IGCSE Mathematics: Algebra and Graphs — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Algebra and Graphs revision notes

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Questions

1. Expand and simplify (x + 5)(x − 3). [2]

2. (Extended) Factorise completely: 6x² − 24. [2]

3. Solve the equation 5x + 4 = 2x + 19. [2]

4. Solve the simultaneous equations: 2x + y = 11 x − y = 1 [3]

5. (Extended) Solve x² − 5x + 6 = 0 by factorisation. [3]

6. (Extended) Make r the subject of the formula A = πr². [2]

7. Solve the inequality 3x + 2 < 17, and show your answer on a number line. [2]

8. The first five terms of a sequence are 4, 9, 16, 25, 36. Find an expression for the nth term. [2]

9. A straight-line graph passes through the points (0, −2) and (3, 10).

(a) Find the gradient of the line. [2] (b) Write the equation of the line in the form y = mx + c. [1]

10. (Extended) Solve x² + 3x − 5 = 0 using the quadratic formula, giving your answers to 2 decimal places. [3]

11. Simplify: (a) a³ × a⁵ (b) a⁸ ÷ a³ (c) a⁻². [3]

12. (Extended) y is directly proportional to x. When x = 4, y = 20. Find y when x = 7. [3]

13. (Extended) Differentiate y = 3x² − 2x + 1, and find the gradient of the curve at x = 2. [3]

14. (Extended) Given f(x) = 2x + 1 and g(x) = x², find fg(3) and the inverse function f⁻¹(x). [3]

15. (Extended) y is inversely proportional to x. When x = 2, y = 15. Find y when x = 5. [3]

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Answers

1. (x + 5)(x − 3) = x² − 3x + 5x − 15 [1] = x² + 2x − 15 [1].

2. 6x² − 24 = 6(x² − 4) [1] = 6(x + 2)(x − 2) [1].

3. 5x − 2x = 19 − 4 [1] → 3x = 15 → x = 5 [1].

4. From equation 2: x = y + 1 [1]. Substitute into equation 1: 2(y + 1) + y = 11 → 2y + 2 + y = 11 → 3y = 9 → y = 3 [1]. Then x = 3 + 1 = x = 4 [1].

5. x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0 [2] → x = 2 or x = 3 [1].

6. A = πr² → r² = A/π [1] → r = √(A/π) [1].

7. 3x < 15 [1] → x < 5 [1] — open circle at 5, shading to the left.

8. These are the square numbers: 4 = 2², 9 = 3², 16 = 4², 25 = 5², 36 = 6² [1] → nth term = (n + 1)² [1].

9. (a) gradient = (10 − (−2)) / (3 − 0) = 12/3 = 4 [2]. (b) c = −2 (the y-intercept, given directly) → y = 4x − 2 [1, allow follow-through from part (a)].

10. a = 1, b = 3, c = −5 [1]. x = (−3 ± √(9 + 20)) / 2 = (−3 ± √29) / 2 [1] → x = 1.19 or x = −4.19 (2 d.p.) [1].

11. (a) a³ × a⁵ = a⁸ [1]. (b) a⁸ ÷ a³ = a⁵ [1]. (c) a⁻² = 1/a² [1].

12. y = kx, so 20 = 4k [1] → k = 5. When x = 7: y = 5 × 7 = 35 [1] [1].

13. dy/dx = 6x − 2 [1] [1]. At x = 2: gradient = 6(2) − 2 = 10 [1]. This uses the rule that for y = axⁿ, dy/dx = anxⁿ⁻¹, applied term by term.

14. fg(3): apply g first, g(3) = 9, then f(9) = 2(9) + 1 = 19 [1]. For f⁻¹(x): let y = 2x + 1, swap x and y to get x = 2y + 1, then solve: f⁻¹(x) = (x − 1)/2 [1] [1].

15. y = k/x, so 15 = k/2 [1] → k = 30. When x = 5: y = 30/5 = 6 [1] [1]. As with direct proportion, the constant k is found first from one known pair of values, then reused to find the rest.

Examiner report insight

• Stopping after removing a numeric common factor without checking for a further structure inside the bracket – e.g. factorising 12m^2 - 75t^2 as 3(4m^2 - 25t^2) and stopping, instead of spotting the remaining difference of two squares.
• In simultaneous equations solved by elimination, using addition where the signs require subtraction (or vice versa) after scaling one equation – check the sign of each term being eliminated, not just its size.
• When a question asks you to solve an equation graphically (e.g. “by drawing a suitable line”), doing so – an algebraic solution to the same equation, even if correct, does not answer what was asked and is not credited the same way.

Source: Cambridge International, 0580 Mathematics Principal Examiner Report, June 2024 series, Papers 11, 21, 22, 23 (verified 2026-09-02).