EDEXCEL GCSE MATHS · FOUNDATION & HIGHER
Formulae
Algebra · Lesson 4 of 7
Teacher copy - includes the notes for whoever is teaching from it.
Last Lesson and Before
Answer each one, then check.
1. If x = 3, what is 2x + 5?
11
2. If y = −2, what is y²?
4
3. What is (−2)³?
−8
4. Last lesson: solve 3x − 4 = 11.
x = 5
5. Last lesson: solve 2(x + 3) = 14.
x = 4
Learning Objectives
1. Tell the difference between an expression, an equation, a formula and an identity.
2. Substitute positive and negative numbers into formulae.
3. Write a formula from a description.
4. Change the subject of a formula.
5. Change the subject when the subject is squared or appears twice. (Higher)
Four Words, Four Meanings
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Expression Terms with no equals sign: 3x + 2. It cannot be solved. |
Equation Two expressions that are equal for particular values: 3x + 2 = 11 is only true when x = 3. |
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Formula A rule linking quantities, each with its own letter: v = u + at, A = πr². |
Identity Two expressions equal for EVERY value of the letter, written with ≡: 2(x + 3) ≡ 2x + 6. |
Substituting into a Formula
Replace each letter with its value, then follow the order of operations.
▸ Use brackets. Put negative numbers in brackets: if a = −2, then a² = (−2)² = 4.
▸ Powers first. In ½at², square t before multiplying.
▸ Show the substitution. Write the formula with the numbers in before working it out: it earns the method mark.
▸ Units. Give the answer in the right units if the question has them.
Substituting Negative Numbers
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s = ut + ½at². Work out s when u = 3, t = 4 and a = −2. |
1. Substitute
s = 3 × 4 + ½ × (−2) × 4²
2. Powers first
4² = 16
3. Multiply
3 × 4 = 12 and ½ × (−2) × 16 = −16
4. Add
12 + (−16) = −4
Answer: s = −4
The Area of a Trapezium
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The area of a trapezium is A = ½(a + b)h, where a and b are the parallel sides and h is the perpendicular height. With a = 5 cm, b = 9 cm and h = 4 cm: A = ½ × (5 + 9) × 4 = ½ × 14 × 4 = 28 cm². |
A = ½(a + b)h: add the parallel sides, halve, multiply by the height. |
PART ONE
Changing the Subject
Rearranging a formula to work out a different letter.
What Is the Subject?
The subject of a formula is the letter on its own on one side: in v = u + at, the subject is v.
▸ The idea. Rearrange the formula so a different letter is on its own.
▸ The method. Exactly like solving an equation: undo what has been done to the new subject, in reverse order, doing the same to both sides.
▸ Why. To find a when you know v, u and t, it is quicker to rearrange once than to solve a new equation every time.
Changing the Subject
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Make t the subject of v = u + at. |
1. Subtract u from both sides
v − u = at
2. Divide both sides by a
(v − u)/a = t
3. Write the subject on the left
t = (v − u)/a
Answer: t = (v − u)/a
Rearranging the Trapezium Formula
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Make h the subject of A = ½(a + b)h. |
1. Multiply both sides by 2
2A = (a + b)h
2. Divide both sides by (a + b)
2A/(a + b) = h
Answer: h = 2A/(a + b)
PART TWO · HIGHER
Harder Rearranging
When the subject is squared, or appears twice.
The Subject Is Squared
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Make r the subject of A = πr². |
1. Divide both sides by π
A/π = r²
2. Square root both sides
√(A/π) = r
Answer: r = √(A/π)
The Subject Appears Twice
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Make x the subject of y = (x + 2)/(x − 3). |
1. Multiply both sides by (x − 3)
y(x − 3) = x + 2
2. Expand
xy − 3y = x + 2
3. Collect the x terms on one side
xy − x = 3y + 2
4. Factorise out x
x(y − 1) = 3y + 2
5. Divide by (y − 1)
x = (3y + 2)/(y − 1)
Answer: x = (3y + 2)/(y − 1)
Case Study
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CASE STUDY Celsius and Fahrenheit The Fahrenheit temperature scale, still used in the USA, is linked to Celsius by the formula F = 1.8C + 32. Rearranging it gives C = (F − 32)/1.8, so a weather forecast of 77 °F becomes (77 − 32)/1.8 = 25 °C. There is one temperature that reads the same on both scales: solve C = 1.8C + 32 and you get C = −40. So −40 °C and −40 °F are exactly the same temperature. |
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−40° The one temperature that is the same on both scales |
100 °C = 212 °F The boiling point of water |
Key Terms
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Formula A rule linking two or more quantities, written with letters. |
Substitute Replace the letters in a formula with numbers. |
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Subject The letter on its own on one side of a formula. |
Change the subject Rearrange a formula so a different letter is the subject. |
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Identity Two expressions that are equal for every value of the letters, written with ≡. |
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Your Task: Rearrange It
12 minutes
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Make the letter in brackets the subject. (a) P = 4s (s) (b) y = mx + c (x) (c) C = 2πr (r) (d) V = IR (I) (e) A = ½bh (b). Higher: (f) E = ½mv² (v) (g) a(x − 2) = x + 5 (x). 1. Undo operations in reverse order. 2. Do the same to both sides. 3. Higher: collect the subject, then factorise. |
A good answer shows: (a) s = P/4 (b) x = (y − c)/m (c) r = C/2π (d) I = V/R (e) b = 2A/h (f) v = √(2E/m) (g) ax − 2a = x + 5, so ax − x = 5 + 2a, x(a − 1) = 5 + 2a and x = (5 + 2a)/(a − 1)
Note: Encourage students to check each rearrangement by substituting numbers into both versions.
Can I...?
☐ Tell a formula from an equation, expression and identity.
☐ Substitute positive numbers into a formula.
☐ Substitute negative numbers into a formula.
☐ Write a formula from words.
☐ Change the subject in one step.
☐ Change the subject in two or more steps.
☐ Change the subject when it is squared. (Higher)
☐ Change the subject when it appears twice. (Higher)
Summary
✓ Expression: no equals sign. Equation: true for some values. Formula: links quantities. Identity: true for all values.
✓ Substitute: numbers in, brackets round negatives, powers first.
✓ Change the subject: undo operations in reverse order, doing the same to both sides.
✓ (Higher) Subject twice: collect those terms, factorise, then divide.
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EXAM FOCUS Make t the subject of the formula v = u + at (2 marks) Write the substitution before calculating, and put negative numbers in brackets. Most lost marks on substitution come from squaring a negative number without brackets. |
Exam Practice: Formulae
Answer all questions. Show your working. Questions 1 to 4 are non-calculator. · 25 minutes
▸ Question 1 · 2 marks · Non-calculator. v = u + at. Work out the value of v when u = 12, a = −3 and t = 5.
▸ Question 2 · 2 marks · Non-calculator. Make t the subject of the formula v = u + at
▸ Question 3 · 2 marks · Non-calculator · Show that. Show that (x + 3)² − 9 ≡ x(x + 6)
▸ Question 4 · 4 marks · Non-calculator · Higher. Make a the subject of 3(a − 2) = ab + 5
▸ Question 5 · 2 marks · Calculator. s = ut + ½at². Work out the value of s when u = 15, a = −9.8 and t = 2.
▸ Question 6 · 2 marks · Calculator · Higher. The volume of a sphere is V = 4/3πr³. Make r the subject of the formula.
Question 1 · 2 marks · Non-calculator
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“v = u + at. Work out the value of v when u = 12, a = −3 and t = 5.” |
HOW TO ANSWER IT Command word: Non-calculator. Worth 2 marks, so plan before writing.
Question 1 · mark scheme
2 marks available. Award a mark for each point made.
▸ 12 + (−3) × 5, or −15 seen. M1
▸ −3. A1
▸ Model answer. v = 12 + (−3) × 5 = 12 − 15 = −3
Question 2 · 2 marks · Non-calculator
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“Make t the subject of the formula v = u + at” |
HOW TO ANSWER IT Command word: Non-calculator. Worth 2 marks, so plan before writing.
Question 2 · mark scheme
2 marks available. Award a mark for each point made.
▸ v − u = at, or a correct first step. M1
▸ t = (v − u)/a. A1
▸ Model answer. v − u = at, so t = (v − u)/a
Question 3 · 2 marks · Non-calculator · Show that
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“Show that (x + 3)² − 9 ≡ x(x + 6)” |
HOW TO ANSWER IT Command word: Non-calculator · Show that. Worth 2 marks, so plan before writing.
Question 3 · mark scheme
2 marks available. Award a mark for each point made.
▸ (x + 3)² expanded to x² + 6x + 9. M1
▸ Correctly simplified to x(x + 6), or both sides shown equal to x² + 6x. A1
▸ Model answer. (x + 3)² − 9 = x² + 6x + 9 − 9 = x² + 6x = x(x + 6)
Question 4 · 4 marks · Non-calculator · Higher
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“Make a the subject of 3(a − 2) = ab + 5” |
HOW TO ANSWER IT Command word: Non-calculator · Higher. Worth 4 marks, so plan before writing.
Question 4 · mark scheme
4 marks available. Award a mark for each point made.
▸ Expands the bracket: 3a − 6. M1
▸ Collects the a terms on one side: 3a − ab = 11. M1
▸ Factorises: a(3 − b) = 11. M1
▸ a = 11/(3 − b). A1
▸ Model answer. 3a − 6 = ab + 5, so 3a − ab = 11, a(3 − b) = 11 and a = 11/(3 − b).
Question 5 · 2 marks · Calculator
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“s = ut + ½at². Work out the value of s when u = 15, a = −9.8 and t = 2.” |
HOW TO ANSWER IT Command word: Calculator. Worth 2 marks, so plan before writing.
Question 5 · mark scheme
2 marks available. Award a mark for each point made.
▸ 15 × 2 + ½ × (−9.8) × 2², or 30 and −19.6 seen. M1
▸ 10.4. A1
▸ Model answer. s = 15 × 2 + ½ × (−9.8) × 2² = 30 − 19.6 = 10.4
Question 6 · 2 marks · Calculator · Higher
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“The volume of a sphere is V = 4/3πr³. Make r the subject of the formula.” |
HOW TO ANSWER IT Command word: Calculator · Higher. Worth 2 marks, so plan before writing.
Question 6 · mark scheme
2 marks available. Award a mark for each point made.
▸ r³ = 3V/4π. M1
▸ r = ∛(3V/4π). A1
▸ Model answer. 3V = 4πr³, so r³ = 3V/4π and r = ∛(3V/4π).