OCR GCSE Mathematics: Solving linear and quadratic equations (J560) – 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 sections 6.03a and 6.03b of the OCR GCSE (9-1) Mathematics specification, Version 2.1 (August 2026): solving linear and quadratic equations. Questions 1 to 6 suit both tiers. Questions 7 to 12 are labelled “(Higher tier only)” because they use the quadratic formula, rearranging, completing the square or algebraic fractions. Learn the methods from the study guide or the revision notes. Course pages: the OCR GCSE Mathematics hub, the checklist and the free 10-minute diagnostics.

Questions

1. Solve each equation.

(a) 5x + 7 = 32 [2] (b) 8 − 3x = −10 [2] (c) 4(x − 3) = 2x + 9 [3]

2. Solve each equation. Show your working.

(a) 7x − 4 = 3(x + 8) [3] (b) 2(3x + 1) − 4(x − 2) = 20 [3]

3. Mia is x years old. Her brother is 5 years older than Mia. Their father is three times as old as Mia. The three ages add up to 80.

(a) Show that 5x + 5 = 80. [2] (b) Work out the age of Mia’s father. [2]

4. Firm P hires out a bicycle for £15 plus £6 per hour. Firm Q charges £9 per hour with no other charge.

(a) Form and solve an equation to find the number of hours for which the two firms charge the same. [3] (b) Tom wants a bicycle for 4 hours and Jo wants one for 7 hours. Which firm is cheaper for each of them? Show your calculations. [2]

5. Solve each equation by factorising.

(a) x² − 7x + 10 = 0 [3] (b) x² + 2x − 24 = 0 [3] (c) x² − 81 = 0 [2] (d) x² − 5x = 0 [3]

6. A rectangle has width x cm and length (x + 4) cm. Its area is 96 cm².

(a) Show that x² + 4x − 96 = 0. [2] (b) Solve the equation to find the value of x. Explain why one solution is rejected. [3] (c) Work out the perimeter of the rectangle. [1]

7. (Higher tier only) Solve (x + 3)(x − 2) = 24. [4]

8. (Higher tier only) Solve 2x² − 5x − 12 = 0 by factorising. [4]

9. (Higher tier only) Solve 2x² − 9x + 5 = 0. Give your answers correct to 3 significant figures. [4]

10. (Higher tier only)

(a) Complete the square for x² − 12x + 13, giving it as (x − a)² − b. [2] (b) Use your result to solve x² − 12x + 13 = 0. Write the solutions as p ± √q. [2]

11. (Higher tier only) Solve 6/x − 2/(x + 2) = 1. Give your answers correct to 3 significant figures. [6]

12. (Higher tier only) Solve (3x − 1)/4 − (x + 2)/3 = 1. [4]

Answers

1. (a) 5x = 25 [1]; x = 5 [1] (b) −3x = −18 [1]; x = 6 [1] (c) 4x − 12 = 2x + 9 [1]; 2x = 21 [1]; x = 10.5 [1] [7] Examiner insight: Show the expanded line in part (c). Subtracting 2x from both sides before moving the 12 keeps the working short and makes slips easy to spot.

2. (a) 7x − 4 = 3x + 24 [1]; 4x = 28 [1]; x = 7 [1] (b) 6x + 2 − 4x + 8 = 20 [1]; 2x + 10 = 20 [1]; x = 5 [1] [6] Examiner insight: In part (b) the second bracket is multiplied by −4, so both terms change sign. A wrong sign here usually makes the final answer wrong, so check by substituting.

3. (a) Brother is x + 5 and father is 3x, so x + (x + 5) + 3x = 80 [1]; collecting terms gives 5x + 5 = 80 [1] (b) 5x = 75, so x = 15 [1]; father is 3 × 15 = 45 years [1] [4] Examiner insight: On a “show that” question, write the unsimplified equation first and then the simplified one. Jumping straight to 5x + 5 = 80 leaves no working to assess.

4. (a) 15 + 6h = 9h [1]; 15 = 3h [1]; h = 5 hours [1] (b) Tom, 4 hours: P costs £39 and Q costs £36, so Q is cheaper [1]. Jo, 7 hours: P costs £57 and Q costs £63, so P is cheaper [1] [5] Examiner insight: The answer 5 hours is where the costs are equal, not where one firm wins. State a conclusion for each person, with both costs shown.

5. (a) (x − 2)(x − 5) = 0 [1]; x = 2 [1]; x = 5 [1] (b) (x + 6)(x − 4) = 0 [1]; x = −6 [1]; x = 4 [1] (c) x² = 81 or (x − 9)(x + 9) = 0 [1]; x = 9 or x = −9 [1] (d) x(x − 5) = 0 [1]; x = 0 [1]; x = 5 [1] [11] Examiner insight: Part (c) has two solutions, and part (d) has the solution x = 0. Dividing by x in part (d) removes it, so factorise instead.

6. (a) Area = x(x + 4) = 96 [1]; expanding gives x² + 4x = 96, so x² + 4x − 96 = 0 [1] (b) (x + 12)(x − 8) = 0 [1]; x = −12 or x = 8 [1]; a width cannot be negative, so x = 8 [1] (c) The sides are 8 cm and 12 cm, so the perimeter is 40 cm [1] [6] Examiner insight: Give the reason for rejecting −12 in words that refer to the context. “It is negative” alone is weaker than “a width cannot be negative”.

7. Expand: x² + x − 6 = 24 [1]; rearrange: x² + x − 30 = 0 [1]; (x + 6)(x − 5) = 0 [1]; x = −6 or x = 5 [1] [4] Examiner insight: A common slip is to set each bracket equal to 24. The product must equal 0 before you split it, so rearrange first.

8. The numbers with product −24 and sum −5 are −8 and +3, so 2x² − 8x + 3x − 12 = 0 [1]; 2x(x − 4) + 3(x − 4) = 0 [1]; (2x + 3)(x − 4) = 0 [1]; x = −3/2 or x = 4 [1] [4] Examiner insight: Expand your brackets back out to check them. It takes a few seconds and confirms the middle term.

9. a = 2, b = −9, c = 5, so x = (9 ± √(81 − 40)) / 4 [1]; b² − 4ac = 41 [1]; x = (9 ± √41) / 4 [1]; x = 3.85 or x = 0.649 [1] [4] Examiner insight: Write the formula and the substituted values before evaluating. Keep the brackets around −9 when squaring so the sign cannot be lost.

10. (a) (x − 6)² − 36 + 13 [1]; (x − 6)² − 23 [1] (b) (x − 6)² = 23, so x − 6 = ±√23 [1]; x = 6 ± √23 [1] [4] Examiner insight: Keep ± when you take the square root, and write the solutions in the form requested. Since 23 has no square factor, √23 cannot be simplified.

11. Multiply every term by x(x + 2), with x not 0 or −2 [1]; 6(x + 2) − 2x = x(x + 2) [1]; 4x + 12 = x² + 2x [1]; x² − 2x − 12 = 0 [1]; x = (2 ± √52) / 2 = 1 ± √13 [1]; x = 4.61 or x = −2.61 [1] [6] Examiner insight: Multiply the right-hand side 1 by the common denominator too. Leaving it out gives an equation that no longer matches the original.

12. Multiply every term by 12: 3(3x − 1) − 4(x + 2) = 12 [1]; 9x − 3 − 4x − 8 = 12 [1]; 5x = 23 [1]; x = 23/5 (4.6) [1] [4] Examiner insight: The minus sign in front of the second fraction applies to the whole bracket (x + 2). Write the bracket explicitly before expanding.

Where marks are usually lost

• Expanding brackets with a negative multiplier and changing only the first term.
• Giving only one root of a quadratic, or losing x = 0 after dividing by x.
• Not setting the right-hand side to 0 before factorising.
• Leaving a negative length as an answer, or failing to say why it is rejected.
• Stopping at the value that makes two costs equal without answering the question.
• Writing the quadratic formula with the wrong sign on −b.
• Rounding before the end, or writing a decimal where an exact form was requested.
• Forgetting to multiply the whole-number term by the common denominator.
• Not substituting the answer back into the original equation to catch a slip, especially when negative numbers are involved.
• Mixing up the signs in a bracket and its root, for example writing x = 2 from the bracket (x + 2).
• Ignoring the accuracy instruction in the question: exact form, fraction, or 3 significant figures.

Next steps

• Review the revision notes and the study guide.
• Tick off the topic on the OCR GCSE Mathematics checklist and visit the course hub.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, GCSE (9-1) Mathematics (J560), Version 2.1 (August 2026), for first assessment in 2017, sections 6.03a and 6.03b.