Algebra & Sequences

What you'll learn

Algebra and sequences form a substantial portion of the CIE IGCSE Mathematics examination, appearing across both Paper 2 (Extended) and Paper 4 (Extended). This topic encompasses algebraic manipulation, solving equations and inequalities, understanding functions, and recognizing patterns in sequences. Proficiency in these areas is essential because algebraic techniques underpin coordinate geometry, graphs, and problem-solving questions worth approximately 30-40% of examination marks.

Key terms and definitions

Expression — a mathematical phrase containing variables, numbers and operations but no equals sign (e.g., 3x + 5).

Equation — a statement showing two expressions are equal, which can be solved to find unknown values (e.g., 3x + 5 = 17).

Term — a single number, variable, or product of numbers and variables separated by + or − signs in an expression.

Coefficient — the numerical factor multiplying a variable in a term (e.g., in 7x², the coefficient is 7).

Linear equation — an equation where the highest power of the variable is 1, producing a straight-line graph.

Quadratic equation — an equation in the form ax² + bx + c = 0, where a ≠ 0.

Sequence — an ordered list of numbers following a specific rule or pattern.

Arithmetic sequence — a sequence where consecutive terms differ by a constant amount called the common difference.

Geometric sequence — a sequence where each term is obtained by multiplying the previous term by a constant ratio.

Core concepts

Simplifying and expanding algebraic expressions

Algebraic manipulation requires systematic application of arithmetic rules to expressions containing variables.

Collecting like terms involves combining terms with identical variable parts:

• 5x + 3y − 2x + 7y = 3x + 10y
• 4a² + 3a − 2a² + 5a = 2a² + 8a

Expanding brackets using the distributive law:

• Single brackets: 3(2x − 5) = 6x − 15
• Double brackets (FOIL method): (x + 4)(x − 3) = x² − 3x + 4x − 12 = x² + x − 12
• Three terms: (x + 2)(x² − 3x + 5) = x³ − 3x² + 5x + 2x² − 6x + 10 = x³ − x² − x + 10

Factorising expressions reverses expansion:

• Common factors: 6x² + 9x = 3x(2x + 3)
• Quadratic expressions: x² + 7x + 12 = (x + 3)(x + 4)
• Difference of two squares: x² − 25 = (x + 5)(x − 5)
• Quadratics with coefficient > 1: 2x² + 7x + 3 = (2x + 1)(x + 3)

Algebraic fractions follow the same rules as numerical fractions:

• Simplification: (x² − 9)/(x + 3) = (x + 3)(x − 3)/(x + 3) = x − 3
• Addition: x/3 + x/4 = 4x/12 + 3x/12 = 7x/12
• Multiplication: (2x/5) × (15/4x²) = 30x/(20x²) = 3/(2x)

Solving linear and simultaneous equations

Linear equations require isolating the variable through inverse operations:

1. Remove brackets and collect like terms
2. Move variable terms to one side, constants to the other
3. Divide by the coefficient

Example: 3(2x − 5) = 4x + 1

• 6x − 15 = 4x + 1
• 2x = 16
• x = 8

Simultaneous equations involve finding values satisfying two equations simultaneously. Two methods appear regularly:

Elimination method:

1. Multiply equations to make coefficients of one variable equal
2. Add or subtract equations to eliminate that variable
3. Solve for remaining variable
4. Substitute back to find the other variable

Substitution method:

1. Rearrange one equation to express one variable in terms of the other
2. Substitute this expression into the second equation
3. Solve the resulting single-variable equation
4. Substitute back to find the other variable

Quadratic equations can be solved by:

• Factorisation: x² + 5x + 6 = 0 → (x + 2)(x + 3) = 0 → x = −2 or x = −3
• Quadratic formula: x = [−b ± √(b² − 4ac)] / 2a for ax² + bx + c = 0
• Completing the square: transforming to (x + p)² = q form

Inequalities and inequality notation

Inequalities use symbols <, >, ≤, ≥ instead of equals signs.

Solving inequalities follows the same steps as equations with one critical rule: when multiplying or dividing by a negative number, reverse the inequality sign.

Example: Solve 3 − 2x < 11

• −2x < 8
• x > −4 (sign reverses)

Number line representation uses open circles (○) for < or > and closed circles (●) for ≤ or ≥.

Regions on coordinate grids appear in Extended paper questions requiring shading or identifying regions satisfying multiple inequalities simultaneously.

Substitution and change of subject

Substitution involves replacing variables with given values to evaluate expressions. Negative numbers and fractions require careful bracketing:

• If a = −3 and b = 2, find 2a² − 3ab
• = 2(−3)² − 3(−3)(2) = 2(9) + 18 = 36

Changing the subject (rearranging formulae) isolates a specified variable:

Example: Make r the subject of V = πr²h

• V/πh = r²
• r = √(V/πh)

For more complex rearrangements involving the subject appearing twice, collect all terms containing that variable on one side before factorising.

Sequences and nth term formulae

Arithmetic sequences have a constant difference between consecutive terms.

For a sequence with first term a and common difference d:

• General term (nth term): Tₙ = a + (n − 1)d
• Sum of first n terms: Sₙ = n/2[2a + (n − 1)d] or Sₙ = n/2(first term + last term)

Example: 5, 8, 11, 14, ...

• First term a = 5, common difference d = 3
• nth term: Tₙ = 5 + (n − 1)3 = 5 + 3n − 3 = 3n + 2
• 20th term: T₂₀ = 3(20) + 2 = 62

Geometric sequences multiply by a constant ratio r:

• General term: Tₙ = arⁿ⁻¹
• Sum of first n terms: Sₙ = a(rⁿ − 1)/(r − 1) or Sₙ = a(1 − rⁿ)/(1 − r)

Example: 3, 6, 12, 24, ...

• First term a = 3, common ratio r = 2
• nth term: Tₙ = 3(2)ⁿ⁻¹
• 6th term: T₆ = 3(2)⁵ = 96

Quadratic sequences have second differences that are constant. Finding the nth term requires identifying the quadratic component:

1. Calculate first differences
2. Calculate second differences
3. Half the second difference gives the coefficient of n²
4. Subtract the n² sequence and find the linear pattern

Functions and function notation

Function notation f(x) represents the output when x is the input.

Examples of operations:

• If f(x) = 3x − 5, find f(4): f(4) = 3(4) − 5 = 7
• If g(x) = x² + 2x, find g(−3): g(−3) = (−3)² + 2(−3) = 9 − 6 = 3

Composite functions combine two functions:

• fg(x) means apply g first, then apply f to the result
• If f(x) = 2x + 1 and g(x) = x², then fg(x) = f(x²) = 2x² + 1

Inverse functions f⁻¹(x) reverse the effect of f(x):

1. Replace f(x) with y
2. Rearrange to make x the subject
3. Swap x and y
4. Replace y with f⁻¹(x)

Worked examples

Example 1: Factorise completely 3x² − 27

Solution:

• First identify common factor: 3(x² − 9)
• Recognise difference of two squares: 3(x + 3)(x − 3)
• Answer: 3(x + 3)(x − 3) [2 marks: 1 for common factor, 1 for difference of squares]

Example 2: Solve the simultaneous equations: 2x + 3y = 13 5x − 2y = 4

Solution using elimination:

• Multiply first equation by 2: 4x + 6y = 26
• Multiply second equation by 3: 15x − 6y = 12
• Add equations: 19x = 38
• x = 2
• Substitute into 2x + 3y = 13: 2(2) + 3y = 13
• 4 + 3y = 13
• 3y = 9
• y = 3
• Answer: x = 2, y = 3 [4 marks: 1 for correct elimination process, 1 for x value, 1 for substitution, 1 for y value]

Example 3: The nth term of a sequence is given by Tₙ = n² − 3n. Find which term in the sequence has value 18.

Solution:

• Set Tₙ = 18: n² − 3n = 18
• Rearrange: n² − 3n − 18 = 0
• Factorise: (n − 6)(n + 3) = 0
• n = 6 or n = −3
• Since n must be positive (term position), n = 6
• Answer: 6th term [3 marks: 1 for forming equation, 1 for solving, 1 for rejecting negative solution]

Common mistakes and how to avoid them

• Mistake: Expanding (x + 3)² as x² + 9, omitting the middle term. Correction: (x + 3)² = (x + 3)(x + 3) = x² + 6x + 9. Always write out both brackets or remember the pattern (a + b)² = a² + 2ab + b².

• Mistake: Incorrectly factorising x² + 7x + 10 as (x + 2)(x + 5) by finding factors of 7 instead of 10. Correction: Find two numbers that multiply to give 10 AND add to give 7 (here 2 and 5), so (x + 2)(x + 5) is correct but through proper method.

• Mistake: Forgetting to reverse the inequality sign when dividing by a negative: −2x < 6 giving x < −3. Correction: When dividing both sides by −2, reverse the sign: x > −3.

• Mistake: In simultaneous equations, substituting back into the manipulated equation rather than an original equation. Correction: Always substitute back into one of the original equations to verify the solution and reduce errors.

• Mistake: Confusing arithmetic and geometric sequences, using wrong formulae. Correction: Check whether consecutive terms have constant difference (arithmetic) or constant ratio (geometric) before selecting the formula.

• Mistake: Writing f⁻¹(x) by simply replacing x with 1/x. Correction: The inverse function is found by swapping subject and value algebraically, not by taking reciprocals.

Exam technique for Algebra & Sequences

• Command words matter: "Simplify" requires a single expression with no brackets; "factorise" requires factored form with brackets; "solve" requires x = ... or y = ... statements. Read carefully to provide the format required.

• Show clear working: Algebra questions award method marks even when final answers are incorrect. Write each step on a separate line, particularly when rearranging equations or expanding brackets. Multi-mark questions typically award 1 mark per logical step.

• Check solutions by substitution: When solving equations, substitute your answer back into the original equation. This catches arithmetic errors and ensures you haven't lost or gained solutions incorrectly.

• Use inverse operations systematically: When changing the subject or solving equations, apply one inverse operation at a time and do the same to both sides. For complex rearrangements, number your steps to avoid confusion under exam pressure.

Quick revision summary

Algebra and sequences require fluent manipulation skills and pattern recognition. Master expanding and factorising (including quadratics and difference of squares), solve linear, simultaneous and quadratic equations using appropriate methods, and handle inequalities with correct notation. For sequences, identify whether arithmetic (constant difference) or geometric (constant ratio), then apply correct nth term formulae. Function notation requires careful substitution and methodical rearrangement for inverse functions. Always show working systematically, check solutions by substitution, and ensure your final answer matches the command word requirement.