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Cambridge O Level · Mathematics 4024

Equations and formulae

Form and solve linear, fractional, simultaneous and quadratic equations, then rearrange formulae to change the subject. This original practice pack targets Cambridge O Level Mathematics 4024 section 2.5. Supporting prerequisites from algebraic manipulation, algebraic fractions and surds are used only where section 2.5 requires them; inequalities, functions, graph methods and general algebraic-fraction manipulation are excluded.

2025 / 2026 / 2027 · Academic review not recorded · Published 2026-09-08

AI-assisted practice — not independently academically reviewed. Answers may contain errors; check important results against your course materials.

Guided tutorial

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19 lesson sections

What you will learn

  • Distinguish an expression, an equation, a formula and an identity, and construct algebraic statements from words.
  • Form equations and simultaneous equations from numerical and contextual information.
  • Solve linear equations in one unknown, including equations with brackets and fractional numerical coefficients.
  • State denominator restrictions and solve fractional equations with numerical or linear algebraic denominators.
  • Solve simultaneous linear equations in two unknowns by elimination or substitution.
  • Solve quadratic equations by factorisation, completing the square and the quadratic formula.
  • Give exact quadratic solutions in simplified surd form when required.
  • Select mathematically valid contextual solutions and reject values that violate stated conditions.
  • Change the subject of a formula when the subject occurs once, occurs twice, or is under a power or root.
  • Check solutions by substitution and check rearrangements by reversing the algebra.

Explanation

Expressions, equations and formulae have different jobs

An expression, such as 3x + 5, represents a value but makes no claim that two quantities are equal. An equation, such as 3x + 5 = 20, states an equality that is true only for particular values. A formula states a relationship between named quantities, while an identity is true for every permitted value of its variables.

When translating words, define each variable with its unit before writing algebra. Words such as sum, difference, product, quotient, consecutive and total determine the operations. Keep the order in phrases such as 'five less than twice x': this is 2x - 5, not 5 - 2x.

Worked example

Worked example: construct an expression and an equation

Let the smaller of two consecutive even integers be 2n, where n is an integer. The next even integer is 2n + 2, so their product is 2n(2n + 2). Using n and n + 2 would also describe consecutive even integers if n itself were explicitly defined as even.

If their product is 168, the equation is 2n(2n + 2) = 168. The expression describes the product; adding '= 168' creates an equation. Defining the variable and its permitted values prevents an algebraically correct expression from misrepresenting the context.

Explanation

Solve linear equations by preserving equality

A linear equation contains the unknown only to the first power after simplification. Perform the same valid operation on both sides, expand brackets, collect unknown terms on one side and constants on the other, then divide by the remaining coefficient. This balance principle is more reliable than memorising that a term 'changes sides'.

Fractions with numerical denominators can be cleared by multiplying every term on both sides by the lowest common multiple. Check the result in the original equation, especially when negative signs or brackets were present.

Worked example

Worked example: a linear equation with brackets

Solve 5 - 2x = 3(x + 7). Expanding the right side gives 5 - 2x = 3x + 21. Subtract 3x from both sides and subtract 5 from both sides to obtain -5x = 16.

Therefore x = -16/5. Substitution checks the answer: the left side is 5 + 32/5 = 57/5, and the right side is 3(-16/5 + 7) = 3(19/5) = 57/5.

Explanation

Fractional equations require restrictions

A fractional equation may have the unknown in a denominator. Before multiplying through, list every value that would make an original denominator zero; those values are excluded. Multiply every term by a common denominator, solve the resulting equation, and test each candidate in the original equation.

Clearing denominators can produce a linear or quadratic equation. A root of the cleared equation is not automatically valid: reject it if it makes any original denominator zero. General simplification of algebraic fractions belongs to section 2.3; here it is used only as supporting technique for solving section 2.5 equations.

Worked example

Worked example: one linear algebraic denominator

Solve x/(2x + 1) = 4. First record 2x + 1 ≠ 0, so x ≠ -1/2. Multiplying both sides by 2x + 1 gives x = 8x + 4, hence -7x = 4 and x = -4/7.

The candidate is allowed because 2(-4/7) + 1 = -1/7, not zero. Substituting gives (-4/7)/(-1/7) = 4, so x = -4/7 is the valid solution.

Worked example

Worked example: two algebraic denominators

Solve 2/(x + 2) + 3/(2x - 1) = 1, with x ≠ -2 and x ≠ 1/2. Multiply every term by (x + 2)(2x - 1): 2(2x - 1) + 3(x + 2) = (x + 2)(2x - 1).

Expanding gives 7x + 4 = 2x^2 + 3x - 2, so 2x^2 - 4x - 6 = 0. Divide by 2 and factorise: (x - 3)(x + 1) = 0. Thus x = 3 or x = -1; neither is excluded, and substitution confirms both.

Explanation

Simultaneous equations locate a common solution

Two simultaneous linear equations in x and y must be true for the same ordered pair. Elimination combines suitable multiples of the equations so that one variable cancels. Substitution expresses one variable in terms of the other and places that expression into the second equation.

Choose the route that keeps arithmetic simple. After finding one variable, substitute into an original equation to find the other, then check the pair in both originals. In a context, define both variables and make sure each equation represents the stated totals rather than unrelated situations.

Worked example

Worked example: solve by elimination

Solve 2x + 3y = 17 and 3x - 2y = 6. Multiply the first equation by 2 to obtain 4x + 6y = 34, and multiply the second by 3 to obtain 9x - 6y = 18.

Adding eliminates y: 13x = 52, so x = 4. Substitute into 2x + 3y = 17 to get 8 + 3y = 17, hence y = 3. The pair (4, 3) also satisfies 3x - 2y = 12 - 6 = 6.

Worked example

Worked example: construct simultaneous equations

Adult tickets cost a dollars and child tickets cost c dollars. Three adult and two child tickets cost 34 dollars, while two adult and five child tickets cost 41 dollars. The model is 3a + 2c = 34 and 2a + 5c = 41.

Multiply the first equation by 5 and the second by 2: 15a + 10c = 170 and 4a + 10c = 82. Subtract to get 11a = 88, so a = 8; then 3(8) + 2c = 34 gives c = 5. Both prices are positive and reproduce both totals.

Explanation

Three exact methods solve quadratic equations

Write a quadratic equation as ax^2 + bx + c = 0 before choosing a method. Factorisation is efficient when factors are visible. Completing the square rewrites the quadratic around a squared bracket. The quadratic formula x = (-b ± √(b^2 - 4ac))/(2a) works whenever a ≠ 0.

The ± symbol usually represents two candidates. Keep exact roots as simplified fractions and surds unless a decimal is requested. Substituting candidates into the original equation checks signs and arithmetic; contextual restrictions may leave only one meaningful root.

Worked example

Worked example: solve a quadratic by factorisation

Solve 2x^2 - 5x - 3 = 0. Factorise the left side as (2x + 1)(x - 3), since expanding gives 2x^2 - 6x + x - 3 = 2x^2 - 5x - 3.

A product is zero when at least one factor is zero. Therefore 2x + 1 = 0 or x - 3 = 0, giving x = -1/2 or x = 3. Both values satisfy the original equation.

Worked example

Worked example: solve by completing the square

Solve x^2 - 6x - 7 = 0. Complete the square: x^2 - 6x - 7 = (x - 3)^2 - 9 - 7 = (x - 3)^2 - 16. The equation becomes (x - 3)^2 = 16.

Taking both square-root possibilities gives x - 3 = ±4. Hence x = 7 or x = -1. Squaring either displacement gives 16, and direct substitution confirms both roots.

Worked example

Worked example: use the quadratic formula and simplify a surd

Solve 3x^2 + 2x - 4 = 0. Here a = 3, b = 2 and c = -4. Substitution into the quadratic formula gives x = [-2 ± √(2^2 - 4(3)(-4))]/6 = [-2 ± √52]/6.

Since √52 = √(4 × 13) = 2√13, divide numerator and denominator by 2 to obtain x = (-1 ± √13)/3. Keeping the surd gives the exact pair of solutions.

Worked example

Worked example: form a quadratic and select the valid root

A rectangle has width x cm and length x + 3 cm. If its area is 40 cm^2, then x(x + 3) = 40, so x^2 + 3x - 40 = 0. Factorising gives (x + 8)(x - 5) = 0.

The algebraic roots are x = -8 and x = 5. A physical width must be positive, so reject -8 and use width 5 cm and length 8 cm. Their product is 40 cm^2, confirming the contextual solution.

Explanation

Changing the subject reverses a formula

To make a chosen variable the subject, isolate it so it appears alone on one side. Undo operations in reverse order while performing equivalent operations on both sides. If the subject occurs in several terms, collect those terms together and factor out the subject before dividing.

A rearrangement may require conditions on denominators and a choice of sign after taking a square root. State the branch appropriate to the context, such as the positive root for a length. Check by substituting the rearranged expression back into the original formula.

Worked example

Worked example: a subject that occurs once

Make t the subject of v = u + at. Subtract u from both sides to obtain v - u = at. Provided a ≠ 0, divide by a to get t = (v - u)/a.

Reversing the steps verifies the rearrangement: multiplying t by a gives v - u, and then adding u gives v. The condition a ≠ 0 matters because division by zero is undefined.

Worked example

Worked example: the subject appears twice

Make x the subject of p = (ax + b)/(cx + d). Multiply by cx + d to obtain pcx + pd = ax + b. Gather the x-terms: pcx - ax = b - pd.

Factor x to get x(pc - a) = b - pd, then divide: x = (b - pd)/(pc - a), provided pc - a ≠ 0 and the original denominator cx + d is non-zero for the values considered. Expanding the factored line recovers the preceding equation.

Worked example

Worked example: remove a power or root

Make r the subject of A = πr^2 for a circle. Divide by π to obtain r^2 = A/π, then take square roots: r = ±√(A/π) algebraically. Because a radius is non-negative, the contextual formula is r = √(A/π).

For T = 2π√(L/g), first divide by 2π, square both sides, and multiply by g: T^2/(4π^2) = L/g, so L = gT^2/(4π^2). Squaring is performed only after the root term has been isolated.

Original practice, not an official examination paper. Readnary is not affiliated with the awarding body. Prepared with AI assistance.