# IGCSE Mathematics Higher: Proportion

## Scheme of work: IGCSE Foundation: Year 11: Term 1: Proportion

#### Prerequisite Knowledge

1. Setting Up and Solving Linear Equations
• Concept: Understanding how to form and solve linear equations from word problems or given conditions.
• Example: $\text{Solve the equation } 3x + 5 = 20.$ $3x + 5 = 20 \rightarrow 3x = 15 \rightarrow x = 5.$
2. Changing the Subject of a Formula
• Concept: Rearranging formulas to make a different variable the subject.
• Example: $\text{Given } y = 3x + 2, \text{ make } x \text{ the subject.}$ $y = 3x + 2 \rightarrow y – 2 = 3x \rightarrow x = \frac{y – 2}{3}.$
3. Plotting Non-Linear Graphs
• Concept: Understanding how to plot and interpret non-linear graphs such as quadratic, cubic, and reciprocal graphs.
• Example: $\text{Plot the graph of } y = x^2 \text{ for } x \text{ values from } -3 \text{ to } 3.$ $\begin{array}{c|c} x & y \\ \hline -3 & 9 \\ -2 & 4 \\ -1 & 1 \\ 0 & 0 \\ 1 & 1 \\ 2 & 4 \\ 3 & 9 \\ \end{array}$

#### Success Criteria

1. Set Up Problems Involving Direct Proportion
• Objective: Formulate equations that represent direct proportionality from given problem statements.
• Example: $\text{If } y \text{ is directly proportional to } x \text{, and } y = 10 \text{ when } x = 2, \text{ find the equation relating } y \text{ and } x.$ $y = kx \rightarrow 10 = 2k \rightarrow k = 5 \rightarrow y = 5x.$
2. Set Up Problems Involving Inverse Proportion
• Objective: Formulate equations that represent inverse proportionality from given problem statements.
• Example: $\text{If } y \text{ is inversely proportional to } x \text{, and } y = 4 \text{ when } x = 3, \text{ find the equation relating } y \text{ and } x.$ $y = \frac{k}{x} \rightarrow 4 = \frac{k}{3} \rightarrow k = 12 \rightarrow y = \frac{12}{x}.$
3. Relate Algebraic Solutions to Graphical Representations for Direct Proportion
• Objective: Plot and interpret graphs representing direct proportionality.
• Example: $\text{Plot the graph of } y = 5x \text{ for } x \text{ values from } -3 \text{ to } 3.$ $\begin{array}{c|c} x & y \\ \hline -3 & -15 \\ -2 & -10 \\ -1 & -5 \\ 0 & 0 \\ 1 & 5 \\ 2 & 10 \\ 3 & 15 \\ \end{array}$
4. Relate Algebraic Solutions to Graphical Representations for Inverse Proportion
• Objective: Plot and interpret graphs representing inverse proportionality.
• Example: $\text{Plot the graph of } y = \frac{12}{x} \text{ for } x \text{ values from } -3 \text{ to } 3 \text{ excluding } 0.$ $\begin{array}{c|c} x & y \\ \hline -3 & -4 \\ -2 & -6 \\ -1 & -12 \\ 1 & 12 \\ 2 & 6 \\ 3 & 4 \\ \end{array}$

#### Key Concepts

1. Set Up Problems Involving Direct Proportion
• Concept: Direct proportion means that as one variable increases, the other variable increases at a constant rate. This relationship can be represented by the equation $$y = kx$$, where $$k$$ is the constant of proportionality.
• Example: $\text{If } y = 10 \text{ when } x = 2, \text{ then } k = \frac{y}{x} = \frac{10}{2} = 5 \rightarrow y = 5x.$
2. Set Up Problems Involving Inverse Proportion
• Concept: Inverse proportion means that as one variable increases, the other variable decreases at a constant rate. This relationship can be represented by the equation $$y = \frac{k}{x}$$, where $$k$$ is the constant of proportionality.
• Example: $\text{If } y = 4 \text{ when } x = 3, \text{ then } k = y \times x = 4 \times 3 = 12 \rightarrow y = \frac{12}{x}.$
3. Relate Algebraic Solutions to Graphical Representations for Direct Proportion
• Concept: The graph of a direct proportion relationship is a straight line that passes through the origin. The gradient of the line represents the constant of proportionality.
• Example: $\text{Plot the graph of } y = 5x \text{ for } x \text{ values from } -3 \text{ to } 3 \text{. The line should pass through points } (-3, -15), (-2, -10), (-1, -5), (0, 0), (1, 5), (2, 10), \text{ and } (3, 15).$
4. Relate Algebraic Solutions to Graphical Representations for Inverse Proportion
• Concept: The graph of an inverse proportion relationship is a hyperbola. The product of the coordinates of any point on the graph is the constant of proportionality.
• Example: $\text{Plot the graph of } y = \frac{12}{x} \text{ for } x \text{ values from } -3 \text{ to } 3 \text{ excluding } 0 \text{. The curve should pass through points } (-3, -4), (-2, -6), (-1, -12), (1, 12), (2, 6), \text{ and } (3, 4).$

#### Common Misconceptions

1. Set Up Problems Involving Direct Proportion
• Common Mistake: Confusing direct proportion with other types of relationships, leading to incorrect formulation of the equation.
• Example: $\text{Incorrect: If } y \text{ is directly proportional to } x \text{, and } y = 10 \text{ when } x = 2, \text{ then incorrectly stating } y = 10x \text{ instead of } y = 5x.$
2. Set Up Problems Involving Inverse Proportion
• Common Mistake: Misunderstanding the concept of inverse proportion, resulting in incorrect formulation of the equation.
• Example: $\text{Incorrect: If } y \text{ is inversely proportional to } x \text{, and } y = 4 \text{ when } x = 3, \text{ then incorrectly stating } y = \frac{4}{x} \text{ instead of } y = \frac{12}{x}.$
3. Relate Algebraic Solutions to Graphical Representations for Direct Proportion
• Common Mistake: Plotting the graph incorrectly by not passing through the origin or not using a consistent scale for the axes.
• Example: $\text{Incorrect: Plotting the graph of } y = 5x \text{ without passing through the origin, resulting in a line that does not accurately represent the relationship.}$
4. Relate Algebraic Solutions to Graphical Representations for Inverse Proportion
• Common Mistake: Incorrectly plotting points or not recognizing the hyperbolic nature of the graph.
• Example: $\text{Incorrect: Plotting the graph of } y = \frac{12}{x} \text{ with points that do not reflect the inverse relationship, such as incorrectly plotting } (2, 12) \text{ instead of } (2, 6).$

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