Algebra and graphs for IGCSE Mathematics: Methods and Worked Examples

Algebra is just a language you already speak

Think about the last time you split a restaurant bill. Someone said "we each owe the total divided by five, plus a tip." You didn't write that as a formula, but your brain treated "the total" as a variable and ran the calculation the moment the number appeared. That's algebra: a shorthand for instructions that work no matter what numbers you plug in. The IGCSE Mathematics syllabus simply asks you to read and write that shorthand fluently, and once you see it as a language rather than a mystery, the whole topic clicks into place.

Building blocks: expressions, terms, and substitution

An algebraic expression is a set of instructions waiting for a number. The expression 3x + 7 says "take a number, multiply it by three, then add seven." The letter x is a placeholder, and substitution is just filling in that placeholder. If x = 4, you get 3(4) + 7 = 19. Simple enough on its own, but this skill underpins everything that follows, so it's worth practising until it feels automatic.

A term is a single chunk of an expression: in 5x² - 3x + 2, the three terms are 5x², -3x, and 2. Like terms share the same variable raised to the same power. You can combine 4x and -x into 3x, but you can't combine 4x and 4x² any more than you'd add apples and oranges.

Algebraic manipulation: tidying the workspace

Most algebra questions come down to one skill: rewriting an expression in a simpler or more useful form. The main tools are collecting like terms, expanding brackets, and factorising.

Collecting like terms

Group terms that match. For 7a + 3b - 2a + b, gather the a-terms (7a - 2a = 5a) and the b-terms (3b + b = 4b) to get 5a + 4b.

Expanding brackets

Multiply every term inside the bracket by whatever sits outside it. For 3(2x - 5), you get 6x - 15. When two brackets meet, each term in the first multiplies each term in the second:

Worked example: Expand (x + 3)(x - 4)

1. x times x = x²
2. x times (-4) = -4x
3. 3 times x = 3x
4. 3 times (-4) = -12
5. Combine: x² - 4x + 3x - 12 = x² - x - 12

Factorising

Factorising reverses expansion: you're looking for the brackets that produced an expression. It comes in several flavours, and the trick is recognising which one applies.

Indices with algebra

The index laws you learned with numbers work identically with letters. The key rules are:

• Multiplying: a^m × aⁿ = a^m+n
• Dividing: a^m ÷ aⁿ = a^m-n
• Power of a power: (a^m)ⁿ = a^mn
• Zero index: a⁰ = 1
• Negative index: a^-n = 1/aⁿ
• Fractional index (Extended): a^1/n = the nth root of a

Worked example: Simplify (2x³)² ÷ 4x

1. Apply the power: (2x³)² = 4x⁶
2. Divide: 4x⁶ ÷ 4x = x⁵

The most common slip here is forgetting to raise the coefficient as well as the variable. 2 squared is 4, not 2.

Equations: finding the unknown

An equation is an expression with an equals sign and a mission: find the value of the unknown. The golden rule is to do the same thing to both sides, peeling away layers until the variable stands alone.

Linear equations

Worked example: Solve 5(x - 2) = 3x + 4

1. Expand: 5x - 10 = 3x + 4
2. Subtract 3x from both sides: 2x - 10 = 4
3. Add 10 to both sides: 2x = 14
4. Divide by 2: x = 7

Simultaneous equations

Two equations, two unknowns. You can eliminate a variable by adding or subtracting the equations, or substitute one equation into the other. Think of it like two witnesses giving overlapping clues: each alone isn't enough, but together they pin down the answer.

Worked example (elimination): Solve 2x + y = 11 and x - y = 1

1. Add the equations: 3x = 12, so x = 4
2. Substitute back: 2(4) + y = 11, so y = 3

Quadratic equations

Quadratics have an x² term and can have zero, one, or two solutions. The three methods are factorising, the quadratic formula, and completing the square (Extended).

Worked example (factorising): Solve x² - 5x + 6 = 0

1. Find two numbers that multiply to 6 and add to -5: that's -2 and -3
2. Write as (x - 2)(x - 3) = 0
3. Either x - 2 = 0 or x - 3 = 0, giving x = 2 or x = 3

If factorising doesn't work cleanly, reach for the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a. It always works, even when the roots aren't whole numbers.

Inequalities

Inequalities follow the same solving steps as equations, with one critical difference: if you multiply or divide by a negative number, you flip the inequality sign. Forget this and you'll get the answer backwards.

Worked example: Solve -3x < 12

1. Divide both sides by -3 and flip the sign: x > -4

On a number line, an open circle means the endpoint isn't included (< or >), while a filled circle means it is (≤ or ≥). Extended candidates also need to shade regions on a coordinate grid where multiple inequalities overlap.

Sequences: spotting and using patterns

A sequence is a list of numbers following a rule. For IGCSE, you need to find the rule and use it to predict any term in the sequence.

For a linear sequence like 5, 8, 11, 14, ..., the common difference is 3, so the nth term is 3n + 2. You can check: when n = 1, you get 3(1) + 2 = 5. Correct.

Extended candidates meet quadratic sequences, where the second differences are constant. If the sequence is 2, 6, 12, 20, 30, ..., the first differences are 4, 6, 8, 10, and the second differences are all 2. A second difference of 2 means the n² coefficient is 1 (half of 2). From there, subtract n² from each term and find the linear part of what remains.

Graphs in practical situations

Distance-time and speed-time graphs turn real journeys into pictures. The key insight is that the gradient tells a story:

• On a distance-time graph, the gradient is speed. A steeper line means faster travel. A flat section means the object has stopped.
• On a speed-time graph, the gradient is acceleration. The area under the curve is the distance travelled.

These graphs love to test whether you can read the story, not just the numbers. A question might say "describe the journey" and expect you to say the cyclist sped up for 10 seconds, travelled at constant speed for 20 seconds, then decelerated to a stop.

Sketching curves

You won't always plot points; sometimes the exam asks you to recognise the shape of a function and sketch it freehand. The main families are:

Being able to match a function to its graph shape is a quick win in the exam. If you see y = 3/x, you should instantly picture two curves sitting in the first and third quadrants (or second and fourth if the coefficient is negative).

Extended-only topics

Algebraic fractions

These work exactly like numerical fractions. To add or subtract, find a common denominator. To simplify, factorise the numerator and denominator and cancel shared factors.

Worked example: Simplify (x² - 9) / (x² + 5x + 6)

1. Factorise numerator: (x + 3)(x - 3)
2. Factorise denominator: (x + 2)(x + 3)
3. Cancel (x + 3): answer is (x - 3) / (x + 2)

Direct and inverse proportion

If y is directly proportional to x, then y = kx for some constant k. If y is inversely proportional to x, then y = k/x. The exam typically gives you one pair of values to find k, then asks you to find y for a different x.

You might also meet y proportional to x² or to the square root of x. The method is identical: write the equation with k, substitute the known values, solve for k, then answer the question.

Differentiation

Differentiation finds the gradient of a curve at any point. For y = axⁿ, the derivative is dy/dx = nax^n-1. Bring the power down as a multiplier, then reduce the power by one.

Worked example: Find dy/dx when y = 3x² + 5x - 2

1. Differentiate each term: 6x + 5
2. The constant -2 vanishes because the derivative of a constant is zero

To find a turning point, set dy/dx = 0 and solve for x. Then substitute back to find the y-coordinate. If the question asks whether the turning point is a maximum or minimum, differentiate again: a positive second derivative means minimum, negative means maximum.

Common mistakes and how to dodge them

Self-check questions

1. Simplify 4a + 3b - a + 5b.
2. Expand and simplify (2x - 1)(x + 4).
3. Factorise completely 3x² - 12.
4. Solve 4(x + 1) = 2(3x - 5).
5. Solve the simultaneous equations: 3x + 2y = 16 and x - y = 2.
6. Solve x² + 2x - 15 = 0 by factorising.
7. Solve -2x + 5 ≥ 11 and represent your answer on a number line.
8. The nth term of a sequence is 4n - 3. Write the first four terms and find the 50th term.
9. A distance-time graph shows a straight line from (0, 0) to (5, 30) and then a horizontal line from (5, 30) to (8, 30). Describe the journey and calculate the speed during the first section.
10. Differentiate y = 2x³ - 4x + 1 and find the gradient of the curve at x = 2. (Extended)

Work through each of these on paper rather than just reading the question and imagining the answer. The difference between knowing a method and being able to execute it under timed IGCSE conditions is practice, and these questions mirror exactly the style Cambridge uses.