GCSE Maths Algebra: From Basics to Complex Problem Solving

Algebra forms the backbone of GCSE Maths, appearing across both papers for all exam boards including AQA, Edexcel, and OCR. This guide is part of our complete GCSE Maths revision series — start there if you want an overview of all topics. Whether you’re aiming for Foundation or Higher tier, mastering algebraic techniques is essential for success. This guide takes you from basic manipulation through to solving complex equations.

Simplifying and Expanding

The foundation of algebra is understanding how to manipulate expressions. When simplifying, collect like terms by adding or subtracting coefficients of terms with identical variables. For example, 3x + 5x - 2x simplifies to 6x, but 3x + 5y cannot be simplified further because x and y are different variables.

Expanding brackets requires multiplying everything inside the bracket by the term outside. For single brackets, 3(2x + 5) = 6x + 15. Watch your signs carefully: -2(3x - 4) = -6x + 8. The minus outside changes both signs inside.

Expanding double brackets uses the grid method or FOIL (First, Outer, Inner, Last). For (x + 3)(x + 5), multiply to get x² + 5x + 3x + 15, which simplifies to x² + 8x + 15. This technique is crucial for later quadratic work.

Factorising

Factorising reverses expansion and is essential for solving quadratic equations. Start by always looking for common factors. For 6x² + 9x, both terms share 3x, so factor it out: 3x(2x + 3).

For quadratic expressions like x² + 7x + 12, find two numbers that multiply to give 12 and add to give 7. That’s 3 and 4, so the expression factorises to (x + 3)(x + 4). Check by expanding back out.

Difference of two squares is a special case: x² - 9 = (x + 3)(x - 3). Recognise the pattern of something squared minus something else squared. This appears frequently on Higher tier papers.

Solving Linear Equations

Linear equations contain variables to the power of 1. The goal is isolating the variable by performing inverse operations to both sides. For 3x + 7 = 19, subtract 7 from both sides (3x = 12), then divide both sides by 3 (x = 4).

When the variable appears on both sides, like 5x + 3 = 2x + 15, collect variables on one side and numbers on the other. Subtract 2x from both sides: 3x + 3 = 15. Subtract 3: 3x = 12. Divide by 3: x = 4.

For equations with fractions, multiply through by the denominator to eliminate fractions first. For (x + 3)/4 = 5, multiply both sides by 4 to get x + 3 = 20, then x = 17.

Solving Quadratic Equations

Quadratics contain x² terms and can be solved by factorising, completing the square, or using the quadratic formula. If the equation is in the form x² + bx + c = 0, try factorising first.

For x² + 5x + 6 = 0, factorise to (x + 2)(x + 3) = 0. For this product to equal zero, either bracket must equal zero, so x + 2 = 0 (giving x = -2) or x + 3 = 0 (giving x = -3). Quadratics have two solutions.

The quadratic formula x = [-b ± √(b² - 4ac)] / 2a works for any quadratic in the form ax² + bx + c = 0. For 2x² + 7x + 3 = 0, a = 2, b = 7, c = 3. Substituting gives x = [-7 ± √(49 - 24)] / 4 = [-7 ± 5] / 4, so x = -0.5 or x = -3. This method is essential when expressions don’t factorise neatly.

Simultaneous Equations

Simultaneous equations involve finding values that satisfy multiple equations. For linear simultaneous equations, use elimination or substitution. The elimination method involves adding or subtracting equations to eliminate one variable.

For example, given 2x + y = 8 and 3x - y = 7, adding the equations eliminates y: 5x = 15, so x = 3. Substitute back into either original equation: 2(3) + y = 8, so y = 2.

When coefficients don’t match, multiply one or both equations first. For 2x + 3y = 13 and 3x + 2y = 12, multiply the first by 3 and the second by 2 to get 6x + 9y = 39 and 6x + 4y = 24. Subtract to eliminate x: 5y = 15, so y = 3.

Sequences

Arithmetic sequences have a constant difference between terms. The nth term formula is a + (n - 1)d, where a is the first term and d is the common difference. For the sequence 5, 8, 11, 14…, a = 5 and d = 3, so the nth term is 5 + (n - 1)3 = 3n + 2.

Geometric sequences multiply by a constant ratio each time. Recognising sequence types and finding nth term expressions is a common exam question.

Higher Tier Topics

Higher tier includes additional content like solving cubic and reciprocal equations graphically, algebraic fractions, proof, and more complex simultaneous equations involving one quadratic equation. Functions and their notation (f(x)) also appear more extensively at Higher tier.

Exam Tips for Algebra

Show all working clearly. Even if your final answer is wrong, you can earn method marks for correct steps. Check solutions by substituting back into the original equation. Watch signs carefully, especially when dealing with negatives and brackets. Time management is crucial, so don’t spend too long on one question.

Related GCSE Maths Guides

• Quadratic formula guide — when to use it, the discriminant, and step-by-step examples
• GCSE Maths formula sheet guide — what’s provided in the exam and what you must memorise
• Geometry and measures guide — angles, shapes, and trigonometry

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