AQA GCSE Mathematics 8300: Solving equations and inequalities – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover subsection 3.2.3, Solving equations and inequalities (A17 to A22), of the AQA GCSE Mathematics (8300) specification, for teaching from September 2015 with exams from May/June 2017 (version 1.0). Questions 1 to 6 and 12 use content for both tiers. Questions 7 to 11 are marked “(Higher)” because they use Higher tier only content: the quadratic formula, completing the square, linear/quadratic simultaneous equations, iteration, quadratic inequalities and regions. Each question says whether it is non-calculator (Paper 1 style) or calculator (Papers 2 and 3 style).

They are different from the questions in the Algebra practice questions, so you can use both. Learn the content first in the study guide and the revision notes. The course hub is AQA GCSE Mathematics and the printable checklist lists every statement.

Questions

1. (non-calculator) Solve 7(x − 2) = 3x + 10. [3]

2. (non-calculator) Solve (3x − 1)/4 = x − 2. [3]

3. (non-calculator)

(a) Solve x² − 5x − 24 = 0. [3] (b) Solve x² − 36 = 0. [2]

4. (non-calculator) Solve the simultaneous equations

4x + 3y = 5 2x − 5y = 9 [4]

5. (non-calculator)

(a) Solve 9 − 4x ≥ 21. [2] (b) List all the integers n that satisfy −7 < 3n − 1 ≤ 8. [3] (c) Describe how you would show −4 ≤ x < 1 on a number line. [1]

6. (calculator) At a café, 4 coffees and 3 muffins cost £13.10. 2 coffees and 5 muffins cost £11.10.

(a) Form two equations and solve them to find the cost of one coffee and the cost of one muffin. [5] (b) Ella has £20. She buys 3 coffees and some muffins. Form and solve an inequality to find the greatest number of muffins she can buy. [3]

7. (calculator) (Higher) Solve 4x(x + 1) = 7. Give your answers to 2 decimal places. [4]

8. (non-calculator) (Higher)

(a) Write x² + 8x + 11 in the form (x + a)² + b. [2] (b) Hence solve x² + 8x + 11 = 0. Give your answers in surd form. [2]

9. (non-calculator) (Higher) Solve algebraically the simultaneous equations

y = 2x − 1 y = x² − 3x + 5 [5]

10. (calculator) (Higher) f(x) = x³ + 3x − 9.

(a) Show that the equation f(x) = 0 has a solution between x = 1 and x = 2. [2] (b) Show that x³ + 3x − 9 = 0 can be rearranged to give x = ∛(9 − 3x). [1] (c) Use xₙ₊₁ = ∛(9 − 3xₙ) with x₀ = 1.5 to find x₁, x₂ and x₃. Give each to 4 decimal places. [3] (d) Show that the solution is 1.61 correct to 2 decimal places. [2]

11. (non-calculator) (Higher)

(a) Solve x² − 3x − 10 > 0. Give your answer in set notation. [4] (b) A region R on a graph is defined by y > x − 1, y ≤ 4 and x ≥ 1. (i) Which boundary line should be drawn dashed? [1] (ii) Does the point (5, 4) lie in R? Give a reason. [1]

12. (calculator) A right-angled triangle has sides of length x cm, (x + 7) cm and (x + 8) cm. The longest side is the hypotenuse.

(a) Show that x² − 2x − 15 = 0. [3] (b) Solve the equation and work out the perimeter of the triangle. [3] (c) Work out the area of the triangle. [1]

Answers

1. 7x − 14 = 3x + 10 [1] 4x = 24 [1] x = 6 [1] Examiner insight: The first mark is for expanding the bracket correctly; 7x − 2 = 3x + 10 loses it, but correct steps after it can still earn a follow-through mark.

2. 3x − 1 = 4(x − 2) [1] 3x − 1 = 4x − 8, so −1 + 8 = 4x − 3x [1] x = 7 [1] Examiner insight: Multiplying only the x by 4 (giving 3x − 1 = 4x − 2) is a method error, so only the first mark is available.

3. (a) (x − 8)(x + 3) = 0 [1] [1] x = 8 or x = −3 [1] (b) (x − 6)(x + 6) = 0, or x² = 36 [1] x = 6 or x = −6 [1] Examiner insight: In (a), brackets that expand to give two of the three terms correctly usually earn the first mark; in (b), giving only x = 6 loses the final mark.

4. Multiply the second equation by 2: 4x − 10y = 18 [1] Subtract from the first: 13y = −13 [1] y = −1 [1] Substitute: 4x − 3 = 5, so x = 2 [1] Examiner insight: A method mark needs a correct attempt to eliminate one variable, with the operation (add or subtract) consistent with the signs; adding here gives 8x − 7y = 23 and earns nothing.

5. (a) −4x ≥ 12 [1] x ≤ −3 [1] (b) −6 < 3n ≤ 9 [1] −2 < n ≤ 3 [1] n = −1, 0, 1, 2, 3 [1] (c) A closed (filled) circle at −4, an open circle at 1, and a line joining them [1] Examiner insight: In (b), including −2 or leaving out 3 loses the final mark; the inequality signs decide which end values count.

6. (a) Let a coffee cost £c and a muffin £m. 4c + 3m = 13.10 [1] 2c + 5m = 11.10 [1] Double the second: 4c + 10m = 22.20, then subtract: 7m = 9.10 [1] m = 1.30, so a muffin costs £1.30 [1] 4c = 13.10 − 3.90 = 9.20, so a coffee costs £2.30 [1] (b) Let k be the number of muffins: 3 × 2.30 + 1.30k ≤ 20, so 6.90 + 1.30k ≤ 20 [1] 1.30k ≤ 13.10, so k ≤ 10.07… [1] 10 muffins [1] Examiner insight: Money answers need two decimal places (£1.30, not £1.3); in (b) the answer must be a whole number that satisfies the inequality, so 10.07 or 11 loses the final mark.

7. 4x² + 4x − 7 = 0 [1] x = (−4 ± √(4² − 4 × 4 × (−7))) / (2 × 4) [1] x = (−4 ± √128) / 8 [1] x = 0.91 or x = −1.91 [1] Examiner insight: The first mark is for rearranging to = 0; using the formula on 4x² + 4x = 7 with c = 7 is a method error and loses all marks that follow.

8. (a) (x + 4)² [1] − 5, so (x + 4)² − 5 [1] (b) (x + 4)² = 5 [1] x = −4 + √5 or x = −4 − √5 [1] Examiner insight: “Hence” means you must use part (a); solving with the formula instead may not earn the marks, and leaving out the ± loses the final mark.

9. x² − 3x + 5 = 2x − 1 [1] x² − 5x + 6 = 0 [1] (x − 2)(x − 3) = 0 [1] x = 2 or x = 3 [1] x = 2, y = 3 and x = 3, y = 5 [1] Examiner insight: The final mark needs both y-values, each correctly paired with its x; a graphical or trial-and-improvement answer scores nothing when the question says “algebraically”.

10. (a) f(1) = 1 + 3 − 9 = −5 [1] f(2) = 8 + 6 − 9 = 5; the sign changes, so there is a solution between 1 and 2 [1] (b) x³ = 9 − 3x, so x = ∛(9 − 3x) [1] (c) x₁ = 1.6510 [1] x₂ = 1.5936 [1] x₃ = 1.6159 [1] (d) f(1.605) = −0.0505 (negative) [1] f(1.615) = 0.0573 (positive); the sign changes, so the solution is 1.61 to 2 d.p. [1] Examiner insight: In (a) and (d) the conclusion must mention the change of sign; correct values with no statement lose the final mark in each part.

11. (a) (x − 5)(x + 2) = 0 [1] Critical values x = 5 and x = −2 [1] x < −2 or x > 5 [1] {x : x < −2} ∪ {x : x > 5} [1] (b)(i) y = x − 1 (strict inequality) [1] (ii) No: on y = x − 1, at x = 5, y = 4, so (5, 4) is on the dashed line and 4 > 4 is false [1] Examiner insight: In (a), writing x < −2 or x > 5 without set notation does not follow the instruction and loses the final mark; −2 > x > 5 is wrong, because no number satisfies it.

12. (a) x² + (x + 7)² = (x + 8)² [1] x² + x² + 14x + 49 = x² + 16x + 64 [1] x² − 2x − 15 = 0 [1] (b) (x − 5)(x + 3) = 0 [1] x = 5, since x = −3 would give a negative length [1] Sides 5, 12 and 13: perimeter 30 cm [1] (c) ½ × 5 × 12 = 30 cm² [1] Examiner insight: “Show that” in (a) needs both expansions written out; jumping from the Pythagoras line straight to the given equation loses the marks for the missing working.

Where marks are usually lost

• Expanding only part of a bracket, such as 7(x − 2) = 7x − 2.
• Dividing a quadratic by x instead of factorising, which loses the root x = 0.
• Adding equations when the signs need subtracting, or the reverse.
• Forgetting to reverse the inequality sign after dividing by a negative.
• Including a boundary value in a list of integers when the inequality is strict.
• Using the quadratic formula before rearranging the equation to = 0.
• Giving one value of x and no y-values in simultaneous equations.
• Rounding iterates during the calculation instead of using the full calculator value.
• Writing a “show that” with steps missing.
• Not rejecting a negative length in a geometry context.

Next steps

• Solving equations and inequalities revision notes
• Solving equations and inequalities study guide
• Algebra practice questions for the rest of Topic 2
• AQA GCSE Mathematics course hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA GCSE Mathematics (8300) specification, for teaching from September 2015, exams from May/June 2017, version 1.0, published by AQA – section 3.2.3 Solving equations and inequalities (A17 to A22), with the Appendix on mathematical formulae.