Welcome to the World of Algebra!

Hello IGCSE student! Don't worry if algebra seems like a puzzle right now. This chapter, "Introduction to Algebra," is your foundation. Think of algebra as a powerful tool that lets you solve problems where numbers are unknown—it turns real-world scenarios into neat mathematical equations.

We are moving from just using numbers (arithmetic) to using letters and symbols (algebra) to represent general situations. Once you master the basics of handling these letters, the rest of mathematics becomes much clearer!

Section 1: Variables, Expressions, and Substitution (C2.1)

Before we start crunching numbers, we need to know the language of algebra.

Key Definitions

  • Variable: A symbol, usually a letter (like \(x\), \(y\), or \(n\)), used to represent an unknown number or a value that can change. (Think of it as a temporary placeholder for a number.)
  • Term: A single number, a single variable, or numbers and variables multiplied together. (Examples: \(5\), \(3x\), \(-2y^2\).)
  • Expression: A collection of terms joined by addition or subtraction signs. It does *not* have an equals sign. (Example: \(3x + 5y - 7\).)
  • Formula: A relationship or rule expressed using symbols, often showing how one quantity depends on others. It *always* has an equals sign. (Example: \(A = \pi r^2\).)
  • Coefficient: The number multiplied by the variable in a term. (In the term \(7x\), the coefficient is 7.)

Substituting Numbers into Expressions and Formulas

Substitution is the process of replacing variables in an expression or formula with their given numerical values.
This is a core skill! When you substitute, you are solving the expression for a specific instance.

Step-by-Step: How to Substitute

Let's use the expression: \(E = 2a + 3b - 5\). If \(a = 4\) and \(b = -2\).

  1. Rewrite with Brackets: Always place the substituted value inside brackets. This helps prevent sign errors, especially when multiplying or dealing with indices.

    \(E = 2(\text{4}) + 3(\text{-2}) - 5\)

  2. Calculate Multiplications:

    \(E = 8 + (-6) - 5\)

  3. Simplify: Use the rules of addition/subtraction.

    \(E = 8 - 6 - 5\)
    \(E = 2 - 5\)
    \(E = -3\)

Quick Tip for Formulas: If you use a formula like the area of a triangle \(A = \frac{1}{2}bh\), and you are given \(b=10\) and \(h=4\), substitute carefully: \(A = \frac{1}{2} \times 10 \times 4\).

⚠ Common Mistake Alert!
When substituting a negative number, like \(x = -3\), into a term like \(4x\), students sometimes write \(4 - 3\).
Correct Way: You must treat it as multiplication: \(4(-3) = -12\). Always use those brackets!

Section 2: Algebraic Manipulation I – Simplifying (C2.2.1)

Simplifying an algebraic expression means making it as short as possible. We do this by collecting like terms.

What are Like Terms?

Like terms are terms that have exactly the same variables and those variables have the same powers.
Analogy: You can add 3 apples (\(3a\)) and 5 apples (\(5a\)) to get 8 apples (\(8a\)). But you cannot add 3 apples and 5 bananas (\(5b\)).

  • \(4x\) and \(-2x\) are LIKE TERMS. (Same variable \(x\), same power 1.)
  • \(5a^2\) and \(a^2\) are LIKE TERMS. (Same variable \(a\), same power 2.)
  • \(7xy\) and \(-xy\) are LIKE TERMS. (Same variables \(x\) and \(y\).)
  • \(4x\) and \(4x^2\) are NOT like terms. (Different powers.)
  • \(5a\) and \(5b\) are NOT like terms. (Different variables.)

Collecting Like Terms: Step-by-Step

Simplify: \(2a + 3b + 5a – 9b\) (Example from syllabus)

  1. Identify Like Terms: Group the terms that are alike. It helps to circle or underline them, including the sign in front of them.

    Group 1 (a terms): \(+2a\) and \(+5a\)
    Group 2 (b terms): \(+3b\) and \(-9b\)

  2. Combine Coefficients: Add or subtract the coefficients in each group.

    For Group 1: \(2 + 5 = 7\). Result: \(7a\)
    For Group 2: \(3 - 9 = -6\). Result: \(-6b\)

  3. Write the Final Expression:

    The simplified expression is: \(7a - 6b\)

Example Practice

Simplify: \(5x + 3y^2 - x + 4y^2\)

x terms: \(5x - x = 4x\)
y² terms: \(3y^2 + 4y^2 = 7y^2\)
Answer: \(4x + 7y^2\)

Key Takeaway: Simplifying relies entirely on correctly identifying and combining like terms. If the variable parts are different, you cannot combine them!

Section 3: Algebraic Manipulation II – Expanding Brackets (C2.2.2)

Expanding means getting rid of the brackets by multiplying every term inside the bracket by the term outside the bracket. This is based on the Distributive Law.

A. Expanding Single Brackets

Multiply the coefficient/term outside the bracket by EACH term inside the bracket.

Example: Expand \(3x(2x - 4y)\) (Syllabus example)

  1. Multiply \(3x\) by \(2x\): \(3x \times 2x = 6x^2\)
  2. Multiply \(3x\) by \(-4y\): \(3x \times (-4y) = -12xy\)
  3. Combine: \(6x^2 - 12xy\)

Another Example: Expand \(-5(4 - 2a)\)

\(-5 \times 4 = -20\)
\(-5 \times (-2a) = +10a\)
Answer: \(-20 + 10a\) or \(10a - 20\)

Memory Aid: Sign Rule Check!
Make sure you multiply the signs as well as the numbers and variables:
\((+) \times (+) = (+)\)
\((-) \times (-) = (+)\)
\((+) \times (-) = (-)\)

B. Expanding Double Brackets (Core Level)

At the Core level, you need to be able to expand products of two brackets involving one variable, like \((2x + 1)(x - 4)\).

We must make sure that every term in the first bracket multiplies every term in the second bracket.

Mnemonic: F.O.I.L.

FOIL is a helpful way to remember the four multiplications needed:

  • First terms: Multiply the first term in each bracket.
  • Outer terms: Multiply the two outermost terms.
  • Inner terms: Multiply the two innermost terms.
  • Last terms: Multiply the last term in each bracket.

Example: Expand \((2x + 1)(x - 4)\)

  1. First: \(2x \times x = 2x^2\)
  2. Outer: \(2x \times (-4) = -8x\)
  3. Inner: \(1 \times x = +x\)
  4. Last: \(1 \times (-4) = -4\)

Intermediate Expression: \(2x^2 - 8x + x - 4\)

Final Step – Simplify (Collect Like Terms): Combine the \(x\) terms (Outer and Inner). \(-8x + x = -7x\)
Final Answer: \(2x^2 - 7x - 4\)

Key Takeaway: Expanding is multiplication. Always simplify by collecting like terms after expanding.

Section 4: Algebraic Manipulation III – Factorising (C2.2.3)

Factorising is the reverse process of expanding. Instead of removing brackets, we introduce them by finding the Highest Common Factor (HCF) of all the terms.

The syllabus requires factorising by extracting common factors (the most basic and essential type of factorisation).

Step-by-Step: Extracting Common Factors

Example: Factorise fully \(9x^2 + 15xy\) (Syllabus example)

  1. Find the HCF of the numbers (Coefficients):

    The numbers are 9 and 15. The HCF of 9 and 15 is 3.

  2. Find the Common Variables:

    The terms are \(9x^2\) and \(15xy\). Both terms have \(x\). The lowest power of \(x\) is \(x^1\).
    Only the first term has \(y\), so \(y\) is not a common factor.

  3. Identify the Overall Common Factor (HCF):

    Combine the results: The overall HCF is \(3x\).

  4. Divide and Write: Write the HCF outside the bracket, and divide each original term by the HCF to find the terms inside the bracket.

    \(9x^2 \div 3x = 3x\)
    \(15xy \div 3x = 5y\)

Final Answer: \(3x(3x + 5y)\)

⚠ Common Mistake Alert!
After factorising, you should always be able to check your answer by expanding the bracket again. If the expansion gives you the original expression, you are correct!

Section 5: Simplifying Basic Algebraic Fractions (C2.3)

An algebraic fraction is simply a fraction that contains variables (algebra) in the numerator, the denominator, or both.

At the Core level, simplifying algebraic fractions involves cancelling out common factors between the numerator (top) and the denominator (bottom). The syllabus states only one step is required, meaning very straightforward cancellation.

Rule: Divide the Top and Bottom by the Same Factor

Example 1: Variables

Simplify \(\frac{x^2}{x}\)

Since \(x^2 = x \times x\), we can cancel one \(x\) from the top and one from the bottom.
\(\frac{x^2}{x} = \frac{x \times x}{x}\)
Answer: \(x\)

Example 2: Numbers and Variables

Simplify \(\frac{3}{6x}\)

  1. Find the HCF of the numbers (3 and 6): 3.
  2. Divide both the numerator and the denominator by 3.

\(\frac{3 \div 3}{6x \div 3} = \frac{1}{2x}\)
Answer: \(\frac{1}{2x}\)

Example 3: Combination

Simplify \(\frac{10a^3}{5a}\)

  1. Simplify the numbers: \(10 \div 5 = 2\).
  2. Simplify the variables: \(\frac{a^3}{a} = a^{3-1} = a^2\).

Answer: \(2a^2\)

Key Takeaway: Simplifying fractions is all about cancellation. You can only cancel factors that are multiplied together, never terms that are added or subtracted!

Quick Review Summary: Introduction to Algebra

You have now mastered the fundamental tools needed for all algebraic work!

Concept What to do Check/Rule
Substitution Replace letters with numbers. Use brackets! Especially for negative numbers.
Simplifying Collect terms with identical variables and powers. Carry the sign (\(+\) or \(-\)) in front of the term.
Expanding Multiply the outside term by every term inside the bracket (Distributive Law). For double brackets, use FOIL (or a similar methodical method).
Factorising Find the Highest Common Factor (HCF) of all terms. Check your answer by expanding the factorised expression.

Keep practicing these four skills—they are the most important building blocks for success in the rest of the Algebra section! You’ve got this!