Edexcel 1MA1: A5
Grade 5-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Changing the subject: brackets, fractions, powers and rootsComing soon

The method is the same as in 8.6: find the subject, see what has been done to it, and undo the steps in reverse order. The new ideas are how to undo brackets, fractions, powers and roots. In each formula in this section the subject appears once.

Brackets

In P = 2(l + w) the bracket (l + w) is multiplied by 2. To get w alone, undo the multiplying first (divide both sides by 2), then undo the adding (subtract l). This leaves w = P2 − l. You could also expand the bracket first (P = 2l + 2w) and then use 8.6. Both give the same answer.

Fractions

If the subject is in the top of a fraction, multiply both sides by the bottom. If the subject is in the bottom, multiply both sides by it to bring it up, then divide. For s = d/t, to make t the subject: multiply both sides by t to get st = d, then divide by s to get t = d/s.

Powers and roots

Squaring and square rooting are inverses. For A = πr², the machine for r to A is “square, then multiply by π”. Run backwards: divide by π, then take the square root.

Top: A = πr². Bottom: undo it in reverse order to get r = √(A/π).

When you take the square root, take it of the whole side, so the whole of A/π must be under the root sign. A square root can be positive or negative, but in these real situations (lengths, times, speeds) the quantity is positive, so we use the positive root. If the subject is under a root, as in v = √(2gh), undo it by squaring both sides: v² = 2gh.

Key ideas

Brackets: undo the multiplying first (divide), then the adding or subtracting inside. Or expand first.
Fractions: subject on top, multiply by the bottom. Subject on the bottom, multiply by it, then divide.
Powers and roots: square root undoes squaring; squaring undoes a square root.
Take the root of, or square, the whole side.
Always check by substituting numbers.

Watch out

√(A/π) is not the same as √A/π. The whole of A/π sits under the root, because the whole of A/π was equal to r². Also, you cannot split √(x² + y²) into x + y: with x = 3 and y = 4 the left side is 5 but x + y = 7.

Worked example 8.7.1

Make w the subject of P = 2(l + w).

Method. Change the subject by undoing the operations one at a time, in reverse order, doing the same to both sides each time, until w is on its own.

Find the letter we want, and see what is done to it. Treat the formula like a function machine. To get w on its own we must undo everything that has been done to it, in the reverse order.
Formula: P = 2(l + w)
Reading from the subject outwards, what happens to w: l is added; then it is multiplied by 2.
To undo it we reverse the order: divide by 2, then subtract l.
Divide both sides by 2. w is being multiplied by 2 (letters or numbers written next to each other are multiplied). The inverse of multiplying is dividing, so divide both sides by 2.
P2 = 2(l + w)2
P2 = l + w
Subtract l from both sides. l is being added to the part containing w. The inverse (opposite) of adding is subtracting, so subtract l from both sides. Doing the same to both sides keeps the formula true.
P2 − l = l + w − l
P2 − l = w
Write it with the subject on the left. The subject is the letter on its own. By convention we write it on the left of the equals sign. Swapping the two sides does not change the meaning.
w = P2 − l

Check. Pick numbers for the other letters: P = 10, l = 4. The new formula gives
w = 102 − 4 = 5 − 4 = 1
Now put w = 1 back into the original formula. Left side: 10. Right side: 2 × (4 + 1) = 2 × 5 = 10. Both sides are equal, so the rearranging is correct. ✓

Watch out. Do the same thing to both sides, and treat every term on a side as a whole. For example, if you divide by 2, the whole side is divided by 2, not just one term.

Worked example 8.7.2

Make r the subject of A = πr².

Method. Change the subject by undoing the operations one at a time, in reverse order, doing the same to both sides each time, until r is on its own.

Find the letter we want, and see what is done to it. Treat the formula like a function machine. To get r on its own we must undo everything that has been done to it, in the reverse order.
Formula: A = πr²
Reading from the subject outwards, what happens to r: it is squared; then it is multiplied by π.
To undo it we reverse the order: divide by π, then take the square root.
Divide both sides by π. r is being multiplied by π (letters or numbers written next to each other are multiplied). The inverse of multiplying is dividing, so divide both sides by π.
Aπ = πr²π
Aπ = r²
Take the square root of both sides. r is being squared. The inverse of squaring is taking the square root, so do that to both sides.
√Aπ = √r²
√Aπ = r
Write it with the subject on the left. The subject is the letter on its own. By convention we write it on the left of the equals sign. Swapping the two sides does not change the meaning.
r = √Aπ

Check. Pick numbers for the other letters (for this check only, let π = 3 to keep the numbers easy): A = 3, π = 3. The new formula gives
r = √33 = √1 = 1
Now put r = 1 back into the original formula. Left side: 3. Right side: 3 × 1² = 3 × 1 = 3. Both sides are equal, so the rearranging is correct. ✓

Watch out. Take the square root of the whole side, not of each term. And a square root can be positive or negative, but in these real-life formulae lengths and times are positive, so we use the positive root.

Worked example 8.7.3

Make v the subject of E = ½mv².

Method. Change the subject by undoing the operations one at a time, in reverse order, doing the same to both sides each time, until v is on its own.

Find the letter we want, and see what is done to it. Treat the formula like a function machine. To get v on its own we must undo everything that has been done to it, in the reverse order.
Formula: E = 12mv²
Reading from the subject outwards, what happens to v: it is squared; then it is multiplied by m; then it is multiplied by 12.
To undo it we reverse the order: multiply by 2, then divide by m, then take the square root.
Multiply both sides by 2. 12 means "divide by 2". The inverse of dividing is multiplying, so multiply both sides by 2.
E × 2 = 12mv² × 2
2E = mv²
Divide both sides by m. v is being multiplied by m (letters or numbers written next to each other are multiplied). The inverse of multiplying is dividing, so divide both sides by m.
2Em = mv²m
2Em = v²
Take the square root of both sides. v is being squared. The inverse of squaring is taking the square root, so do that to both sides.
√2Em = √v²
√2Em = v
Write it with the subject on the left. The subject is the letter on its own. By convention we write it on the left of the equals sign. Swapping the two sides does not change the meaning.
v = √2Em

Check. Pick numbers for the other letters: E = 10, m = 5. The new formula gives
v = √2 × 105 = √205 = √4 = 2
Now put v = 2 back into the original formula. Left side: 10. Right side: 12 × 5 × 2² = 12 × 5 × 4 = 2.5 × 4 = 10. Both sides are equal, so the rearranging is correct. ✓

Worked example 8.7.4

Make x the subject of y = 12x + 1.

Method. Change the subject by undoing the operations one at a time, in reverse order, doing the same to both sides each time, until x is on its own.

Find the letter we want, and see what is done to it. Treat the formula like a function machine. To get x on its own we must undo everything that has been done to it, in the reverse order.
Formula: y = 12x + 1
Reading from the subject outwards, what happens to x: 1 is added; then it is the divisor in 12 ÷ (x + 1).
To undo it we reverse the order: multiply by (x + 1), then divide by y, then subtract 1.
Multiply both sides by (x + 1). x is on the bottom of a fraction. To bring it up, multiply both sides by (x + 1). (Anything divided by (x + 1), then multiplied by (x + 1), is back where it started.)
y × (x + 1) = 12x + 1 × (x + 1)
y(x + 1) = 12
Divide both sides by y. x is being multiplied by y (letters or numbers written next to each other are multiplied). The inverse of multiplying is dividing, so divide both sides by y.
12y = y(x + 1)y
12y = x + 1
Subtract 1 from both sides. 1 is being added to the part containing x. The inverse (opposite) of adding is subtracting, so subtract 1 from both sides. Doing the same to both sides keeps the formula true.
12y − 1 = x + 1 − 1
12y − 1 = x
Write it with the subject on the left. The subject is the letter on its own. By convention we write it on the left of the equals sign. Swapping the two sides does not change the meaning.
x = 12y − 1

Check. Pick numbers for the other letters: y = 4. The new formula gives
x = 124 − 1 = 3 − 1 = 2
Now put x = 2 back into the original formula. Left side: 4. Right side: 122 + 1 = 123 = 4. Both sides are equal, so the rearranging is correct. ✓

Watch out. Do the same thing to both sides, and treat every term on a side as a whole. For example, if you divide by 2, the whole side is divided by 2, not just one term.

Worked example 8.7.5

Make l the subject of T = 2π√(l/g).

Method. Change the subject by undoing the operations one at a time, in reverse order, doing the same to both sides each time, until l is on its own.

Find the letter we want, and see what is done to it. Treat the formula like a function machine. To get l on its own we must undo everything that has been done to it, in the reverse order.
Formula: T = 2π√lg
Reading from the subject outwards, what happens to l: it is divided by g; then the square root is taken; then it is multiplied by 2π.
To undo it we reverse the order: divide by 2π, then square, then multiply by g.
Divide both sides by 2π. l is being multiplied by 2π (letters or numbers written next to each other are multiplied). The inverse of multiplying is dividing, so divide both sides by 2π.
T2π = 2π√lg2π
T2π = √lg
Square both sides. l is under a square root. The inverse of taking a root is squaring, so do that to both sides.
(T2π)² = (√lg)²
(T2π)² = lg
Multiply both sides by g. The part containing l is being divided by g. The inverse of dividing is multiplying, so multiply both sides by g.
(T2π)² × g = lg × g
g(T2π)² = l
Write it with the subject on the left. The subject is the letter on its own. By convention we write it on the left of the equals sign. Swapping the two sides does not change the meaning.
l = g(T2π)²

Check. Pick numbers for the other letters (for this check only, let π = 3 to keep the numbers easy): T = 3, g = 4, π = 3. The new formula gives
l = 4 × (32 × 3)² = 4 × (36)² = 4 × 0.5² = 4 × 0.25 = 1
Now put l = 1 back into the original formula. Left side: 3. Right side: 2 × 3 × √14 = 2 × 3 × √0.25 = 2 × 3 × 0.5 = 6 × 0.5 = 3. Both sides are equal, so the rearranging is correct. ✓

Watch out. Do the same thing to both sides, and treat every term on a side as a whole. For example, if you divide by 2, the whole side is divided by 2, not just one term.

8.8

Higher: the subject appears twice

Edexcel 1MA1: A5
Grade 6-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Higher: the subject appears twiceComing soon

In 8.6 and 8.7 the subject appeared once, so we could peel the operations off one at a time. When the subject appears twice, as in ax + b = cx + d or y = x + 3x − 2, no single inverse operation will isolate it. We need a different plan: gather all the terms that contain the subject, factorise the subject out so that it appears only once, and then divide.

The five-step plan. Not every question needs every step: skip any step that is not needed.

Why does factorising work? Look at xy − x = 3. The letter x appears twice, but both terms contain it: xy is x × y and x is x × 1. So we can write x(y − 1) = 3. Now x appears once, multiplied by the bracket (y − 1), and we can divide both sides by that bracket. Remember from Chapter 7 that factorising is the reverse of expanding: x(y − 1) = xy − x.

Two details matter. First, when you multiply to clear a fraction, multiply every term on both sides (the whole of each side). Second, when a term crosses the equals sign its sign changes, which is the same as adding or subtracting it from both sides.

Key ideas

If the subject appears more than once, collect every term containing it on one side and everything else on the other.
Factorise the subject out: xy − x = x(y − 1).
Then divide both sides by the whole bracket.
Clear fractions first by multiplying both sides by the denominator, and expand brackets before collecting.

Watch out

After factorising you must divide by the whole bracket. From x(y − 1) = 3 + 2y we get x = (3 + 2y) ÷ (y − 1), with the whole of (y − 1) on the bottom. It is wrong to cancel the y in x(y − 1) = ... to leave x − 1.

Worked example 8.8.1

Make x the subject of ax + b = cx + d.

Key words. When the subject (the letter we want) appears twice, we cannot simply undo operations one at a time. Instead we collect every term containing the subject on one side, factorise it out (write it as a product, with the subject outside a bracket), and then divide.

Method. Collect the terms containing the subject on one side, factorise the subject out, then divide.

Write the formula. The subject x appears on both sides, so we cannot undo one operation at a time.
ax + b = cx + d
Collect every term containing x on the left and every other term on the right. Subtract cx from both sides and subtract b from both sides. Each time a term crosses the equals sign its sign changes.
ax − cx = d − b
Factorise the left side. Both terms contain x. Take x outside a bracket and put what is left inside.
x(a − c) = d − b
Divide both sides by the bracket. x is multiplied by (a − c), so divide both sides by it.
x = d − ba − c

Check. Pick numbers for the other letters: a = 4, b = 10, c = 5, d = 6. The new formula gives
x = 6 − 104 − 5 = −4−1 = 4
Now put x = 4 back into the original formula. Left side: 4 × 4 + 10 = 16 + 10 = 26. Right side: 5 × 4 + 6 = 20 + 6 = 26. Both sides are equal, so the rearranging is correct. ✓

Watch out. Terms with the subject go on one side and all other terms on the other. When you move a term across the equals sign, its sign changes. Remember to multiply every term when you clear a fraction.

Worked example 8.8.2

Make x the subject of a(x + b) = c(x + d).

Method. Expand the brackets, collect the terms containing the subject, factorise, then divide.

Write the formula. x is inside brackets on both sides.
a(x + b) = c(x + d)
Expand both brackets. Multiply everything inside each bracket by the number outside.
ax + ab = cx + cd
Collect the x terms on the left and the others on the right. Subtract cx and ab from both sides.
ax − cx = cd − ab
Factorise the left side. Take x outside a bracket.
x(a − c) = cd − ab
Divide both sides by the bracket. Divide by (a − c).
x = cd − aba − c

Check. Pick numbers for the other letters: a = 8, b = 3, c = 6, d = 5. The new formula gives
x = 6 × 5 − 8 × 38 − 6 = 30 − 242 = 62 = 3
Now put x = 3 back into the original formula. Left side: 8 × (3 + 3) = 8 × 6 = 48. Right side: 6 × (3 + 5) = 6 × 8 = 48. Both sides are equal, so the rearranging is correct. ✓

Worked example 8.8.3

Make x the subject of y = x + 3x − 2.

Method. Clear the fraction, expand, collect the subject terms, factorise, then divide.

Write the formula. x appears on the top and the bottom of the fraction.
y = x + 3x − 2
Multiply both sides by the bottom, (x − 2). This removes the fraction. The whole of the bottom is multiplied.
y(x − 2) = x + 3
Expand the bracket. Multiply both terms inside by y.
yx − 2y = x + 3
Collect the x terms on the left and the others on the right. Subtract x from both sides and add 2y to both sides.
yx − x = 3 + 2y
Factorise the left side. Take x outside a bracket.
x(y − 1) = 3 + 2y
Divide both sides by the bracket. Divide by (y − 1).
x = 3 + 2yy − 1

Check. Pick numbers for the other letters: y = 6. The new formula gives
x = 3 + 2 × 66 − 1 = 3 + 125 = 155 = 3
Now put x = 3 back into the original formula. Left side: 6. Right side: 3 + 33 − 2 = 61 = 6. Both sides are equal, so the rearranging is correct. ✓

Worked example 8.8.4

Make x the subject of y = ax + bcx + d.

Method. Clear the fraction, expand, collect the subject terms, factorise, then divide.

Write the formula. x appears on the top and the bottom of the fraction.
y = ax + bcx + d
Multiply both sides by the bottom. This removes the fraction.
y(cx + d) = ax + b
Expand the bracket. Multiply both terms inside by y.
ycx + yd = ax + b
Collect the x terms on the left and the others on the right. Subtract ax and yd from both sides.
ycx − ax = b − yd
Factorise the left side. Take x outside a bracket.
x(yc − a) = b − yd
Divide both sides by the bracket. Divide by the whole bracket.
x = b − ydyc − a

8.9

Using formulae: problems and reasoning

Edexcel 1MA1: A2, A5, A21
Grade 4-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Using formulae: problems and reasoningComing soon

Exam questions often use a formula in a short story, and you must decide what to do. Here is a reliable plan.

A five-step plan for formula problems.

There are two sound ways to find an unknown that is not the subject. Suppose A = ½(a + b)h and we know A, a and b but want h.

Substitute first, then solve the equation you get: 54 = ½ × 18 × h = 9h, so h = 54 ÷ 9. This is quick for one calculation.
Rearrange first, then substitute: h = 2A ÷ (a + b). This is better when you need to do many calculations with different numbers.

Other useful skills in this section:

Checking a formula by putting in simple test values to see if it gives the right answer.
Showing two expressions are equal by expanding or substituting, and ending with a clear conclusion.
Combining formulae (Higher): if one formula gives r in terms of C, substitute it into another formula to get A in terms of C.

A cylinder of radius r and height h has volume V = πr²h.

Key ideas

Read carefully and write down what you know and what you want.
Either substitute first and solve, or rearrange first and substitute.
Check units (convert minutes to hours, pence to pounds) before you start.
To check a formula, try simple numbers. To show something, give a calculation and a conclusion.
Does the answer make sense? A speed of 600 km/h for a bike does not.

Watch out

Average speed is total distance ÷ total time. It is not the average of the two speeds. If you ride 12 km at 15 km/h and 8 km at 10 km/h, the average speed is not (15 + 10) ÷ 2 = 12.5 by luck: work out each time first.

Worked example 8.9.1

The area of a trapezium is A = ½(a + b)h. A trapezium has area 54 cm², a = 7 cm and b = 11 cm. Work out h.

Method. Substitute the known values, simplify, then solve the simple equation for h.

Write the formula and substitute. Put in A = 54, a = 7 and b = 11.
54 = ½ × (7 + 11) × h
Work out the bracket, then ½ of it. Brackets first. 7 + 11 = 18 and ½ of 18 is 9.
54 = 9h
Divide both sides by 9. h is multiplied by 9, so the inverse is dividing by 9.
h = 54 ÷ 9 = 6 cm

Check. Put h = 6 back: A = 12 × (7 + 11) × 6 = 12 × 18 × 6 = 9 × 6 = 54 cm². ✓

Watch out. Do not forget the ½. A common slip is to forget it and get h = 3.

Worked example 8.9.2

A cyclist rides 12 km at 15 km/h and then 8 km at 10 km/h. Work out her average speed for the whole journey.

Method. Average speed = total distance ÷ total time. Use t = d/s to find each time first.

Find the time for the first part. Time = distance ÷ speed.
t = 1215 = 0.8 hours (= 0.8 h)
Find the time for the second part. Same formula.
t = 810 = 0.8 hours (= 0.8 h)
Add the distances and the times. The whole journey is both parts together.
distance = 12 + 8 = 20 km
time = 0.8 + 0.8 = 1.6 hours
Work out the average speed. Use s = d/t with the totals.
s = 201.6 = 12.5 km/h

Check. In 1.6 hours at 12.5 km/h she would travel 1.6 × 12.5 = 20 km. ✓

Watch out. The average of the two speeds is 12.5 here only by chance; try 12 km at 12 km/h and 4 km at 4 km/h to see it fail.

Worked example 8.9.3

A circular lawn has circumference 31.4 m. Use π = 3.14 and the formulae C = 2πr and A = πr² to work out its area.

Method. The area needs the radius. Use the circumference formula to find r first, then use it in the area formula.

Make r the subject of C = 2πr. Divide both sides by 2π.
r = C2π
Substitute to find r. 2 × 3.14 = 6.28 and 31.4 ÷ 6.28 = 5.
r = 31.42 × 3.14 = 31.46.28 = 5 m
Use r in the area formula. Square first, then multiply by π.
A = 3.14 × 5² = 3.14 × 25 = 78.5 m²

Check. Estimate: π ≈ 3 and 3 × 25 = 75, close to 78.5. ✓

Watch out. Find the radius first. Do not put the circumference straight into the area formula.

Worked example 8.9.4

A circle has circumference C and area A. Show that A = C²4π.

Method. Find r from the circumference formula, then substitute it into the area formula and simplify.

Make r the subject of C = 2πr. Divide both sides by 2π.
r = C2π
Substitute into A = πr². Replace r by C/(2π). The whole fraction is squared.
A = π × (C2π)²
Square the fraction. Square the top and the bottom: (2π)² = 4π².
A = π × C²4π²
Cancel one π. π on the top cancels with one of the two π factors in π² on the bottom.
A = C²4π

Check. Test with r = 1 and π = 3: C = 6, A = 3. Then C²/(4π) = 36/12 = 3. ✓

Watch out. Square the whole of 2π, including the 2: (2π)² = 4π², not 2π².

Chapter summary

Skill	Method
Vocabulary	Expression: no =. Equation: true for some values. Formula: a rule linking quantities. Identity (≡): true for all values. A term is one part of an expression.
Writing algebra from words	Choose a letter; “more than” is +, “less than” is − (start from the quantity), “times” is ×, “shared between” is ÷. Test with a number.
Substituting	Replace each letter with its number (put the × sign back, brackets round negatives), then use BIDMAS.
Negative numbers	Subtracting a negative = adding. Same signs multiply to give a positive. (−3)² = 9 but −3² = −9.
Using a formula	Write the formula, substitute, work out, write the answer with units. Keep units consistent. Round only at the end.
Writing a formula	Say what each letter stands for. Typical shape: total = fixed part + rate × amount. Test with a simple number.
Change the subject: one or two steps	Undo the operations in reverse order using inverse operations, doing the same to both sides each time.
Brackets, fractions, powers, roots	Divide to undo a bracket multiplier; multiply to undo a division; take the square root of the whole side to undo a square; square to undo a root.
Higher: subject appears twice	Clear fractions, expand brackets, collect subject terms on one side, factorise the subject out, divide by the bracket.
Checking	Test the formula or the rearranged formula with simple numbers. A rearrangement should take you back to where you started.

Can you…?

I can say whether something is an expression, an equation, a formula or an identity.
I can write an expression or a formula from words and from a diagram.
I can substitute positive and negative numbers into an expression, using BIDMAS.
I can use formulae from science and geometry and give answers with units.
I can write a formula for a real situation and test it.
I can change the subject of a formula in one or two steps.
I can change the subject when the formula has brackets, fractions, powers or roots.
Higher: I can change the subject when the subject appears twice, by factorising.
I can check a rearranged formula by substituting numbers.

Key terms

Variable: A letter that stands for a number that can change.
Expression: Numbers and letters combined with operations, with no equals sign, such as 3n + 5.
Equation: Two expressions joined by an equals sign, true only for some values of the letter.
Formula: A rule, written with letters and an equals sign, that links quantities, such as A = lw.
Identity: An equation that is true for every value of the letters, written with ≡.
Term: One part of an expression, separated from the others by + or − signs.
Substitute: Replace a letter with its number value.
Subject: The letter on its own on one side of a formula.
Change the subject: Rearrange a formula so that a different letter is on its own.
Inverse operation: An operation that undoes another: adding and subtracting, multiplying and dividing, squaring and square rooting.
BIDMAS: The order of operations: Brackets, Indices, Division and Multiplication, Addition and Subtraction.
Factorise: Write an expression as a product, for example xy − x = x(y − 1).

Total 48 marks. Allow about 48 minutes.

← Chapter 7: Algebra basics Chapter 9: Linear equations →

GCSE Maths Substitution and Formulae (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

You should already know

Where this leads

Exercise P Do you remember?

Letters, expressions and formulae

Writing algebra in short form

Algebra shorthand

Turning words into algebra

Watch out

Exercise 8A Words, expressions and formulae

Substituting into expressions

The order of working: BIDMAS

Powers

Brackets, fractions and roots

Key ideas

Watch out

Exercise 8B Substituting into expressions

Substituting negative numbers

Squaring a negative number

Key ideas

Watch out

Exercise 8C Substituting negative numbers

Substituting into formulae

Using a formula: four steps

Key ideas

Watch out

Exercise 8D Using formulae (non-calculator)

Exercise 8E Using formulae with a calculator

Writing formulae

Formulae from diagrams

Key ideas

Watch out

Exercise 8F Writing formulae

Changing the subject: one step and two steps

Think of a function machine

Key ideas

Watch out

Exercise 8G Changing the subject: one and two steps

Changing the subject: brackets, fractions, powers and roots

Brackets

Fractions

Powers and roots

Key ideas

Watch out

Exercise 8H Brackets, fractions, powers and roots

Higher: the subject appears twice

Key ideas

Watch out

Exercise 8I Subject appears twice

Using formulae: problems and reasoning

Key ideas

Watch out

Exercise 8J Using formulae: problems and reasoning

Chapter summary

Can you…?

Key terms

Exercise 8Z Check-up (non-calculator)

Exercise 8X Exam-style questions