CIE IGCSE Mathematics (0580) Equations Study Notes - New Syllabus
CIE IGCSE Mathematics (0580) Equations Study Notes
LEARNING OBJECTIVE
- Constructing Simple Expressions, Equations, and Formulas
Key Concepts:
- Constructing Simple Expressions, Equations
- Solving Linear Equations
- Changing the Subject of a Formula
Constructing Simple Expressions, Equations, and Formulas
Constructing Simple Expressions, Equations, and Formulas
An expression is a combination of numbers, variables (letters), and operations (like +, −, ×, ÷) without an equals sign. It represents a value.
An equation is a mathematical statement that shows two expressions are equal using an equals sign (=).
A formula is a special type of equation that shows the relationship between different variables.
Steps to Construct:
- Identify the quantities involved (use letters for unknown values).
- Decide the mathematical operation(s) required.
- Write the correct form — expression, equation, or formula.
Example :
A bag contains \( x \) apples. Each apple costs $\$3$. Write an expression for the total cost.
▶️ Answer/Explanation
Total cost = \( 3x \)
This is a simple expression (no equals sign).
Example :
A number is multiplied by 4 and the result is 28. Write an equation.
▶️ Answer/Explanation
Let the number be \( x \). Then:
\( 4x = 28 \)
This is an equation because it has an equals sign.
Formulas in Real-World Contexts
A formula is a rule written as an equation that shows how one quantity depends on others. Formulas are commonly used in geometry, physics, finance, and everyday life.
They involve variables (letters) that represent quantities. You substitute values into the formula to calculate results.
Examples of Common Formulas:
- Perimeter of a rectangle: \( P = 2(l + w) \)
- Area of a triangle: \( A = \frac{1}{2} b h \)
- Speed: \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \)
- Simple Interest: \( \text{SI} = \frac{P \times R \times T}{100} \)
Example :
Construct a formula for the area of a rectangle where \( A \) is the area, \( l \) is the length, and \( w \) is the width.
▶️ Answer/Explanation
The area of a rectangle is calculated by multiplying its length and width:
\( A = l \times w \)
This is a formula that can be used to calculate area if length and width are known.
Solving Linear Equations
Solving Linear Equations in One Unknown
A linear equation in one unknown is an equation that involves a variable (usually \( x \)) raised only to the power of 1. The goal is to isolate the variable on one side of the equation.
Steps to Solve:
- Simplify both sides of the equation if needed (expand brackets, combine like terms).
- Move all variable terms to one side and constant terms to the other using inverse operations.
- Solve for the unknown by dividing or multiplying both sides appropriately.
Example :
Solve \( x + 7 = 13 \)
▶️ Answer/Explanation
Subtract 7 from both sides:
\( x = 13 – 7 = 6 \)
Answer: \( x = 6 \)
Example :
Solve \( 3x – 4 = 11 \)
▶️ Answer/Explanation
Add 4 to both sides: \( 3x = 15 \)
Divide both sides by 3: \( x = \frac{15}{3} = 5 \)
Answer: \( x = 5 \)
Example :
Solve \( 2(x – 3) = 5x + 1 \)
▶️ Answer/Explanation
Expand left side: \( 2x – 6 = 5x + 1 \)
Bring all terms to one side: \( 2x – 6 – 5x – 1 = 0 \Rightarrow -3x – 7 = 0 \)
Add 7: \( -3x = 7 \)
Divide: \( x = \frac{-7}{3} \)
Answer: \( x = -\frac{7}{3} \)
Solving Fractional Equations
A fractional (or rational) equation contains fractions with variables in the numerators and/or denominators. To solve these:
- Step 1: Identify the denominators in the equation.
- Step 2: Find the Least Common Denominator (LCD).
- Step 3: Multiply the entire equation by the LCD to eliminate denominators.
- Step 4: Solve the resulting linear equation.
- Step 5: Check for restrictions — values that make any denominator zero must be excluded.
Example :
Solve: \( \dfrac{x}{3} + \dfrac{1}{2} = 5 \)
▶️ Answer/Explanation
LCD of 3 and 2 is 6. Multiply entire equation by 6:
\( 6 \left( \dfrac{x}{3} \right) + 6 \left( \dfrac{1}{2} \right) = 6 \times 5 \)
\( 2x + 3 = 30 \)
\( 2x = 27 \Rightarrow x = 13.5 \)
Answer: \( x = 13.5 \)
Example:
Solve: \( \dfrac{1}{x – 1} = \dfrac{3}{x + 2} \)
▶️ Answer/Explanation
Cross multiply:
\( (x – 1)(3) = (x + 2)(1) \)
\( 3x – 3 = x + 2 \)
\( 2x = 5 \Rightarrow x = \dfrac{5}{2} \)
Restrictions: \( x \ne 1, -2 \)
Answer: \( x = \dfrac{5}{2} \)
Example:
Solve: \( \dfrac{2}{x} + \dfrac{3}{x + 1} = 1 \)
▶️ Answer/Explanation
LCD = \( x(x + 1) \). Multiply through:
\( x(x + 1)\left( \dfrac{2}{x} + \dfrac{3}{x + 1} \right) = x(x + 1) \cdot 1 \)
\( 2(x + 1) + 3x = x(x + 1) \)
\( 2x + 2 + 3x = x^2 + x \)
\( 5x + 2 = x^2 + x \)
\( 0 = x^2 – 4x – 2 \)
Use quadratic formula: \( x = \dfrac{4 \pm \sqrt{16 + 8}}{2} = \dfrac{4 \pm \sqrt{24}}{2} = \dfrac{4 \pm 2\sqrt{6}}{2} = 2 \pm \sqrt{6} \)
Answer: \( x = 2 \pm \sqrt{6} \)
Solving Simultaneous Linear Equations in Two Unknowns
Simultaneous equations involve two linear equations with two unknowns (usually \( x \) and \( y \)). The solution is the pair \((x, y)\) that satisfies both equations.
Methods:
- Substitution: Solve one equation for one variable, substitute into the other.
- Elimination: Add or subtract equations to eliminate one variable.
- Graphical: Plot both equations; their intersection gives the solution.
Example (By Substitution):
Solve:
\( x + y = 10 \)
\( x = 2y \)
▶️ Answer/Explanation
Substitute \( x = 2y \) into the first equation:
\( 2y + y = 10 \Rightarrow 3y = 10 \Rightarrow y = \frac{10}{3} \)
Then \( x = 2y = 2 \times \frac{10}{3} = \frac{20}{3} \)
Answer: \( x = \frac{20}{3},\ y = \frac{10}{3} \)
Example (By Elimination):
Solve:
\( 2x + 3y = 12 \)
\( 4x – 3y = 6 \)
▶️ Answer/Explanation
Add the two equations:
\( (2x + 3y) + (4x – 3y) = 12 + 6 \)
\( 6x = 18 \Rightarrow x = 3 \)
Substitute into first equation: \( 2(3) + 3y = 12 \Rightarrow 6 + 3y = 12 \Rightarrow 3y = 6 \Rightarrow y = 2 \)
Answer: \( x = 3,\ y = 2 \)
Example (By Graphical Method – concept):
Solve:
\( y = 2x + 1 \)
\( y = -x + 7 \)
▶️ Answer/Explanation
Set the two expressions for \( y \) equal:
\( 2x + 1 = -x + 7 \Rightarrow 3x = 6 \Rightarrow x = 2 \)
Substitute \( x = 2 \) into either equation: \( y = 2(2) + 1 = 5 \)
Answer: \( x = 2,\ y = 5 \)
This means the lines intersect at the point (2, 5).
Simultaneous Equations: One Linear and One Non-Linear
Simultaneous equations with one linear and one non-linear equation involve solving for values that satisfy both equations at the same time.
Key Concepts:
- One equation is linear (e.g. \( y = 2x + 1 \))
- One equation is non-linear — usually quadratic (e.g. \( y = x^2 + 3 \))
- Use substitution: replace the linear expression for \( y \) into the non-linear equation
- Solve the resulting quadratic equation and substitute back to find the corresponding \( y \)
Example:
Solve the system:
\( y = 2x + 1 \) (linear)
\( y = x^2 + x \) (non-linear)
▶️ Answer/Explanation
Substitute the linear into the non-linear:
\( 2x + 1 = x^2 + x \)
\( 0 = x^2 + x – 2x – 1 = x^2 – x – 1 \)
Use quadratic formula: \( x = \dfrac{1 \pm \sqrt{1^2 + 4(1)(1)}}{2} = \dfrac{1 \pm \sqrt{5}}{2} \)
Now find \( y \):
\( y = 2x + 1 = 2 \cdot \dfrac{1 \pm \sqrt{5}}{2} + 1 = (1 \pm \sqrt{5}) + 1 = 2 \pm \sqrt{5} \)
Solutions: \( \left( \dfrac{1 + \sqrt{5}}{2}, 2 + \sqrt{5} \right) \), \( \left( \dfrac{1 – \sqrt{5}}{2}, 2 – \sqrt{5} \right) \)
Example:
Solve:
\( y = 4 – x^2 \) (non-linear)
\( y = x + 2 \) (linear)
▶️ Answer/Explanation
Substitute: \( x + 2 = 4 – x^2 \)
\( x^2 + x + 2 – 4 = 0 \Rightarrow x^2 + x – 2 = 0 \)
Factor: \( (x + 2)(x – 1) = 0 \Rightarrow x = -2 \) or \( x = 1 \)
Now find \( y \):
For \( x = -2 \), \( y = -2 + 2 = 0 \); For \( x = 1 \), \( y = 1 + 2 = 3 \)
Solutions: \( (-2, 0) \), \( (1, 3) \)
Solving Quadratic Equations
Solving Quadratic Equations
A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \), where \( a \ne 0 \). There are three common methods to solve such equations:
Methods:
- Factorisation: Express the quadratic as a product of two linear factors.
- Completing the square: Convert the equation into the form \( (x + p)^2 = q \).
- Quadratic formula: Use the formula:
\( x = \dfrac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
Example:
Solve: \( x^2 – 5x + 6 = 0 \)
▶️ Answer/Explanation
Factor the quadratic: \( (x – 2)(x – 3) = 0 \)
Set each factor to 0:
\( x – 2 = 0 \Rightarrow x = 2 \)
\( x – 3 = 0 \Rightarrow x = 3 \)
Solution: \( x = 2, 3 \)
Example:
Solve: \( x^2 + 6x + 5 = 0 \)
▶️ Answer/Explanation
Write in complete square form:
\( x^2 + 6x + 5 = (x + 3)^2 – 4 \)
So equation becomes: \( (x + 3)^2 – 4 = 0 \)
\( (x + 3)^2 = 4 \Rightarrow x + 3 = \pm 2 \)
\( x = -3 \pm 2 \Rightarrow x = -1, -5 \)
Solution: \( x = -1, -5 \)
Example:
Solve: \( 2x^2 + 3x – 2 = 0 \)
▶️ Answer/Explanation
Here, \( a = 2 \), \( b = 3 \), \( c = -2 \)
Use the formula: \( x = \dfrac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
\( x = \dfrac{-3 \pm \sqrt{3^2 – 4(2)(-2)}}{2 \cdot 2} = \dfrac{-3 \pm \sqrt{9 + 16}}{4} = \dfrac{-3 \pm \sqrt{25}}{4} \)
\( x = \dfrac{-3 \pm 5}{4} \Rightarrow x = \dfrac{2}{4} = 0.5 \) or \( x = \dfrac{-8}{4} = -2 \)
Solution: \( x = 0.5, -2 \)
Changing the Subject of a Formula
Changing the Subject of a Formula
To change the subject of a formula means to rearrange it so that a different variable stands alone on one side of the equation. This is done using inverse operations (addition ↔ subtraction, multiplication ↔ division).
Key Steps:
- Identify the new subject you want to make the subject.
- Use inverse operations to isolate that variable.
- Keep the equation balanced: perform the same operation on both sides.
Example :
Make \( x \) the subject: \( y = 2x + 3 \)
▶️ Answer/Explanation
Step 1: Subtract 3 from both sides: \( y – 3 = 2x \)
Step 2: Divide both sides by 2: \( x = \frac{y – 3}{2} \)
Answer: \( x = \frac{y – 3}{2} \)
Example :
Make \( h \) the subject: \( A = \frac{1}{2} b h \)
▶️ Answer/Explanation
Step 1: Multiply both sides by 2: \( 2A = b h \)
Step 2: Divide both sides by \( b \): \( h = \frac{2A}{b} \)
Answer: \( h = \frac{2A}{b} \)
Example :
Make \( x \) the subject: \( y = 3(x – 2) \)
▶️ Answer/Explanation
Step 1: Divide both sides by 3: \( \frac{y}{3} = x – 2 \)
Step 2: Add 2 to both sides: \( x = \frac{y}{3} + 2 \)
Answer: \( x = \frac{y}{3} + 2 \)