OCR GCSE Mathematics: Simultaneous equations, graphical solutions and iteration (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 follow sections 6.03c, 6.03d and 6.03e of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026). Questions 1 to 7 are open to both tiers, apart from part (c) of question 7. Questions 8 to 12 are Higher tier only. Show all working, because the working shows how you reached each value.

Links: OCR GCSE Mathematics hub, printable checklist, study guide, revision notes, 10-minute diagnostics.

Questions

1. Solve 3x + y = 17 and x − y = 3. [3]

2. Solve 3x + 4y = −6 and 2x − 5y = 19. [4]

3. A café sells coffees at £c each and teas at £t each. Four coffees and three teas cost £17.70. Two coffees and five teas cost £16.20. Find the price of a coffee and the price of a tea. [5]

4. Solve y = 3x − 11 and 2x + 5y = −4. [3]

5. Complete a table of values for y = 2x − 3 for x = 0, 1, 2, 3, 4. Draw the graph and use it to solve 2x − 3 = 4. [3]

6. Two lines have equations x + 2y = 7 and y = 3x − 2.

(a) Draw both lines for values of x from 0 to 4 and use your graph to estimate the solution of the pair. [3] (b) Solve the pair algebraically, giving exact values. [3]

7. Here is a table for y = x² − 5x + 2.

x	−1	0	1	2	3	4	5	6
y	8	2	−2		−4	−2		8

(a) Fill in the two missing values. [2] (b) Draw the curve and use it to estimate the roots of x² − 5x + 2 = 0. [2] (c) (Higher tier only) By drawing a suitable line, estimate the solutions of x² − 5x + 2 = 3. [2]

8. (Higher tier only) Solve x² + y² = 20 and x + y = 6. [5]

9. (Higher tier only) A curve has equation y = x² − 6x + 3 and a line has equation y = 2x − 1. Find the coordinates of the points where they meet. Give exact values first, then decimals to 2 decimal places. [5]

10. (Higher tier only) f(x) = x³ − 2x − 6.

(a) Show that f(x) = 0 has a root between 2 and 3. [2] (b) Use a decimal search to find the root to 1 decimal place. [3]

11. (Higher tier only) Use interval bisection on f(x) = x³ + 2x − 5, starting with the interval 1 to 2, to find the root of f(x) = 0 to 1 decimal place. Show each midpoint and the sign of f. [6]

12. (Higher tier only) Curve C has equation y = x² + x − 6.

(a) Find the points where C meets the line y = 3x + 2. [4] (b) Show that the line y = 3x − 7 touches C at exactly one point and find that point. [3] (c) Explain why the line y = 3x − 8 does not meet C. [2]

Answers

1. Add the equations: 4x = 20 [1], so x = 5 [1]. Then y = 5 − 3 = 2 [1]. x = 5, y = 2 [3] Examiner insight: A check in the other equation (15 + 2 = 17) is a quick way to catch a sign error before you leave the question.

2. Multiply the first by 2 and the second by 3: 6x + 8y = −12 and 6x − 15y = 57 [1]. Subtract: 23y = −69 [1], so y = −3 [1]. Then 2x + 15 = 19, so x = 2 [1]. x = 2, y = −3 [4] Examiner insight: When one coefficient is negative, subtracting 57 from −12 is where slips happen, so write the subtraction as its own line.

3. 4c + 3t = 17.70 [1] and 2c + 5t = 16.20 [1]. Double the second: 4c + 10t = 32.40 [1]. Subtract: 7t = 14.70, so t = 2.10 [1]. Then 4c = 17.70 − 6.30 = 11.40, so c = 2.85 [1]. Coffee £2.85, tea £2.10 [5] Examiner insight: State what each letter stands for and give the answer with the pound sign, because this is a context question where the answer is a price rather than a bare number.

4. Substitute: 2x + 5(3x − 11) = −4 [1]. This gives 17x − 55 = −4, so 17x = 51 and x = 3 [1]. Then y = 9 − 11 = −2 [1]. x = 3, y = −2 [3] Examiner insight: Put the substituted expression in brackets. Writing 2x + 5 × 3x − 11 is a common way to lose the whole method.

5. The values are −3, −1, 1, 3, 5 [1]. Plot the points and draw a straight line with a ruler [1]. Draw y = 4 and read across: x = 3.5 [1] [3] Examiner insight: Show the horizontal line on your graph. A bare reading is harder to credit than one that visibly comes from the line y = 4.

6. (a) Points on x + 2y = 7 include (1, 3) and (3, 2); points on y = 3x − 2 include (0, −2) and (2, 4) [1]. Both lines drawn accurately [1]. They cross near (1.6, 2.7) [1]. (b) Substitute: x + 2(3x − 2) = 7 [1]. So 7x = 11 [1]. x = 11/7, y = 19/7 [1]. [6] Examiner insight: Your graph estimate should agree with the algebra to the accuracy of the grid (11/7 is about 1.57). If it does not, recheck the plotting.

7. (a) At x = 2, y = −4 [1]. At x = 5, y = 2 [1]. (b) A smooth curve through all points [1]. It crosses the x-axis near x ≈ 0.4 and x ≈ 4.6 [1]. (c) Draw the line y = 3 across the graph [1]. The crossings are near x ≈ −0.2 and x ≈ 5.2 [1]. [6] Examiner insight: A curve drawn as straight segments, or with a flat bottom, suggests careless plotting and makes a root reading unreliable. Draw one smooth curve.

8. y = 6 − x [1]. x² + (6 − x)² = 20 [1]. This gives 2x² − 12x + 16 = 0, so x² − 6x + 8 = 0 [1]. This factorises as (x − 2)(x − 4), giving x = 2 and x = 4 [1]. Using y = 6 − x: (2, 4) and (4, 2) [1] [5] Examiner insight: Each value of x belongs with one value of y. Write the final answer as two coordinate pairs.

9. Equate: x² − 6x + 3 = 2x − 1 [1]. So x² − 8x + 4 = 0 [1]. The quadratic formula gives x = (8 ± √48)/2 = 4 ± 2√3 [1]. Then y = 2x − 1 = 7 ± 4√3 [1]. The points are (4 + 2√3, 7 + 4√3) ≈ (7.46, 13.93) and (4 − 2√3, 7 − 4√3) ≈ (0.54, 0.07) [1] [5] Examiner insight: Keep surds until the final step. Rounding x early and then doubling it can shift the second decimal place in y.

10. (a) f(2) = 8 − 4 − 6 = −2 [1] and f(3) = 27 − 6 − 6 = 15. The signs differ and the graph of f has no breaks, so there is a root between 2 and 3 [1]. (b) f(2.1) = −0.939 and f(2.2) = 0.248, so the root is between 2.1 and 2.2 [1]. f(2.15) = −0.362, which is negative, so the root is between 2.15 and 2.2 [1]. The root is 2.2 to 1 decimal place [1]. [5] Examiner insight: For a rounded answer, testing the half-way value is the step that proves which way the root rounds. Without it, 2.1 and 2.2 are both still possible.

11. f(1) = −2 and f(2) = 7, so the start is a sign change [1].

Interval	Midpoint	f(midpoint)	Keep
1 to 2	1.5	1.375	1 to 1.5
1 to 1.5	1.25	−0.547	1.25 to 1.5
1.25 to 1.5	1.375	0.350	1.25 to 1.375
1.25 to 1.375	1.3125	−0.114	1.3125 to 1.375
1.3125 to 1.375	1.34375	0.114	1.3125 to 1.34375

First two midpoints correct [1]. Third midpoint and interval [1]. Fourth midpoint and interval [1]. Fifth midpoint and interval [1]. Both ends of the final interval round to 1.3, so the root is 1.3 to 1 decimal place [1] [6] Examiner insight: Keep the ends of the interval and the sign of f in a table. If a later arithmetic slip happens, a table shows exactly which line it started in.

12. (a) x² + x − 6 = 3x + 2 [1]. So x² − 2x − 8 = 0 [1], giving (x − 4)(x + 2) = 0, so x = 4 or −2 [1]. Using y = 3x + 2: (4, 14) and (−2, −4) [1]. (b) x² + x − 6 = 3x − 7 gives x² − 2x + 1 = 0 [1]. The left side is the perfect square (x − 1)², so x = 1 is the only solution [1]. One solution means one point of contact: (1, −4) [1]. (c) x² + x − 6 = 3x − 8 gives x² − 2x + 2 = 0 [1]. The discriminant is 4 − 8 = −4, which is negative, so there are no real solutions and the line does not meet C [1]. [9] Examiner insight: For a “show that” or “explain why” part, state the equation you get and what it implies about the number of solutions, not only the final conclusion.

Where marks are usually lost

• Not multiplying every term when scaling an equation, especially the right-hand side.
• Choosing to add or subtract the wrong way round, given the signs.
• Giving only one of the two unknowns.
• Skipping the check in the second equation.
• Substituting without brackets, as in 2x + 5 × 3x − 11.
• Giving the x-values and y-values of a quadratic pair as separate lists.
• Using the quadratic equation to find y, which creates false pairs.
• Drawing a curve in segments, or a line without a ruler, and then trusting a reading from it.
• Showing only one sign in a sign-change argument.
• Rounding a root without testing the half-way value.

Next steps

• Revisit the revision notes for the methods in brief.
• Read the study guide for full explanations.
• Open the OCR GCSE Mathematics hub and the checklist.
• 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, published by OCR.