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Maths · Algebra
Formulae
A formula is a rule linking several quantities. Substituting puts numbers in; changing the subject rearranges the rule so it works out a different quantity - using exactly the same balancing as solving an equation.
Last Lesson and Before
Answer each one, then check.
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1
If \(x = 3\), what is \(2x + 5\)?
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\(11\)
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2
If \(y = -2\), what is \(y^2\)?
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\(4\)
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3
What is \((-2)^3\)?
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\(-8\)
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4
Last lesson: solve \(3x - 4 = 11\).
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\(x = 5\)
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5
Last lesson: solve \(2(x + 3) = 14\).
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\(x = 4\)
Learning Objectives
- 1Tell the difference between an expression, an equation, a formula and an identity.
- 2Substitute positive and negative numbers into formulae.
- 3Write a formula from a description.
- 4Change the subject of a formula.
- 5Change 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.
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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: \(v = u + at\), \(A = \pi r^2\).
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Identity
Two expressions equal for EVERY value of the letter, written with \(\equiv\): \(2(x + 3) \equiv 2x + 6\).
Substituting into a Formula
Replace each letter with its value, then follow the order of operations.
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Use brackets
Put negative numbers in brackets: if \(a = -2\), then \(a^2 = (-2)^2 = 4\).
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Powers first
In \(\frac{1}{2}at^2\), square \(t\) before multiplying.
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Show the substitution
Write the formula with the numbers in before working it out: it earns the method mark.
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Units
Give the answer in the right units if the question has them.
Substituting Negative Numbers
\(s = ut + \frac{1}{2}at^2\). Work out \(s\) when \(u = 3\), \(t = 4\) and \(a = -2\).
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- 1 Substitute \(s = 3 \times 4 + \frac{1}{2} \times (-2) \times 4^2\)
- 2 Powers first \(4^2 = 16\)
- 3 Multiply \(3 \times 4 = 12\) and \(\frac{1}{2} \times (-2) \times 16 = -16\)
- 4 Add \(12 + (-16) = -4\)
Answer\(s = -4\)
The Area of a Trapezium
The area of a trapezium is \(A = \frac{1}{2}(a + b)h\), 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: \(A = \frac{1}{2} \times (5 + 9) \times 4 = \frac{1}{2} \times 14 \times 4 = 28\) cm².
\(A = \frac{1}{2}(a + b)h\): add the parallel sides, halve, multiply by the height.
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 \(v\).
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The idea
Rearrange the formula so a different letter is on its own.
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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.
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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
Make \(t\) the subject of \(v = u + at\).
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- 1 Subtract \(u\) from both sides \(v - u = at\)
- 2 Divide both sides by \(a\) \(\dfrac{v - u}{a} = t\)
- 3 Write the subject on the left \(t = \dfrac{v - u}{a}\)
Answer\(t = \dfrac{v - u}{a}\)
Rearranging the Trapezium Formula
Make \(h\) the subject of \(A = \frac{1}{2}(a + b)h\).
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- 1 Multiply both sides by 2 \(2A = (a + b)h\)
- 2 Divide both sides by \((a + b)\) \(\dfrac{2A}{a + b} = h\)
Answer\(h = \dfrac{2A}{a + b}\)
The Subject Is Squared
Make \(r\) the subject of \(A = \pi r^2\).
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- 1 Divide both sides by \(\pi\) \(\dfrac{A}{\pi} = r^2\)
- 2 Square root both sides \(\sqrt{\dfrac{A}{\pi}} = r\)
Answer\(r = \sqrt{\dfrac{A}{\pi}}\)
The Subject Appears Twice
Make \(x\) the subject of \(y = \dfrac{x + 2}{x - 3}\).
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- 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 \((y - 1)\) \(x = \dfrac{3y + 2}{y - 1}\)
Answer\(x = \dfrac{3y + 2}{y - 1}\)
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 \(C = \dfrac{F - 32}{1.8}\), so a weather forecast of 77 °F becomes \(\dfrac{77 - 32}{1.8} = 25\) °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.
Rearrange It
Make the letter in brackets the subject. (a) \(P = 4s\) (\(s\)) (b) \(y = mx + c\) (\(x\)) (c) \(C = 2\pi r\) (\(r\)) (d) \(V = IR\) (\(I\)) (e) \(A = \frac{1}{2}bh\) (\(b\)). Higher: (f) \(E = \frac{1}{2}mv^2\) (\(v\)) (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) \(s = \dfrac{P}{4}\) (b) \(x = \dfrac{y - c}{m}\) (c) \(r = \dfrac{C}{2\pi}\) (d) \(I = \dfrac{V}{R}\) (e) \(b = \dfrac{2A}{h}\) (f) \(v = \sqrt{\dfrac{2E}{m}}\) (g) \(ax - 2a = x + 5\), so \(ax - x = 5 + 2a\), \(x(a - 1) = 5 + 2a\) and \(x = \dfrac{5 + 2a}{a - 1}\)
Can I...?
- 1Tell a formula from an equation, expression and identity.
- 2Substitute positive numbers into a formula.
- 3Substitute negative numbers into a formula.
- 4Write a formula from words.
- 5Change the subject in one step.
- 6Change the subject in two or more steps.
- 7Change the subject when it is squared. (Higher)
- 8Change the subject when it appears twice. (Higher)
Summary & Exam Focus
- 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.
Exam focus
Make \(t\) the subject of the formula \(v = u + at\) (2 marks) (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.
Key terms
The vocabulary this lesson expects you to use. Each one is linked from the first place it appears above.
- Formula
- A rule linking two or more quantities, written with letters.
- Substitute
- Replace the letters in a formula with numbers.
- 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.
- Identity
- Two expressions that are equal for every value of the letters, written with \(\equiv\).
Practice questions
Have a go at each one before you open its answer.
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Question 1 Non-calculator 2 marks
\(v = u + at\). Work out the value of \(v\) when \(u = 12\), \(a = -3\) and \(t = 5\).
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Model answer
\(v = 12 + (-3) \times 5 = 12 - 15 = -3\)
Mark scheme
- \(12 + (-3) \times 5\), or \(-15\) seen — M1
- \(-3\) — A1
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Question 2 Non-calculator 2 marks
Make \(t\) the subject of the formula \(v = u + at\)
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Model answer
\(v - u = at\), so \(t = \dfrac{v - u}{a}\)
Mark scheme
- \(v - u = at\), or a correct first step — M1
- \(t = \dfrac{v - u}{a}\) — A1
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Question 3 Non-calculator · Show that 2 marks
Show that \((x + 3)^2 - 9 \equiv x(x + 6)\)
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Model answer
\((x + 3)^2 - 9 = x^2 + 6x + 9 - 9 = x^2 + 6x = x(x + 6)\)
Mark scheme
- \((x + 3)^2\) expanded to \(x^2 + 6x + 9\) — M1
- Correctly simplified to \(x(x + 6)\), or both sides shown equal to \(x^2 + 6x\) — A1
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Question 4 Non-calculator · Higher 4 marks
Make \(a\) the subject of \(3(a - 2) = ab + 5\)
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Model answer
\(3a - 6 = ab + 5\), so \(3a - ab = 11\), \(a(3 - b) = 11\) and \(a = \dfrac{11}{3 - b}\).
Mark scheme
- Expands the bracket: \(3a - 6\) — M1
- Collects the \(a\) terms on one side: \(3a - ab = 11\) — M1
- Factorises: \(a(3 - b) = 11\) — M1
- \(a = \dfrac{11}{3 - b}\) — A1
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Question 5 Calculator 2 marks
\(s = ut + \frac{1}{2}at^2\). Work out the value of \(s\) when \(u = 15\), \(a = -9.8\) and \(t = 2\).
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Model answer
\(s = 15 \times 2 + \frac{1}{2} \times (-9.8) \times 2^2 = 30 - 19.6 = 10.4\)
Mark scheme
- \(15 \times 2 + \frac{1}{2} \times (-9.8) \times 2^2\), or 30 and \(-19.6\) seen — M1
- \(10.4\) — A1
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Question 6 Calculator · Higher 2 marks
The volume of a sphere is \(V = \frac{4}{3}\pi r^3\). Make \(r\) the subject of the formula.
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Model answer
\(3V = 4\pi r^3\), so \(r^3 = \dfrac{3V}{4\pi}\) and \(r = \sqrt[3]{\dfrac{3V}{4\pi}}\).
Mark scheme
- \(r^3 = \dfrac{3V}{4\pi}\) — M1
- \(r = \sqrt[3]{\dfrac{3V}{4\pi}}\) — A1
Quick check
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\(P = 2l + 2w\). What is \(P\) when \(l = 6\) and \(w = 4\)?
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B: 20
\(P = 2 \times 6 + 2 \times 4 = 12 + 8 = 20\).
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Make \(x\) the subject of \(y = 3x - 4\).
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C: \(x = \dfrac{y + 4}{3}\)
Add 4: \(y + 4 = 3x\). Divide by 3: \(x = \dfrac{y + 4}{3}\).
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Which of these is an identity?
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A: \(2(x + 3) \equiv 2x + 6\)
\(2(x + 3)\) and \(2x + 6\) are equal for every value of \(x\).
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