# IGCSE Higher: Linear Equations

## Scheme of work: IGCSE Higher: Year 10: Term 1: Linear Equations

#### Prerequisite Knowledge

1. Solving Simple Linear Equations with the Unknown on One Side
• Concept: Understanding how to isolate the unknown on one side of the equation.
• Example: Solve $$2x + 3 = 9$$. $2x + 3 = 9 \Rightarrow 2x = 9 – 3 \Rightarrow 2x = 6 \Rightarrow x = 3$
2. Solving Simple Linear Equations with the Unknown on Both Sides
• Concept: Understanding how to manipulate the equation to isolate the unknown by moving terms from one side to the other.
• Example: Solve $$3x + 4 = 2x – 1$$. $3x + 4 = 2x – 1 \Rightarrow 3x – 2x = -1 – 4 \Rightarrow x = -5$
3. Understanding Basic Algebraic Manipulation
• Concept: Understanding how to perform basic algebraic operations such as addition, subtraction, multiplication, and division.
• Example: Simplify $$5x – 3x + 2$$. $5x – 3x + 2 = 2x + 2$
4. Knowledge of Fractional Coefficients
• Concept: Understanding how to handle equations with fractional coefficients.
• Example: Solve $$\frac{1}{2}x + \frac{1}{3} = 1$$. $\frac{1}{2}x + \frac{1}{3} = 1 \Rightarrow \frac{1}{2}x = 1 – \frac{1}{3} \Rightarrow \frac{1}{2}x = \frac{2}{3} \Rightarrow x = \frac{2}{3} \times 2 = \frac{4}{3}$

#### Success Criteria

1. Solve Linear Equations with Fractional Coefficients
• Objective: Students should be able to solve linear equations where the unknown appears on both sides of the equation and includes fractional coefficients.
• Example: Solve $$\frac{4x + 5}{2} = 3$$. $\frac{4x + 5}{2} = 3 \Rightarrow 4x + 5 = 6 \Rightarrow 4x = 1 \Rightarrow x = \frac{1}{4}$
2. Set Up Simple Linear Equations from Given Data
• Objective: Students should be able to set up and solve linear equations from word problems or given data.
• Example: The three angles of a triangle are $$a^\circ$$, $$(a + 10)^\circ$$, and $$(a + 20)^\circ$$. Find the value of $$a$$. $a + (a + 10) + (a + 20) = 180 \Rightarrow 3a + 30 = 180 \Rightarrow 3a = 150 \Rightarrow a = 50$
3. Calculate the Exact Solution of Two Simultaneous Equations in Two Unknowns
• Objective: Students should be able to solve simultaneous linear equations algebraically.
• Example: Solve the simultaneous equations $$x + y = 14$$ and $$x – y = 2$$. $x + y = 14 \\ x – y = 2 \\ \text{Adding these equations: } 2x = 16 \Rightarrow x = 8 \\ \text{Substituting } x \text{ into the first equation: } 8 + y = 14 \Rightarrow y = 6$
4. Solve Linear Equations with Variables on Both Sides
• Objective: Students should be able to solve linear equations where the unknown appears on both sides of the equation.
• Example: Solve $$7(x + 3) = 5x – 8$$. $7(x + 3) = 5x – 8 \Rightarrow 7x + 21 = 5x – 8 \Rightarrow 7x – 5x = -8 – 21 \Rightarrow 2x = -29 \Rightarrow x = -\frac{29}{2}$

#### Key Concepts

1. Balance Method in Solving Equations
• Concept: Understanding that whatever operation is done to one side of the equation must be done to the other side to maintain balance.
• Example: To solve $$2x + 3 = 9$$: $2x + 3 = 9 \Rightarrow 2x = 9 – 3 \Rightarrow 2x = 6 \Rightarrow x = 3$
2. Combining Like Terms
• Concept: Combining terms that have the same variable part to simplify equations.
• Example: Simplify $$3x + 4 – 2x + 1$$: $3x + 4 – 2x + 1 = (3x – 2x) + (4 + 1) = x + 5$
3. Elimination Method for Simultaneous Equations
• Concept: Using the elimination method to solve simultaneous equations by adding or subtracting equations to eliminate one of the variables.
• Example: Solve the simultaneous equations $$x + y = 14$$ and $$x – y = 2$$: $x + y = 14 \\ x – y = 2 \\ \text{Adding these equations: } 2x = 16 \Rightarrow x = 8 \\ \text{Substituting } x \text{ into the first equation: } 8 + y = 14 \Rightarrow y = 6$
4. Substitution Method for Simultaneous Equations
• Concept: Using the substitution method to solve simultaneous equations by expressing one variable in terms of the other and substituting it into the second equation.
• Example: Solve the simultaneous equations $$2a + 5b = 12$$ and $$3a + b = 5$$: $3a + b = 5 \Rightarrow b = 5 – 3a \\ \text{Substituting } b \text{ into the first equation: } 2a + 5(5 – 3a) = 12 \Rightarrow 2a + 25 – 15a = 12 \Rightarrow -13a + 25 = 12 \Rightarrow -13a = -13 \Rightarrow a = 1 \\ \text{Substituting } a = 1 \text{ into } b = 5 – 3a \Rightarrow b = 5 – 3(1) = 2$

#### Common Misconceptions

1. Incorrectly Combining Like Terms
• Misconception: Students often make mistakes when combining like terms, especially when dealing with negative coefficients or multiple terms on both sides of the equation.
• Example: Incorrectly simplifying $$3x + 4 – 2x + 1$$ as $$5x + 5$$ instead of the correct simplification: $3x + 4 – 2x + 1 = (3x – 2x) + (4 + 1) = x + 5$
2. Misunderstanding the Balance Method
• Misconception: Students may not consistently apply the same operation to both sides of the equation, leading to incorrect solutions.
• Example: Incorrectly solving $$2x + 3 = 9$$ by subtracting 3 only from one side: $2x + 3 = 9 \Rightarrow 2x + 3 – 3 = 9 \Rightarrow 2x = 9 – 3 \Rightarrow 2x = 6 \Rightarrow x = 3$
3. Errors in the Elimination Method for Simultaneous Equations
• Misconception: Students might incorrectly add or subtract equations without aligning the terms properly, leading to incorrect solutions.
• Example: Incorrectly solving the system $$x + y = 14$$ and $$x – y = 2$$ by subtracting incorrectly: $x + y = 14 \\ x – y = 2 \\ \text{Incorrectly subtracting:} \\ (x + y) – (x – y) = 14 – 2 \Rightarrow 0 + 2y = 12 \Rightarrow y = 6 \quad (\text{incorrect})$
4. Mistakes in the Substitution Method for Simultaneous Equations
• Misconception: Students may make errors when substituting one equation into the other, especially if they do not correctly isolate the variable first.
• Example: Incorrectly solving the system $$2a + 5b = 12$$ and $$3a + b = 5$$ by misinterpreting the substitution step: $3a + b = 5 \Rightarrow b = 5 – 3a \\ \text{Substituting } b \text{ into the first equation:} \\ 2a + 5(5 – 3a) = 12 \Rightarrow 2a + 25 – 15a = 12 \Rightarrow -13a + 25 = 12 \Rightarrow -13a = -13 \Rightarrow a = 1 \\ \text{Substituting } a = 1 \text{ into } b = 5 – 3a \Rightarrow b = 5 – 3(1) = 2$

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