GCSE Foundation — Algebra
Every algebra statement a GCSE Foundation student may be asked is here; together they are 20% of the assessment. There are 21 algebra statements to cover. Working through them one at a time beats mixed practice while a topic is still shaky — mixed practice tells you that something is wrong, not what. Expect algebraic notation, substitution into formulae, expressions, equations, formulae, identities and inequalities, simplifying, expanding and factorising, rearranging formulae and identities, equivalence and algebraic proof. None of this is Higher-only material: it is the shared content both tiers sit. Only 50% of the paper is straight technique (AO1); 50% asks you to reason, interpret or solve a problem in context, and those marks are won by setting the work out clearly. The bulk of it sits comfortably on the non-calculator paper; a calculator quietly hides the gap until the exam.
- A1 — Algebraic notation
- A2 — Substitution into formulae
- A3 — Expressions, equations, formulae, identities and inequalities
- A4 — Simplifying, expanding and factorising
- A5 — Rearranging formulae
- A6 — Identities, equivalence and algebraic proof
- A7 — Functions, inverse and composite functions
- A8 — Coordinates in all four quadrants
- A9 — Straight-line graphs and y = mx + c
- A10 — Gradients and intercepts of linear functions
- …and 11 more algebra statements at this tier
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Algebra at GCSE Foundation: the 21 DfE statements
Every question on this site is filed against one of these statements. Pick one to practise it on its own.
Sample algebra questions for GCSE Foundation
- Factorise fully 5x + 5y − 5(a)5(xy − 1)(b)5(x + y + 1)(c)5(x + y − 1)(d)5(x + y − 5)
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5(x + y − 1) — Method: take out the highest common factor of all three terms and divide every term by it, the number term included. Working: the highest common factor of 5x, 5y and −5 is 5; dividing gives 5x ÷ 5 = x, 5y ÷ 5 = y and −5 ÷ 5 = −1, so the bracket holds x + y − 1. Answer: 5(x + y − 1), which multiplies back out to 5x + 5y − 5. The distractors: 5(x + y + 1) comes from dividing −5 by 5 and losing the minus sign; 5(x + y − 5) comes from dividing only the terms containing a letter by 5 and carrying the −5 into the bracket unchanged; 5(xy − 1) comes from collecting the unlike terms 5x and 5y as 5xy before factorising. - Factorise fully 56x − 24(a)8(7x − 16)(b)8(56x − 3)(c)8(7x + 3)(d)8(7x − 3)
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8(7x − 3) — Method: find the highest common factor of the two terms, write it in front of a bracket and divide each term by it. Working: 56 = 8 × 7 and 24 = 8 × 3, so the highest common factor is 8; dividing gives 56x ÷ 8 = 7x and 24 ÷ 8 = 3, and the subtraction sign stays between them. Answer: 8(7x − 3), which multiplies back out to 56x − 24. The distractors: 8(7x + 3) comes from dropping the minus sign of −24 while dividing; 8(56x − 3) comes from dividing only the number term by 8 and leaving 56x untouched inside the bracket; 8(7x − 16) comes from subtracting 8 from 24 instead of dividing 24 by 8. - Expand and simplify (x − 3)²(a)x² + 6x + 9(b)x² − 9(c)x² − 3x + 9(d)x² − 6x + 9
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x² − 6x + 9 — Method: squaring a bracket means multiplying that bracket by itself, so expand (x − 3)(x − 3) term by term and then collect like terms. Working: x × x = x², x × (−3) = −3x, (−3) × x = −3x and (−3) × (−3) = 9, giving x² − 3x − 3x + 9, and the two middle terms collect to −6x. Answer: x² − 6x + 9. The distractors: x² − 3x + 9 comes from writing down only one of the two middle products instead of both; x² + 6x + 9 comes from treating (−3) × x as +3x, so the middle terms are added rather than subtracted; x² − 9 comes from treating the square as the difference of two squares (x − 3)(x + 3). - Simplify (5x² + 3x − 2) − (2x² − x + 5)(a)3x² + 4x + 3(b)3x² + 2x + 3(c)7x² + 2x + 3(d)3x² + 4x − 7
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3x² + 4x − 7 — Method: the minus sign in front of the second bracket changes the sign of every term inside it; then collect like terms. Working: removing the brackets gives 5x² + 3x − 2 − 2x² + x − 5; the squared terms give 5x² − 2x² = 3x², the x terms give 3x + x = 4x, and the number terms give −2 − 5 = −7. Answer: 3x² + 4x − 7. The distractors: 7x² + 2x + 3 comes from adding the two brackets instead of subtracting, giving 5x² + 2x², 3x − x and −2 + 5; 3x² + 2x + 3 comes from applying the minus sign to 2x² only, leaving −x and +5 unchanged so that 3x − x = 2x and −2 + 5 = 3; 3x² + 4x + 3 comes from changing the signs of the terms with letters but leaving +5 as it stood, so the number terms give −2 + 5 = 3. - Simplify 2a + 3b − a + 2b − b(a)a + 4b(b)a + 6b(c)5ab(d)3a + 4b
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a + 4b — Method: collect the terms in a and the terms in b separately, keeping the sign that stands in front of each term. Working: the a terms give 2a − a = a; the b terms give 3b + 2b − b = 4b; a and b are unlike terms, so the two results cannot be joined. Answer: a + 4b. The distractors: 3a + 4b comes from reading −a as +a, so that 2a + a = 3a; a + 6b comes from reading −b as +b, so that 3b + 2b + b = 6b; 5ab comes from combining the unlike terms a and 4b into one term by adding 1 and 4 and joining the letters.
Frequently asked questions
How much of the GCSE Foundation paper is algebra?
20% of the total marks, and the three papers each carry an equal share of the qualification, so it is spread across all of them rather than concentrated in one.
How many topics are there in algebra at GCSE Foundation?
21 DfE content statements, coded A1 to A25 within this area. Every question in the bank is tagged to one of them.
Is any algebra content on this page Higher only?
No. Foundation pages only show statements a Foundation entry may be asked. The Higher-only algebra material is on the GCSE Higher page for this area.
Can I use a calculator for algebra questions?
Paper 1 is non-calculator; Papers 2 and 3 allow one. Of the 21 statements in this area, 18 are naturally non-calculator and 0 naturally calculator, with the rest workable either way — so practise both.
What kind of marks does algebra carry?
The GCSE Foundation split is 50% AO1 (use and apply standard techniques), 25% AO2 (reason, interpret and communicate) and 25% AO3 (solve problems in and out of context). Questions in this area are set across all three.
Are these past paper questions?
No. Every question at /gcse-foundation is original, written to the DfE content statements and checked before it is published. We do not host past papers or mark schemes.
Do I need an account?
No — you can start practising straight away. Progress is saved automatically in your browser.
Is it free?
Yes. The questions, worked answers and printable worksheets are all free, with no adverts.
Tips for algebra at GCSE Foundation
- Factorising is the central algebra skill: always look for a common factor first, and only then try a difference of two squares or a quadratic in the form x² + bx + c.
- Know the three expansions by heart: (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b², and (a + b)(a − b) = a² − b².
- To solve a quadratic, rearrange so that one side is 0, factorise, and set each bracket equal to 0. A product is 0 only when one of its factors is 0.
- For simultaneous equations write the two equations one under the other with the x terms aligned. Use substitution when one unknown is already the subject; use elimination when the coefficients of one unknown are equal or opposite.
- After finding one unknown, substitute it back into an original equation to find the other, then check both values in both equations.
Common mistakes (and how to avoid them)
- Writing (a − b)² = a² − b² and forgetting the middle term −2ab. Squaring a bracket always gives three terms.
- Losing a solution: from x² = 9 writing only x = 3, when x = −3 is also a solution.
- Adding the equations in elimination when the coefficients are equal rather than opposite, so the unknown does not cancel. To eliminate you need +3y and −3y.
- Finding x and stopping — the solution of a pair of simultaneous equations is a pair of values (x, y), not a single number.
Worked examples
- Look for two numbers that multiply to 6 and add to −5.
- The numbers are −2 and −3: (−2) × (−3) = 6 and −2 + (−3) = −5.
- Write the factorised form using those numbers.
- Two numbers that multiply to 10 and add to −7 are −2 and −5.
- Factorise: (x − 2)(x − 5) = 0.
- A product is 0 when a factor is 0, so x − 2 = 0 or x − 5 = 0.
- Solve each: x = 2 or x = 5.
- The y coefficients are +1 and −1, so adding the equations eliminates y.
- (x + y) + (x − y) = 10 + 4, so 2x = 14 and x = 7.
- Substitute x = 7 into the first equation: 7 + y = 10, so y = 3.
- Check in the second equation: 7 − 3 = 4. Correct.
Algebra is a fifth of the Foundation paper and the language every later topic is written in. Factorising, the standard expansions and solving quadratics by factorising are the skills the grade 4-5 questions turn on, and simultaneous equations teach the habit that makes them all work: do the same thing to both sides, and substitute your answer back to check it. Practise a little every day, write every step, and treat the check as part of the method rather than an optional extra.
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