AQA GCSE Mathematics: Algebra — 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 study guide | Algebra revision notes

---

Section A

1. Simplify fully: (a) 3a + 5b − a + 2b (b) 4x × 3x² (c) expand and simplify 2(3x − 1) + 3(x + 4) [4]

2. Factorise fully: (a) 6x + 9 (b) x² − 5x + 6 [3]

Section B

3. Solve the equation 5x − 3 = 2x + 12. [3]

4. Rearrange the formula v = u + at to make a the subject. [2]

5. A line has equation 3y = 6x − 9.

(a) Find its gradient and y-intercept. [2] (b) State the equation of a line parallel to it that passes through (0, 4). [2]

6. Solve the quadratic equation x² + 2x − 15 = 0 by factorising. [3]

7. Solve the quadratic equation 2x² − 3x − 4 = 0 using the quadratic formula, giving your answers to 2 decimal places. [4]

8. Solve the simultaneous equations y = x + 2 and y = x² − 4 algebraically. [5]

9. Solve the inequality 3x − 4 ≤ 11, and represent your answer on a number line. [3]

10. (a) Find the nth term of the sequence 4, 7, 10, 13, … [2] (b) Find the nth term of the sequence 2, 5, 10, 17, 26, …, showing your method. [4]

---

Answers

1. (a) 2a + 7b [1]. (b) 12x³ [1]. (c) 2(3x − 1) + 3(x + 4) = 6x − 2 + 3x + 12 = 9x + 10 [1] [1].

2. (a) 6x + 9 = 3(2x + 3) [1]. (b) x² − 5x + 6 = (x − 2)(x − 3) [1] [1].

3. 5x − 3 = 2x + 12 → 3x = 15 [1] [1] → x = 5 [1].

4. v = u + at → v − u = at → a = (v − u) / t [1] [1].

5. (a) 3y = 6x − 9 → y = 2x − 3, so gradient = 2, y-intercept = −3 [1] [1]. (b) A parallel line has the same gradient (2) and passes through (0, 4), so its equation is y = 2x + 4 [1] [1].

6. x² + 2x − 15 = 0 → (x + 5)(x − 3) = 0 [1] [1] → x = −5 or x = 3 [1].

7. Using x = (−b ± √(b² − 4ac)) / 2a with a = 2, b = −3, c = −4: x = (3 ± √(9 + 32)) / 4 = (3 ± √41) / 4 [1] [1]. This gives x = 2.35 (3 s.f. rounded to 2 d.p.) or x = −0.85 (2 d.p.) [1] [1].

8. Substituting y = x + 2 into y = x² − 4: x + 2 = x² − 4 [1] → 0 = x² − x − 6 [1] → 0 = (x − 3)(x + 2) [1] → x = 3 or x = −2 [1]. Substituting back: x = 3 gives y = 5; x = −2 gives y = 0, so the solutions are (3, 5) and (−2, 0) [1].

9. 3x − 4 ≤ 11 → 3x ≤ 15 [1] → x ≤ 5 [1]. On a number line: a filled circle at 5, with the line shaded to the left, indicating all values equal to or less than 5 [1].

10. (a) The common difference is 3, so the coefficient of n is 3: 3n gives 3, 6, 9, 12, each 1 less than the sequence, so the nth term is 3n + 1 [1] [1]. (b) First differences: 3, 5, 7, 9 (not constant) [1]; second differences: 2, 2, 2 (constant, so quadratic) [1]. Coefficient of n² = 2 ÷ 2 = 1, so compare n² (1, 4, 9, 16, 25) to the sequence (2, 5, 10, 17, 26): the difference is always +1 [1], so the nth term is n² + 1 [1].

---

Where marks are usually lost

• Leaving an answer not in simplest form, such as “2 × x + 3 × x” instead of “5x.”
• Misreading the gradient from an equation that has not yet been rearranged into y = mx + c form.
• Forgetting a step when rearranging a formula, particularly when the subject appears on both sides of an intermediate line.
• Applying the quadratic formula with a sign error on b or c, especially when the original equation already contains a negative coefficient.
• Attempting to spot a quadratic sequence’s nth term directly rather than using the structured second-differences method.

Approaching algebra questions

Treat notation and manipulation fluency as the prerequisite skill for every other part of this topic: an error collecting like terms or expanding brackets early in a multi-step question propagates into every later line, so it is worth double-checking simplification steps even in questions that are primarily about equations, graphs or sequences. For any straight-line graph question, always rearrange the given equation into y = mx + c form before reading off the gradient or intercept, since the specification deliberately tests this rearrangement rather than allowing the gradient to be read directly from an unrearranged equation. For quadratic sequences, always find the second differences first and use them to identify the n² coefficient before attempting to match the remaining linear part by comparison — this structured method is reliable in a way that spotting the pattern directly usually is not under exam time pressure.