Equations Explained: IGCSE Mathematics Complete Breakdown

A universal language with ancient roots

From the mathematicians of medieval Baghdad who first formalised the balance method to the engineers in Tokyo designing earthquake-resistant towers, equations sit at the heart of how the world solves problems. Every culture, every era, every discipline that deals with unknowns eventually arrives at the same principle: if two expressions are equal, you can manipulate both sides in identical ways to uncover what you don't yet know. For IGCSE Mathematics candidates, equations represent one of the most frequently examined topics across all four papers, and mastering them unlocks performance across algebra, geometry, and applied problem-solving alike.

What exactly is an equation?

An equation is a mathematical statement asserting that two expressions have the same value. It always contains an equals sign, and typically contains one or more unknowns (variables) whose values make the statement true. The act of solving an equation means finding those values.

This distinguishes an equation from an expression (which has no equals sign and cannot be "solved") and from an identity (which is true for all values of the variable, such as 2(x + 3) = 2x + 6). The IGCSE syllabus expects candidates to recognise all three and to handle equations of increasing complexity from Core through to Extended tier.

Linear equations: the foundation

A linear equation is one in which the variable appears only to the first power. Solving it requires isolating the variable on one side of the equals sign through a sequence of inverse operations. Think of it as peeling away layers: if x has been multiplied by 3 and then had 5 added, you reverse the process by subtracting 5 first and then dividing by 3.

Worked example 1: a straightforward linear equation

Solve 4x - 7 = 13.

Verification: 4(5) - 7 = 20 - 7 = 13. Correct.

Equations with brackets and unknowns on both sides

When brackets appear, expand them first. When the variable appears on both sides, collect variable terms on one side and constants on the other. The principle remains the same regardless of how the equation looks at first glance.

Worked example 2: unknowns on both sides

Solve 3(2x + 1) = 5x + 9.

1. Expand the bracket: 6x + 3 = 5x + 9
2. Subtract 5x from both sides: x + 3 = 9
3. Subtract 3 from both sides: x = 6

Verification: Left side = 3(2(6) + 1) = 3(13) = 39. Right side = 5(6) + 9 = 39. Both sides match.

Simultaneous equations: two unknowns, two constraints

When a problem involves two unknowns, a single equation isn't enough to pin down unique values. You need two independent equations that the unknowns must satisfy simultaneously. The IGCSE syllabus tests two methods for solving these: elimination and substitution. Both arrive at the same answer; the choice depends on which is more efficient for the particular pair of equations.

The elimination method

The idea is to add or subtract the equations so that one variable disappears entirely. If the coefficients of one variable don't match, multiply one or both equations to make them match first.

Worked example 3: elimination

Solve the simultaneous equations: 2x + 3y = 12 and 5x - 3y = 9.

The y-coefficients are already equal in magnitude (3 and -3). Adding the two equations eliminates y:

1. Add: (2x + 3y) + (5x - 3y) = 12 + 9, giving 7x = 21
2. Divide: x = 3
3. Substitute x = 3 into the first equation: 2(3) + 3y = 12, so 6 + 3y = 12, so 3y = 6, so y = 2

Solution: x = 3, y = 2.

The substitution method

Rearrange one equation to express one variable in terms of the other, then substitute that expression into the second equation. This method is particularly useful when one equation already has a variable with coefficient 1.

Worked example 4: substitution

Solve: y = 2x - 1 and 3x + 2y = 12.

1. The first equation already gives y in terms of x. Substitute into the second: 3x + 2(2x - 1) = 12
2. Expand: 3x + 4x - 2 = 12
3. Simplify: 7x - 2 = 12, so 7x = 14, so x = 2
4. Substitute back: y = 2(2) - 1 = 3

Solution: x = 2, y = 3.

Quadratic equations: when the power rises to two

A quadratic equation takes the general form ax^2 + bx + c = 0, where a is not zero. Unlike linear equations, quadratics can have two solutions, one solution, or (in the real number system) no solutions at all. The IGCSE syllabus requires candidates to solve quadratics by three methods: factorisation, the quadratic formula, and completing the square.

Method 1: factorisation

If the quadratic can be written as a product of two brackets, setting each bracket equal to zero gives the solutions. This is the fastest method when it works, and examiners will often design questions so that the numbers factor neatly.

Worked example 5: factorisation

Solve x^2 - 5x + 6 = 0.

We need two numbers that multiply to give +6 and add to give -5. Those numbers are -2 and -3.

1. Factorise: (x - 2)(x - 3) = 0
2. Set each factor to zero: x - 2 = 0 or x - 3 = 0
3. Solutions: x = 2 or x = 3

Method 2: the quadratic formula

When factorisation isn't obvious, the formula x = (-b +/- sqrt(b^2 - 4ac)) / (2a) will always work. This formula is given on the IGCSE formula sheet, so there's no need to memorise it, but candidates should practise using it fluently under exam conditions.

Worked example 6: quadratic formula

Solve 2x^2 + 3x - 5 = 0, giving answers to 2 decimal places.

Here a = 2, b = 3, c = -5.

1. Calculate the discriminant: b^2 - 4ac = 9 - 4(2)(-5) = 9 + 40 = 49
2. Apply the formula: x = (-3 +/- sqrt(49)) / (2 x 2) = (-3 +/- 7) / 4
3. Two solutions: x = (-3 + 7) / 4 = 4/4 = 1, or x = (-3 - 7) / 4 = -10/4 = -2.50

Solutions: x = 1 or x = -2.50.

Method 3: completing the square (Extended tier)

Completing the square rewrites ax^2 + bx + c in the form a(x + p)^2 + q, from which the solutions and the turning point of the corresponding graph can be read directly. This technique connects algebra to the geometry of parabolas, a link that Cambridge examiners test regularly.

Worked example 7: completing the square

Write x^2 + 6x + 2 in the form (x + p)^2 + q and hence solve x^2 + 6x + 2 = 0.

1. Take half the coefficient of x: half of 6 is 3
2. Write: (x + 3)^2 - 9 + 2 = (x + 3)^2 - 7
3. So the equation becomes: (x + 3)^2 - 7 = 0
4. Rearrange: (x + 3)^2 = 7
5. Take the square root: x + 3 = +/-sqrt(7)
6. Solutions: x = -3 + sqrt(7) or x = -3 - sqrt(7)

As decimals: x = -0.35 (2 d.p.) or x = -5.65 (2 d.p.).

Inequalities: equations with a direction

Inequalities follow the same manipulation rules as equations with one critical exception: multiplying or dividing both sides by a negative number reverses the inequality sign. This single rule catches more IGCSE candidates than any other aspect of the topic.

Worked example 8: solving a linear inequality

Solve 7 - 2x < 3.

1. Subtract 7 from both sides: -2x < -4
2. Divide by -2 (and reverse the sign): x > 2

The solution set is all values of x greater than 2. On a number line, this is represented by an open circle at 2 with a ray extending to the right.

Forming equations from context

Some of the most challenging equation questions on the IGCSE don't begin with algebra at all. They begin with a description: a rectangle whose length is 3 more than twice its width, a journey where speed and time produce a total distance, or a geometric pattern that grows according to a rule. The candidate's task is to translate words into symbols, set up the equation, and then solve it.

This is where equations connect to the world beyond the examination hall. An architect in Barcelona calculating load distributions, a data analyst in Nairobi modelling population growth, a pharmacist in Montreal calibrating drug dosages: all are forming and solving equations derived from real constraints.

Worked example 9: forming and solving

The perimeter of a rectangle is 38 cm. The length is 5 cm more than the width. Find the dimensions.

1. Let the width be w cm. Then the length is (w + 5) cm.
2. Perimeter = 2(length + width), so 2(w + 5 + w) = 38
3. Simplify: 2(2w + 5) = 38, so 4w + 10 = 38
4. Solve: 4w = 28, so w = 7
5. Width = 7 cm, length = 12 cm

Verification: perimeter = 2(7 + 12) = 2(19) = 38 cm. Correct.

Simultaneous equations with one quadratic (Extended tier)

Extended candidates face a more demanding variant: one linear equation and one quadratic equation solved simultaneously. The approach is always substitution. Rearrange the linear equation to express one variable in terms of the other, substitute into the quadratic, and solve the resulting quadratic equation.

Worked example 10: linear and quadratic simultaneous equations

Solve: y = x + 1 and x^2 + y^2 = 25.

1. Substitute the linear equation into the quadratic: x^2 + (x + 1)^2 = 25
2. Expand: x^2 + x^2 + 2x + 1 = 25
3. Simplify: 2x^2 + 2x + 1 = 25, so 2x^2 + 2x - 24 = 0
4. Divide through by 2: x^2 + x - 12 = 0
5. Factorise: (x + 4)(x - 3) = 0
6. x = -4 or x = 3
7. Find corresponding y values: when x = -4, y = -3; when x = 3, y = 4

Solutions: (-4, -3) and (3, 4). Geometrically, these are the two points where the straight line y = x + 1 intersects the circle x^2 + y^2 = 25.

The discriminant and what it reveals

The expression b^2 - 4ac inside the quadratic formula is called the discriminant, and it tells you how many real solutions a quadratic equation has before you solve it.

• If b^2 - 4ac > 0, there are two distinct real solutions.
• If b^2 - 4ac = 0, there is exactly one repeated real solution.
• If b^2 - 4ac < 0, there are no real solutions.

Extended tier questions sometimes ask candidates to find the value of a constant that makes the discriminant equal to zero (the boundary condition for a tangent line touching a curve at exactly one point). This connects the abstract algebra to geometric intuition.

Common errors and how to avoid them

Self-check questions

Test your understanding with these practice problems. Work through each one fully before checking your reasoning.

1. Solve 5(x - 2) = 3x + 4. (Answer: x = 7)
2. Solve the simultaneous equations 3x + 2y = 16 and x - y = 2. (Answer: x = 4, y = 2)
3. Factorise and solve x^2 + 2x - 15 = 0. (Answer: x = 3 or x = -5)
4. Solve the inequality 3 - 4x > 11. (Answer: x < -2)
5. Use the quadratic formula to solve 3x^2 - 2x - 4 = 0, giving answers to 2 decimal places. (Answer: x = 1.54 or x = -0.87)
6. Write x^2 - 8x + 3 in the form (x + p)^2 + q. (Answer: (x - 4)^2 - 13)
7. The sum of two consecutive integers is 47. Form and solve an equation to find them. (Answer: 23 and 24)

Equations are, at their core, a conversation between what you know and what you seek. The balance scale on a medieval merchant's table in Venice, the structural load calculations for a bridge in Osaka, the economic models built in offices across Zurich: all rest on the same elegant principle that what holds true on one side of the equals sign must hold true on the other. For IGCSE candidates, fluency with equations isn't just an exam skill. It's the entry point to mathematical reasoning that spans cultures, centuries, and disciplines.