# IGCSE Higher: Expressions and Formulae

## Scheme of work: IGCSE Higher: Year 10: Term 3: Expressions and Formulae

#### Prerequisite Knowledge

1. Basic Algebraic Manipulation
• Concept: Understanding how to perform basic algebraic operations such as addition, subtraction, multiplication, and division on equations.
• Example: $2x + 5 = 15 \rightarrow 2x = 10 \rightarrow x = 5$
2. Solving Linear Equations
• Concept: Knowing how to solve linear equations for a single variable.
• Example: $3x – 4 = 11 \rightarrow 3x = 15 \rightarrow x = 5$

#### Success Criteria

1. Change the Subject of a Formula Where the Subject Appears Once
• Objective: Students should be able to manipulate a formula to make a different variable the subject when the subject appears only once.
• Example: Make $$r$$ the subject of $$V = \frac{4}{3} \pi r^3$$: $V = \frac{4}{3} \pi r^3 \rightarrow \frac{3V}{4\pi} = r^3 \rightarrow r = \sqrt[3]{\frac{3V}{4\pi}}$
2. Change the Subject of a Formula Where the Subject Appears Twice
• Objective: Students should be able to manipulate a formula to make a different variable the subject when the subject appears more than once.
• Example: Make $$a$$ the subject of $$3a + 5 = \frac{4 – a}{r}$$: $3a + 5 = \frac{4 – a}{r} \rightarrow 3a + 5 = \frac{4}{r} – \frac{a}{r} \rightarrow 3ar + 5r = 4 – a \rightarrow 3ar + a = 4 – 5r \rightarrow a(3r + 1) = 4 – 5r \rightarrow a = \frac{4 – 5r}{3r + 1}$
3. Change the Subject of a Formula Involving Powers
• Objective: Students should be able to manipulate a formula to make a different variable the subject when the subject involves powers or roots.
• Example: Make $$l$$ the subject of $$T = 2\pi \sqrt{\frac{l}{g}}$$: $T = 2\pi \sqrt{\frac{l}{g}} \rightarrow \frac{T}{2\pi} = \sqrt{\frac{l}{g}} \rightarrow \left(\frac{T}{2\pi}\right)^2 = \frac{l}{g} \rightarrow l = g \left(\frac{T}{2\pi}\right)^2$

#### Key Concepts

1. Working with Fractions in Equations
• Concept: Knowing how to handle equations involving fractions by multiplying through by the denominator to clear fractions.
• Example: Make $$a$$ the subject of $$3a + 5 = \frac{4 – a}{r}$$: $3a + 5 = \frac{4 – a}{r} \rightarrow 3ar + 5r = 4 – a \rightarrow 3ar + a = 4 – 5r \rightarrow a(3r + 1) = 4 – 5r \rightarrow a = \frac{4 – 5r}{3r + 1}$
2. Knowing When to Factorise Out the Variable
• Concept: Recognising when it is necessary to factorise out the variable to isolate it.
• Example: Make $$a$$ the subject of $$3a + 5 = \frac{4 – a}{r}$$: $3a + 5 = \frac{4 – a}{r} \rightarrow 3ar + 5r = 4 – a \rightarrow 3ar + a = 4 – 5r \rightarrow a(3r + 1) = 4 – 5r \rightarrow a = \frac{4 – 5r}{3r + 1}$
3. Handling Powers and Roots
• Concept: Understanding how to manipulate equations involving powers and roots by using appropriate inverse operations.
• Example: Make $$l$$ the subject of $$T = 2\pi \sqrt{\frac{l}{g}}$$: $T = 2\pi \sqrt{\frac{l}{g}} \rightarrow \frac{T}{2\pi} = \sqrt{\frac{l}{g}} \rightarrow \left(\frac{T}{2\pi}\right)^2 = \frac{l}{g} \rightarrow l = g \left(\frac{T}{2\pi}\right)^2$
4. Changing the Subject When the Term is Raised to a Fractional Power
• Concept: Knowing how to handle equations where the subject is raised to a fractional power.
• Example: Make $$r$$ the subject of $$V = \frac{4}{3} \pi r^{3/2}$$: $V = \frac{4}{3} \pi r^{3/2} \rightarrow \frac{3V}{4\pi} = r^{3/2} \rightarrow r = \left(\frac{3V}{4\pi}\right)^{2/3}$

#### Common Misconceptions

1. Incorrect Application of Inverse Operations
• Common Mistake: Students might apply inverse operations incorrectly, leading to incorrect manipulation of the formula.
• Example: Making $$r$$ the subject of $$V = \frac{4}{3} \pi r^3$$ incorrectly:
$\text{Incorrect: } V = \frac{4}{3} \pi r^3 \rightarrow r^3 = \frac{4V}{3\pi} \rightarrow r = \sqrt[3]{\frac{4V}{3\pi}}$ $\text{Correct: } V = \frac{4}{3} \pi r^3 \rightarrow \frac{3V}{4\pi} = r^3 \rightarrow r = \sqrt[3]{\frac{3V}{4\pi}}$
2. Errors in Manipulating Fractions
• Common Mistake: Students might fail to correctly clear fractions by not multiplying through by the denominator.
• Example: Making $$a$$ the subject of $$3a + 5 = \frac{4 – a}{r}$$ incorrectly:
$\text{Incorrect: } 3a + 5 = \frac{4 – a}{r} \rightarrow 3a + 5r = 4 – a \rightarrow 4 – 5r = 3a – a \rightarrow a = \frac{4 – 5r}{3}$ $\text{Correct: } 3a + 5 = \frac{4 – a}{r} \rightarrow 3ar + 5r = 4 – a \rightarrow 3ar + a = 4 – 5r \rightarrow a(3r + 1) = 4 – 5r \rightarrow a = \frac{4 – 5r}{3r + 1}$
3. Missteps in Handling Powers and Roots
• Common Mistake: Students might incorrectly handle powers and roots, leading to incorrect solutions.
• Example: Making $$l$$ the subject of $$T = 2\pi \sqrt{\frac{l}{g}}$$ incorrectly:
$\text{Incorrect: } T = 2\pi \sqrt{\frac{l}{g}} \rightarrow \frac{T}{2\pi} = \sqrt{\frac{l}{g}} \rightarrow \frac{T}{2\pi}^2 = \frac{l}{g} \rightarrow l = g \frac{T}{2\pi}^2$ $\text{Correct: } T = 2\pi \sqrt{\frac{l}{g}} \rightarrow \frac{T}{2\pi} = \sqrt{\frac{l}{g}} \rightarrow \left(\frac{T}{2\pi}\right)^2 = \frac{l}{g} \rightarrow l = g \left(\frac{T}{2\pi}\right)^2$

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