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GCSE Maths · EdexcelChapter 8 · Substitution and formulae
Chapter 8 · Foundation tier

GCSE Maths Substitution and Formulae (Foundation): 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)

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, Foundation 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.

Exercise P  Do you remember?

Non-calculator
1
FWork out each of these.
(a)−3 + 7
(b)4 − (−5)
(c)−2 × 6
(d)(−3)2
2
FWork out each of these, using the order of operations (BIDMAS).
(a)2 + 3 × 42
(b)(2 + 3) × 4
(c)20 − 12 ÷ 4
3
FSimplify by collecting like terms.
(a)3a + 2a
(b)5x − 2x + x
(c)4p + 3q − p
4
FSolve each equation.
(a)x + 7 = 12
(b)3x = 21
(c)x4 = 5
5
FWrite an expression for “3 more than twice n”.
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.

WordMeaningExample
ExpressionNumbers and letters joined by +, −, ×, ÷. There is no equals sign.3n + 5
EquationTwo expressions joined by an equals sign. It is true for only some values of the letter.3n + 5 = 20 (true when n = 5)
FormulaA rule linking real quantities, written with an equals sign. You use it again and again with different numbers.A = lw (area = length × width)
IdentityTwo 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.

n5Tia: n + 5
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 questionOperationExample in algebra
more than, plus, sum, total, increased byadd (+)7 more than n is n + 7
less than, minus, fewer, decreased by, take awaysubtract (−)7 less than n is n − 7
times, multiplied by, product, twice, double, triplemultiply (×)double n is 2n
divided by, shared equally between, half ofdivide (÷)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”.

n×3+23n + 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.

  1. “7 less than” “Less than” means take away. The 7 is taken away from the number.
    − 7
  2. “n” The letter n stands for the number we start with. It comes first because we start from it.
    n
  3. 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.

  1. “A pen costs p pence” We let the letter p stand for the price of one pen, in pence.
    p
  2. “4 pens” 4 pens cost 4 lots of p, which is 4 × p. In algebra shorthand that is 4p.
    4p
  3. “and a notebook costing 90 pence” “And” tells us to add the notebook. The notebook is a fixed 90 pence.
    + 90
  4. 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.

  1. (a) 5n − 2 There is no equals sign.
    It is an expression. It is a phrase made of the terms 5n and −2.
  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.
  3. (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).

Exercise 8A  Words, expressions and formulae

Non-calculator
1
FWrite down whether each of these is an expression, an equation, a formula or an identity.
(a)4n − 7
(b)4n − 7 = 21
(c)3(n + 2) ≡ 3n + 6
(d)2(l + w)
(e)P = 2l + 2w
(f)5 + 2x
2
FWrite each of these in the shortest algebra form.
(a)3 × n
(b)n × n
(c)4 × a × b
(d)b × a × 3
(e)x ÷ 5
(f)2 × (n + 4)
3
FWrite an expression for each statement. Use the letter n for the number.
(a)5 more than n
(b)6 times n
(c)9 less than n
(d)n shared equally between 4 people
(e)double n, then add 3
(f)take n away from 12
4
FPens cost 35 pence each and a ruler costs 60 pence. Write an expression, in pence, for the cost of:
(a)n pens
(b)n pens and one ruler
(c)the change from £5 when buying n pens and one ruler
5
FAli is a years old. Ben is 3 years older than Ali. Cara is half as old as Ali. Write an expression for:
(a)Ben’s age
(b)Cara’s age
(c)the total of their three ages (you do not need to simplify)
6
RDan says, “4 more than n is 4n.” Show, using n = 5, that Dan is wrong, and write the correct expression.
7
PSA taxi charges £3 to start, plus £2 for every mile. The cost is £C for a journey of m miles.
(a)Write a formula for C in terms of m.
(b)Use your formula to work out the cost of a 5 mile journey.
8
FWrite an expression for each description.
(a)Add 3 to n, then double the result
(b)Double n, then add 3
(c)Multiply n by itself, then add 3
(d)Add 3 to n, then multiply the result by itself
9
FThe diagram shows a triangle. The lengths are in centimetres. Write down and simplify an expression for the perimeter.
xx + 32x − 1
10
PSAmir has £a. Bea has £5 more than Amir. Cal has twice as much as Amir. Write an expression for the amount that:
(a)Bea has
(b)Cal has
(c)all three have altogether, in its simplest form
11
PSn is a whole number. Write an expression for:
(a)the next whole number after n
(b)the whole number just before n
(c)the next even number after the even number 2n
12
RTia says that 3(n + 2) is the same as 3n + 2. Test her idea with n = 1 and n = 4. Is she right?
13
PSA cinema sells adult tickets for £a each and child tickets for £c each. A family buys 3 adult tickets and 2 child tickets. Write an expression for the total cost in pounds.
14
Rn is a whole number. Explain why 2n is always even and why 2n + 1 is always odd.
15
RA rectangle has width w cm. Its length is 3 cm more than twice its width. Write and simplify an expression for the perimeter of the rectangle. Explain why the perimeter is always a multiple of 6.
w2w + 3
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.

BBracketsinside brackets first (also top and bottom of fractions)IIndicespowers such as 5² and roots such as √DMDivision and Multiplicationwork from left to rightASAddition and Subtractionwork from left to right1stlast
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.

  1. Write down the expression. Start by copying the expression exactly.
    3x + 2
  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
  3. 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
  4. 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.

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.

  1. Write down the expression. Start by copying the expression exactly.
    2a2 − b
  2. 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
  3. 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
  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
  5. 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.

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.

  1. Write down the expression. Start by copying the expression exactly.
    2a + bc
  2. 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
  3. 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
  4. 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
  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.
    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.

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.

  1. Write down the expression. Start by copying the expression exactly.
    √x2 + y2
  2. Substitute: x = 6, y = 8. Substitute means replace each letter with its number.
    √x2 + y2 = √62 + 82
  3. 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
  4. 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
  5. 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.

Exercise 8B  Substituting into expressions

Non-calculator
1
FWork out the value of each expression when x = 4.
(a)3x
(b)x + 7
(c)10 − x
(d)x2
(e)5x − 3
(f)2x + 9
2
FWork out the value of each expression when a = 3 and b = 5.
(a)a + b
(b)ab
(c)2a + b
(d)3b − a
(e)ab + 4
(f)4a − 2b + 1
3
FWork out the value of each expression when p = 2 and q = 10.
(a)qp
(b)pq
(c)q − p
(d)3p + q
(e)2q − 5p
4
FWork out the value of each expression when x = 3.
(a)x2
(b)2x2
(c)x2 + 1
(d)x3
(e)5x2 − x
(f)10 − x2
5
FWork out the value of each expression when n = 5.
(a)2(n + 3)
(b)3(n − 1)
(c)4(2n − 3)
(d)(n − 2)2
(e)(n + 2)(n − 1)
6
FWork out the value of each expression when x = 8 and y = 4.
(a)x + y3
(b)xy
(c)3xy
(d)x + 4y − 2
(e)x2y
7
RTia is asked to work out 5x when x = 3. She writes “5x = 53”. Explain her mistake and write the correct value.
8
PSA gym charges a joining fee of £12 plus £8 for each class. The cost in pounds of n classes is C = 8n + 12. Work out the cost of:
(a)3 classes
(b)6 classes
(c)10 classes
9
FComplete this table of values for the expression 4x − 3. Work out the value for each x.
(a)x = 1
(b)x = 2
(c)x = 3
(d)x = 4
(e)x = 5
10
PSA rectangle has length l = 12 cm and width w = 7.5 cm. Use the formulae to work out
12 cm7.5 cm
(a)the perimeter, P = 2(l + w)
(b)the area, A = lw
11
FWork out the value of each expression.
(a)√x  when x = 49
(b)√2x + 1  when x = 12
(c)√x2 + y2  when x = 5, y = 12
12
PSThe cost in pounds to hire a bike for h hours is C = 5h + 3. Sam hires a bike for 4 hours and Dev hires one for 7 hours.
(a)How much does Sam pay?
(b)How much does Dev pay?
(c)How much more does Dev pay than Sam?
13
R(a) Work out 2x and x2 when x = 1, 2, 3 and 4. (b) For which of these values of x are 2x and x2 equal? (c) Is 2x always bigger than x2 after that value? Use your results to explain.
14
RThe expression 4n + 1 is worked out for n = 4, n = 5 and n = 6. Which value of n gives 25?
15
Rx and y are whole numbers with x + y = 9 and xy = 20. Work out the value of x2 + y2.
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.

-5-4-3-2-1012345+ (−3)-5-4-3-2-1012345− (−3)
Top: 2 + (−3) = −1. Adding a negative moves left. Bottom: 2 − (−3) = 5. Subtracting a negative moves right.

These are the rules you need.

CalculationRuleExample
Add a negativesame as subtracting5 + (−3) = 5 − 3 = 2
Subtract a negativesame as adding5 − (−3) = 5 + 3 = 8
Positive × negative (or negative × positive)answer is negative4 × (−3) = −12
Negative × negativeanswer is positive(−4) × (−3) = 12
Dividingfollows 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:

ExpressionSubstituteWorkingValue
x2(−3)2(−3) × (−3)9
−x2−(−3)2square first: 9, then the minus sign in front makes it negative−9
(−x)2(−(−3))2−(−3) = 3, then 329

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.

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.

  1. Write down the expression. Start by copying the expression exactly.
    3x + 7
  2. 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
  3. 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
  4. 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.

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.

  1. Write down the expression. Start by copying the expression exactly.
    a − 2b
  2. 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)
  3. 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)
  4. 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.

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.

  1. Write down the expression. Start by copying the expression exactly.
    x2 − 3x
  2. 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)
  3. 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)
  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)
  5. 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.

Exercise 8C  Substituting negative numbers

Non-calculator
1
FWork out each of these. They revise the rules for negative numbers.
(a)5 + (−3)
(b)−4 − (−6)
(c)3 × (−4)
(d)−2 × (−7)
(e)12 ÷ (−3)
(f)−20 ÷ (−4)
2
FWork out the value of each expression when x = −2.
(a)x + 5
(b)3x
(c)4 − x
(d)x − 1
(e)2x + 7
(f)10 + 3x
3
FWork out the value of each expression when a = −3 and b = 4.
(a)a + b
(b)ab
(c)a − b
(d)2a + 3b
(e)b − a
4
FWork out the value of each expression when x = −5.
(a)x2
(b)−x2
(c)(−x)2
(d)x2 + 3
(e)3x2
5
PSThe formula to change a temperature from degrees Celsius to degrees Fahrenheit is F = 9C5 + 32. Work out F when
(a)C = −10
(b)C = −5
(c)C = −40
6
FWork out the value of each expression when a = −2 and b = −3.
(a)ab
(b)a2b
(c)a2 + b2
(d)a − b
(e)ab2
(f)(a + b)2
7
RLena says, “When x = −3, x2 = −9.” Is she correct? Explain.
8
PSJo has £40 in her bank account and spends £25 each month with no money coming in. After t months her balance is B = 40 − 25t pounds.
(a)Work out B when t = 1.
(b)Work out B when t = 2.
(c)Work out B when t = 3.
(d)What does your answer to part (c) mean for Jo?
9
FWork out the value of each expression when x = −4 and y = 6.
(a)x + y2
(b)yx
(c)3xy
(d)x2y + 2
10
PSWhen x = −2, put these expressions in order, starting with the smallest value: x2,   2x,   x3,   −x.
11
FWork out the value of each expression when x = −1.
(a)(x − 3)2
(b)3(2 − x)
(c)(x + 4)(x − 2)
(d)5 − (x + 2)
12
RShow that x2 + 2x + 1 = 0 when x = −1.
13
RTwo different numbers x satisfy x2 = 16. Find both numbers and show that each one works.
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.

QuantityFormulaWhat the letters mean (usual units)
Area of a rectangleA = lwl length, w width (cm², m²)
Perimeter of a rectangleP = 2l + 2wdistance round the outside (cm, m)
Area of a triangleA = 12bhb base, h perpendicular height
Area of a trapeziumA = 12(a + b)ha and b the parallel sides, h the distance between them
Volume of a cuboidV = lwhlength, width, height (cm³, m³)
Circumference of a circleC = πdd diameter
Area of a circleA = πr²r radius
Speeds = dtdistance d and time t (km/h, m/s)
DensityD = mVmass m and volume V (g/cm³)
PressureP = FAforce F and area A (N/m²)
ForceF = mamass m (kg), acceleration a (m/s²), force in newtons (N)
Final speedv = u + atu starting speed, a acceleration, t time
Kinetic energyE = 12mv²mass m (kg), speed v (m/s), energy in joules (J)
Electric potential differenceV = IRcurrent I (amps), resistance R (ohms), volts (V)
abhlhw
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

  1. Write the formula.
  2. Substitute each known value (with its × sign, and brackets round negatives).
  3. Work out the answer in the correct order (BIDMAS).
  4. 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.

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

  1. Write down the formula. This is the rule we are going to use.
    s = dt
  2. Substitute: d = 150, t = 2.5. Substitute means replace each letter with its number.
    s = 1502.5
  3. 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
  4. 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.

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

  1. Write down the formula. This is the rule we are going to use.
    A = 12(a + b)h
  2. 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
  3. 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
  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.
    12 × 18 = 9
    A = 9 × 6
  5. 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
  6. 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.

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

  1. Write down the formula. This is the rule we are going to use.
    E = 12mv2
  2. 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
  3. 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
  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.
    12 × 4 = 2
    E = 2 × 100
  5. 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
  6. 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.

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

  1. Write down the formula. This is the rule we are going to use.
    A = πr2
  2. 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
  3. 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…
  4. 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)
  5. 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.

Exercise 8D  Using formulae (non-calculator)

Non-calculator
1
FUse each formula to work out the answer.
(a)Area of a rectangle with length 12 cm and width 5 cm: A = lw
(b)Perimeter of the same rectangle: P = 2l + 2w
(c)Volume of a cuboid 6 cm by 4 cm by 3 cm: V = lwh
(d)Area of a triangle with base 10 cm and height 7 cm: A = ½bh
2
FWork out each answer. Remember the units.
(a)A cyclist travels 240 km in 3 hours. Speed: s = d/t
(b)A metal block has mass 540 g and volume 200 cm³. Density: D = m/V
(c)A force of 600 N acts on an area of 20 m². Pressure: P = F/A
(d)A mass of 12 kg has acceleration 4 m/s². Force: F = ma
3
FA car has starting speed u, acceleration a and time t. Its final speed is v = u + at. Work out v when
(a)u = 5, a = 3, t = 4
(b)u = 12, a = 2, t = 6
(c)u = 0, a = 10, t = 3
4
FThe electrical formula V = IR gives the potential difference V volts from the current I amps and the resistance R ohms. Work out V when
(a)I = 3 and R = 8
(b)I = 0.5 and R = 12
5
PSThe time, T minutes, needed to cook a joint of meat is T = 40m + 20, where m is the mass of the joint in kg. Work out the cooking time for a joint of mass
(a)1.5 kg
(b)2.5 kg
(c)3 kg
6
PSA formula to change degrees Celsius (C) to degrees Fahrenheit (F) is F = 1.8C + 32. Work out F when
(a)C = 20
(b)C = 100
(c)C = −40
7
FWork out the area of the trapezium using A = ½(a + b)h.
5 cm9 cm4 cm
8
FWork out the kinetic energy E = ½mv² (in joules) when
(a)m = 2 and v = 6
(b)m = 10 and v = 3
(c)m = 0.5 and v = 8
9
RZoe uses v = u + at with u = 4, a = 3 and t = 2. She writes v = 7 × 2 = 14. Explain her mistake and work out the correct value of v.
10
PSCar A travels 180 km in 2.5 hours. Car B travels 150 km in 2 hours. Use s = d/t.
(a)Work out the average speed of car A.
(b)Work out the average speed of car B.
(c)Which car has the greater average speed?
11
PSBlock X has mass 840 g and volume 300 cm³. Block Y has mass 620 g and volume 250 cm³. Use D = m/V.
(a)Work out the density of block X.
(b)Work out the density of block Y.
(c)Which block is denser?
12
PSThe final speed v m/s of an object is v = √(u² + 2as). Work out v when
(a)u = 4, a = 3 and s = 8
(b)u = 6, a = 2 and s = 7
13
RUse P = F/A to show that a force of 500 N on an area of 25 m² gives the same pressure as a force of 100 N on an area of 5 m².

Exercise 8E  Using formulae with a calculator

Calculator allowed
1
FUse A = πr² to work out the area of each circle. Give your answers to 1 decimal place.
(a)radius 4.5 cm
(b)radius 12 cm
(c)radius 0.8 cm
2
FUse C = πd to work out the circumference of each circle. Give your answers to 1 decimal place.
(a)diameter 7.3 cm
(b)diameter 15.6 cm
3
FThe volume of a cylinder is V = πr²h. Work out the volume of each cylinder, to 1 decimal place.
(a)radius 3.5 cm, height 10 cm
(b)radius 2.4 cm, height 7.5 cm
4
PSUsain Bolt ran 100 m in 9.58 seconds. Use s = d/t to work out his average speed in m/s, to 1 decimal place.
5
FA metal sample has mass 84.6 g and volume 12.5 cm³. Use D = m/V to work out its density. Give your answer to 3 significant figures.
6
PSA stone falls from a height of 12 m. Its speed on landing is v = √(2gh), where g = 9.8 and h is the height in metres. Work out v to 1 decimal place.
7
PSThe time for a pendulum of length l metres to swing is T = 2π√(l/g) seconds, where g = 9.8. Work out T when l = 1.5, to 2 decimal places.
8
FThe volume of a sphere is V = 43πr³. Work out the volume of a sphere of radius 2.4 cm, to 1 decimal place.
9
PSA circular ring has outer radius R = 8.5 cm and inner radius r = 6.2 cm. Its area is A = π(R² − r²). Work out A to 1 decimal place.
10
PSA cylindrical tin has radius 4.2 cm and height 11.5 cm. Work out its volume, to the nearest cm³.
11
FUse F = 1.8C + 32 to change a body temperature of 36.6°C into degrees Fahrenheit. Give your answer to 1 decimal place.
12
RSam works out the area of a circle with radius 5 cm by typing (π × 5)² on his calculator. He gets 246.7. Explain his mistake and work out the correct area, to 1 decimal place.
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:

  1. Choose letters and say what each stands for, including units (for example C = cost in pounds, n = number of tickets).
  2. Split the rule into a fixed part (paid once, never changes) and a changing part (a rate multiplied by the amount).
  3. Write the formula with the subject on its own on the left.
  4. 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.

1 triangle2 triangles3 triangles
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.

  1. “£8 a month” A fixed amount, paid once. It does not depend on the minutes.
    8
  2. “4p for each minute” 4p = £0.04. With m minutes the cost is 0.04 × m.
    0.04m
  3. “plus” The fixed charge and the minute charge are added.
    +
  4. 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.

x + 32x − 1
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.

  1. Write the perimeter. There are 2 lengths and 2 widths.
    P = 2(x + 3) + 2(2x − 1)
  2. 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
  3. 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.

  1. “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
  2. “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
  3. 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.

Exercise 8F  Writing formulae

Non-calculator
1
FWrite a formula for each situation. Use the letters given.
(a)The cost C pounds of n tickets at £7 each
(b)The total T pounds charged by a plumber who charges £40 call-out plus £25 per hour for h hours
(c)The perimeter P of a square with side s
(d)The area A of a square with side s
(e)The mean m of two numbers a and b
(f)The change C pounds from £20 after buying n books at £4 each
2
PSA plumber charges a £40 call-out fee plus £25 for each hour of work.
(a)Write a formula for the total charge T pounds for h hours.
(b)Use your formula to find the charge for 3 hours.
(c)A customer is charged £140. For how many hours did the plumber work?
3
FThe diagram shows a rectangle with length (x + 5) cm and width 3 cm. Write simplified expressions for
x + 53
(a)the perimeter, P cm
(b)the area, A cm²
4
FA square has side s cm. Write a formula for
ss
(a)its perimeter P
(b)its area A
5
PSMatchsticks are used to make a row of n triangles. The formula is m = 2n + 1, where m is the number of matchsticks.
1 triangle2 triangles3 triangles
(a)How many matchsticks are needed for 10 triangles?
(b)Meg has exactly 100 matchsticks. Explain why she cannot use them all to make a row of triangles.
6
PSA bike shop charges £6 for each hour plus a £10 deposit.
(a)Write a formula for the total cost C pounds for h hours.
(b)Work out the cost for 4 hours.
7
PSAn electricity company charges £15 a month plus 20p for each unit used. Write a formula for the monthly cost C pounds for u units and use it to find the cost of 250 units.
(a)Formula for C
(b)Cost of 250 units
8
RMatt writes a formula for the cost, in pounds, of n pencils at 45p each and a ruler costing £2: C = 45 + 2n. Test his formula with n = 2. Explain what is wrong and write a correct formula.
9
PSThe shape is made from two rectangles. The lengths are in centimetres. Write simplified expressions for the perimeter P and the area A of the whole shape.
3x4x5
(a)perimeter, P cm
(b)area, A cm²
10
PSA hall costs £120 to hire plus £3.50 for each guest. Write a formula for the total cost C pounds for g guests, and work out the cost for 60 guests.
(a)Formula for C
(b)Cost for 60 guests
11
PSA farmer fences a rectangular field on three sides. The fourth side is a wall. The field has length l metres and width w metres.
walllww
(a)Write a formula for the length of fence, F metres.
(b)Work out F when l = 40 and w = 15.
12
PSShirts cost £p each. A shop takes 20% off the total price when you buy n shirts. Write a formula for the amount paid, £C, and use it to find the cost of 3 shirts at £10 each.
(a)Formula for C
(b)Cost of 3 shirts at £10
13
RTaxi firm A charges £2 plus £1.50 per mile. Taxi firm B charges £5 plus £1 per mile. For a journey of m miles the costs are A = 2 + 1.5m and B = 5 + m pounds.
(a)Work out both costs for a 6 mile journey. What do you notice?
(b)Which firm is cheaper for a 10 mile journey? Show your working.
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.

x×3+2yy−2÷3x
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:

OperationInverse (undoes it)
add a numbersubtract the same number
subtract a numberadd the same number
multiply by a numberdivide by the same number
divide by a numbermultiply by the same number
squaresquare root
square rootsquare

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.

Many students use a formula triangle for formulae like speed = distance ÷ time. Cover the letter you want, and the triangle shows the formula. It is a useful memory aid for s = d/t, but it only works for formulae shaped like this, so make sure you can also rearrange using inverse operations.

dst
Cover d: d = s × t
Cover s: s = d ÷ t
Cover t: t = d ÷ s
The speed–distance–time triangle. Letters side by side are multiplied; a letter above another is divided by it.

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.

  1. 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.
  2. 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
  3. 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.

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 x is on its own.

  1. 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.
  2. 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
  3. 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
  4. 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.

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.

  1. 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.
  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
  3. 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
  4. 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.

Exercise 8G  Changing the subject: one and two steps

Non-calculator
1
FMake x the subject of each formula.
(a)Make x the subject of y = x + 5
(b)Make x the subject of y = x − 3
(c)Make x the subject of y = 4x
(d)Make x the subject of y = x3
(e)Make x the subject of y = x + a
(f)Make x the subject of y = kx
2
FMake x the subject of each formula.
(a)Make x the subject of y = 2x + 3
(b)Make x the subject of y = 5x − 4
(c)Make x the subject of y = x2 + 1
(d)Make x the subject of y = x4 − 2
(e)Make x the subject of y = 7 + 3x
(f)Make x the subject of y = 2x − 9
3
FRearrange each formula to make the letter in brackets the subject.
(a)Make a the subject of F = ma
(b)Make m the subject of F = ma
(c)Make I the subject of V = IR
(d)Make R the subject of V = IR
(e)Make d the subject of s = dt
(f)Make t the subject of s = dt
4
FMake the letter shown the subject.
(a)Make u the subject of v = u + at
(b)Make a the subject of v = u + at
(c)Make d the subject of C = πd
(d)Make h the subject of A = 12bh
(e)Make w the subject of P = 2l + 2w
(f)Make s the subject of P = 4s
5
FMake x the subject of each formula.
(a)Make x the subject of y = mx + c
(b)Make m the subject of y = mx + c
(c)Make c the subject of y = mx + c
6
RCallum is asked to make x the subject of y = 3x + 2. He writes x = y − 23. Use x = 5 to show that he is wrong, and give the correct answer.
7
PSThe cost, £C, of hiring a van for d days is C = 35d + 20.
(a)Make d the subject of the formula.
(b)Use your formula to find how many days the van was hired for when the cost was £195.
8
PSThe area of a triangle is A = ½bh.
(a)Make h the subject of the formula.
(b)A triangle has area 30 cm² and base 12 cm. Work out its height.
9
PSA car has mass m kg and a force of F newtons acts on it. The acceleration a m/s² is given by F = ma.
(a)Make a the subject of the formula.
(b)Work out a when F = 3000 and m = 1200.
10
PSA train travels a distance d km at an average speed s km/h. It takes t hours, where s = d/t.
(a)Make t the subject.
(b)How long does a journey of 360 km take at an average speed of 90 km/h?
11
FThe simple interest, I, earned on a sum of money P at R% a year for T years is I = PRT100. Make P the subject.
12
FMake y the subject of each formula.
(a)Make y the subject of x = 2y + 1
(b)Make y the subject of x = y5 − 3
(c)Make y the subject of a = 3y − b
(d)Make y the subject of p = q + 4y
13
PSThe cooking time T minutes for a turkey of mass m kg is T = 40m + 20. Jess cooks a turkey for 180 minutes. Make m the subject and work out the mass of her turkey.
(a)Make m the subject.
(b)Work out m when T = 180.
14
RStarting from s = d/t, show that d = st.
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.

r²×πAA÷π√r
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).

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 w is on its own.

  1. 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.
  2. 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
  3. 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
  4. 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.

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 r is on its own.

  1. 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.
  2. 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
  3. 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
  4. 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.

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 v is on its own.

  1. 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.
  2. 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
  3. 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
  4. 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
  5. 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. ✓

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.

Exercise 8H  Brackets, fractions, powers and roots

Non-calculator
1
FMake x the subject of each formula.
(a)Make x the subject of y = 3(x + 2)
(b)Make x the subject of y = 5(x − 1)
(c)Make x the subject of y = 4(x + 3) − 1
(d)Make x the subject of y = 2(3x − 1)
2
FMake the letter shown the subject of each formula.
(a)Make l the subject of P = 2(l + w)
(b)Make b the subject of A = 12(a + b)h
(c)Make C the subject of F = 95C + 32
(d)Make a the subject of S = 2(ab + bc)
3
FMake x the subject of each formula.
(a)Make x the subject of y = x5
(b)Make x the subject of y = x + 32
(c)Make x the subject of y = 2x − 13
(d)Make x the subject of M = x + y2
4
FMake r (or the letter shown) the subject of each formula.
(a)Make s the subject of A = s2
(b)Make r the subject of A = πr2
(c)Make x the subject of y = x2 + 3
(d)Make v the subject of E = 12mv2
(e)Make x the subject of y = 2x2
5
FMake x (or the letter shown) the subject of each formula.
(a)Make x the subject of y = √x
(b)Make x the subject of y = √x + 3
(c)Make h the subject of v = √2gh
(d)Make x the subject of y = √5x
6
RBen says that A = πr² can be rearranged to r = √A/π. Use r = 2 to show that he is wrong, and write the correct formula.
7
PSThe area of a circular pond is A = πr².
(a)Make r the subject.
(b)The pond has area 28.26 m². Use π = 3.14 to work out its radius.
8
PSAn object dropped from rest falls a distance d metres in t seconds, where d = ½gt² and g = 10.
(a)Make t the subject.
(b)How long does it take to fall 80 m?
9
PSThe kinetic energy of an object is E = ½mv².
(a)Make v the subject.
(b)Work out v when E = 100 joules and m = 2 kg.
10
PSMake h the subject of A = h(a + b)2 and work out h when A = 48, a = 5 and b = 7.
(a)Make h the subject
(b)Work out h
11
R(a) Make x the subject of y = 2x². (b) Work out x when y = 50. (c) There is a second value of x that also gives y = 50. What is it?
(a)Make x the subject
(b)x when y = 50
(c)The second value of x
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.

1. Readlist what youknow and want2. Choosethe formula3. Rearrange(if the subjectis wrong)4. Substituteand work out5. Checkunits andsense
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.

Other useful skills in this section:

rh
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.

  1. Write the formula and substitute. Put in A = 54, a = 7 and b = 11.
    54 = ½ × (7 + 11) × h
  2. Work out the bracket, then ½ of it. Brackets first. 7 + 11 = 18 and ½ of 18 is 9.
    54 = 9h
  3. 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.

  1. Find the time for the first part. Time = distance ÷ speed.
    t = 1215 = 0.8 hours (= 0.8 h)
  2. Find the time for the second part. Same formula.
    t = 810 = 0.8 hours (= 0.8 h)
  3. 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
  4. 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.

  1. Make r the subject of C = 2πr. Divide both sides by 2π.
    r = C2π
  2. 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
  3. 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.

Exercise 8J  Using formulae: problems and reasoning

Non-calculator
1
PSThe volume of a cylinder is V = πr²h.
rh
(a)Make h the subject.
(b)A cylinder has volume 157 cm³ and radius 5 cm. Use π = 3.14 to work out its height.
2
PSA bicycle wheel has circumference 220 cm. Use C = πd with π = 227 to work out the diameter of the wheel.
3
PSA car travels 30 km at 60 km/h and then 60 km at 40 km/h. Use t = d/s to find the time for each part, then work out the average speed for the whole journey.
4
R(a) Make C the subject of F = 95C + 32. (b) Show that −40°C is the same temperature as −40°F.
(a)Make C the subject
(b)Show that −40°C = −40°F
5
RThe area of a circle of radius r is A = πr². Show that doubling the radius makes the area four times as big.
6
PSA car of mass 1200 kg has kinetic energy E = ½mv² joules when its speed is v m/s.
(a)Work out E when v = 20.
(b)Work out E when v = 40.
(c)How many times bigger is the energy when the speed is doubled?
7
PSDensity is D = m/V.
(a)Make m the subject.
(b)Steel has density 7800 kg/m³. Work out the mass of 2.5 m³ of steel.
8
RShow that the expression 3(x + 2) − 3x − 6 is always equal to 0, whatever the value of x.
9
PSA model for the stopping distance d metres of a car travelling at v metres per second is d = 0.3v + v²50.
(a)Work out d when v = 20.
(b)Work out d when v = 40.
(c)Tom says, “If the speed doubles, the stopping distance doubles.” Use your answers to show that Tom is wrong.
10
RThe formula y = 3x + 2 is rearranged to give x = y − 23. Show that putting this value of x back into 3x + 2 gives y.

Chapter summary

SkillMethod
VocabularyExpression: 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 wordsChoose a letter; “more than” is +, “less than” is − (start from the quantity), “times” is ×, “shared between” is ÷. Test with a number.
SubstitutingReplace each letter with its number (put the × sign back, brackets round negatives), then use BIDMAS.
Negative numbersSubtracting a negative = adding. Same signs multiply to give a positive. (−3)² = 9 but −3² = −9.
Using a formulaWrite the formula, substitute, work out, write the answer with units. Keep units consistent. Round only at the end.
Writing a formulaSay what each letter stands for. Typical shape: total = fixed part + rate × amount. Test with a simple number.
Change the subject: one or two stepsUndo the operations in reverse order using inverse operations, doing the same to both sides each time.
Brackets, fractions, powers, rootsDivide 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 twiceClear fractions, expand brackets, collect subject terms on one side, factorise the subject out, divide by the bracket.
CheckingTest 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).

Exercise 8Z  Check-up (non-calculator)

Non-calculator
1
Work out the value of 5x + 2 when x = −3.
2
Use v = u + at to work out v when u = 4, a = 3 and t = 5.
3
Books cost £6 each and postage is £2. Write a formula for the total cost £C of n books.
4
Make x the subject of y = 4x − 1.
5
Make w the subject of P = 2l + 2w.
6
Work out the value of x2 − 3x when x = −2.
7
Use E = ½mv² to work out E when m = 6 and v = 4.
8
Make t the subject of v = u + at.
9
A rectangle has length (2x − 1) cm and width (x + 4) cm. Write a simplified expression for its perimeter.
2x − 1x + 4
10
Make r the subject of A = πr².

Total 32 marks. Allow about 32 minutes.

Exercise 8X  Exam-style questions

Non-calculator
1
FWork out the value of 5a − 2b when a = 4 and b = −3.[2 marks]
2
Fv = u + at. Work out the value of v when u = 6, a = −2 and t = 4.[3 marks]
3
PSAn electrician charges a £35 call-out fee plus £20 for each hour of work.[4 marks]
(a)Write a formula for the total charge £C for h hours.
(b)Work out the charge for 3.5 hours.
4
FMake x the subject of y = 5x − 7.[2 marks]
5
FA = πr². Make r the subject of the formula.[2 marks]
6
PSThe diagram shows a rectangle with length (3x + 2) cm and width (x − 1) cm. Write a simplified expression for the perimeter of the rectangle.[3 marks]
3x + 2x − 1
7
PSAmy runs 5 km in 25 minutes. Use s = d/t to work out her average speed in km/h.[3 marks]
8
RShow that 3(x + 4) − 2(x + 1) = x + 10.[2 marks]
9
PSThe kinetic energy E joules of an object of mass m kg moving at v m/s is E = ½mv².[4 marks]
(a)Make m the subject of the formula.
(b)Work out m when E = 80 and v = 4.
10
FWork out the value of 2x2 − x when x = −4.[2 marks]
11
RMaya says, “When x = −2, the value of −x2 is 4.” Is Maya correct? You must show your working.[2 marks]
12
RFirm A hires a ladder for £20 plus £5 per hour. Firm B hires the same ladder for £8 plus £8 per hour. Show that the two firms charge the same for 4 hours.[3 marks]
← Chapter 7: Algebra basicsChapter 9: Linear equations →

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