OxfordAQA International GCSE Mathematics: Simultaneous equations, linear and linear-quadratic (9260) – 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 statement A21 in section 3.2.3 of the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards). Questions 1 to 6 are Core content, which is also assessable at Extension. Questions 7 to 11 are marked (Extension only) because they involve a linear and a quadratic equation together.

Read the study guide first, or use the revision notes for a quick recap. The course hub and printable checklist list the rest of the course, and all free 10-minute diagnostics help you decide what to practise next.

Questions

1. Solve the simultaneous equations by adding the equations first.

4x + y = 19 2x − y = 5 [3]

2. Solve the simultaneous equations. Show your working.

3x + 4y = 1 5x + 3y = 9 [4]

3. Solve the simultaneous equations using substitution.

y = 5 − 2x 4x − 3y = 25 [3]

4. In a stationery shop, n is the cost in dollars of one notebook and p is the cost in dollars of one pen. The prices satisfy these equations.

4n + 3p = 16.5 2n + 5p = 13.5

Work out the total cost of one notebook and one pen. [4]

5. Here are two equations.

5x + 2y = 10 y = x + 1

(a) Copy and complete the table of values for 5x + 2y = 10.

x	0	2	4
y

[2]

(b) On a grid, draw the line 5x + 2y = 10 using your table, and draw the line y = x + 1. Use your graph to estimate the solution of both equations together. [2]

(c) Solve the equations algebraically, giving your answers as fractions. [3]

6. (a) Explain why 3x + 6y = 10 and x + 2y = 5 have no solution. [2]

(b) Find the value of k for which 2x + 3y = 7 and 4x + 6y = k have infinitely many solutions. [1]

7. (Extension only) Solve y = 3x − 2 and y = x² − 2x + 2. [5]

8. (Extension only) Solve 3x + y = 5 and y = x² − 2x − 7. [5]

9. (Extension only) Solve 2x − y = 1 and y = 2x² − 3x − 2. Give the x-values exactly, and all values of x and y to 3 significant figures. [6]

10. (Extension only) The curve y = x² − 4x + 7 and the line y = 2x + c are drawn, where c is a constant.

(a) Show that the x-coordinates of any intersection satisfy x² − 6x + (7 − c) = 0. [2]

(b) Find the value of c for which the line touches the curve at exactly one point. [2]

(c) Show that the line does not meet the curve when c = −5. [2]

11. (Extension only) The curve y = x² − 4x + 3 meets the line y = x − 1 at points A and B.

(a) Find the coordinates of A and B. [5]

(b) Show that the line y = x − 2 meets the curve at two points. [2]

(c) Explain why the line y = x − 4 does not meet the curve. [2]

Answers

1. Add the equations: 6x = 24 [1]. So x = 4 [1]. Substitute: 16 + y = 19, so y = 3 [1]. Check in the second equation: 8 − 3 = 5 ✓. x = 4, y = 3 [3] Examiner insight: A substitution check in the equation you did not use takes a few seconds and catches a slip in the last step before you move on.

2. Multiply the first equation by 5 and the second by 3: 15x + 20y = 5 and 15x + 9y = 27 [1]. Subtract: 11y = −22 [1]. So y = −2 [1]. Substitute: 3x − 8 = 1, so x = 3 [1]. Check: 15 − 6 = 9 ✓. x = 3, y = −2 [4] Examiner insight: Subtracting 9y from 20y gives 11y, but a common slip is to subtract a negative result wrongly. Writing the subtraction as its own line makes the signs visible and gives the examiner something to credit.

3. Substitute: 4x − 3(5 − 2x) = 25 [1]. Expand and collect: 4x − 15 + 6x = 25, so 10x = 40 [1]. So x = 4 and y = 5 − 8 = −3 [1]. x = 4, y = −3 [3] Examiner insight: The bracket around 5 − 2x matters. Without it, the −3 multiplies only the 5, and the whole solution goes wrong from the first line.

4. Multiply the second equation by 2: 4n + 10p = 27 [1]. Subtract the first equation: 7p = 10.5, so p = 1.5 [1]. Substitute: 4n + 4.5 = 16.5, so n = 3 [1]. Total for one notebook and one pen: 3 + 1.5 = 4.50 dollars [1]. 4.50 dollars [4] Examiner insight: The question asks for a total, not for n and p. Read the last line of a context question again before you write the final answer.

5. (a) At x = 0: 2y = 10, so y = 5 [1]. At x = 2: y = 0, and at x = 4: y = −5 [1].

x	0	2	4
y	5	0	−5

(b) Plot (0, 5), (2, 0), (4, −5) and join them with a ruled line. The second line passes through (0, 1), (1, 2) and (2, 3) [1]. The lines cross at about x = 1.1, y = 2.1 [1]. x ≈ 1.1, y ≈ 2.1 [2]

(c) Substitute y = x + 1: 5x + 2(x + 1) = 10 [1]. So 7x = 8 and x = 8/7 [1]. Then y = 8/7 + 1 = 15/7 [1]. x = 8/7, y = 15/7 [3] Examiner insight: Your graph reading of 1.1 and 2.1 is a useful check on the algebra, since 8/7 is 1.14 and 15/7 is 2.14. A graph reading that is far from the algebra usually signals a plotting slip.

6. (a) Multiplying the second equation by 3 gives 3x + 6y = 15 [1]. This cannot be true at the same time as 3x + 6y = 10, so no pair (x, y) satisfies both. The lines are parallel [1]. (b) Doubling 2x + 3y = 7 gives 4x + 6y = 14, so k = 14 [1]. [3] Examiner insight: Say why there is no solution. The statement “they are parallel” is stronger when it is supported by the contradiction you found.

7. Equate the expressions for y: x² − 2x + 2 = 3x − 2 [1]. Rearrange: x² − 5x + 4 = 0 [1]. Factorise: (x − 1)(x − 4) = 0, so x = 1 or x = 4 [1]. Use the linear equation: y = 3(1) − 2 = 1 [1] and y = 3(4) − 2 = 10 [1]. (1, 1) and (4, 10) [5] Examiner insight: Finding y from the linear equation keeps each pair matched. Quote the pairs as coordinates, so it is clear which y belongs to which x.

8. Rearrange the line: y = 5 − 3x [1]. Equate: x² − 2x − 7 = 5 − 3x [1]. Rearrange: x² + x − 12 = 0 [1]. Factorise: (x + 4)(x − 3) = 0, so x = −4 or x = 3 [1]. Use the linear equation: y = 17 and y = −4 [1]. (−4, 17) and (3, −4) [5] Examiner insight: The sign change when you rearrange 3x + y = 5 is the easy place to slip. Write y = 5 − 3x on its own line before substituting.

9. Rearrange the line: y = 2x − 1 [1]. Equate: 2x² − 3x − 2 = 2x − 1 [1]. Rearrange: 2x² − 5x − 1 = 0 [1]. Formula: x = (5 ± √(25 + 8))/4 = (5 ± √33)/4 [1]. So x = 2.69 or x = −0.186 [1]. Use y = 2x − 1: y = 4.37 and y = −1.37 [1]. x = (5 ± √33)/4: (2.69, 4.37) and (−0.186, −1.37) [6] Examiner insight: Keep the surd form through the working and round only at the end. Work out each y-value from the unrounded x-value so that rounding errors do not build up.

10. (a) Set the equations equal: x² − 4x + 7 = 2x + c [1]. Rearrange: x² − 6x + 7 − c = 0 [1]. (b) One point of contact means b² − 4ac = 0 [1]: 36 − 4(7 − c) = 8 + 4c = 0, so c = −2 [1]. (c) With c = −5 the equation is x² − 6x + 12 = 0, so b² − 4ac = 36 − 48 = −12 [1]. This is negative, so there are no real solutions and the line does not meet the curve [1]. [6] Examiner insight: In a “show that” part, finish with the sentence that links your calculation to the claim. The number −12 alone does not say what it means.

11. (a) Equate: x² − 4x + 3 = x − 1 [1]. Rearrange: x² − 5x + 4 = 0 [1]. Factorise: (x − 1)(x − 4) = 0, so x = 1 or x = 4 [1]. Use y = x − 1: y = 0 and y = 3 [1]. So A(1, 0) and B(4, 3) [1]. (b) Equate: x² − 4x + 3 = x − 2, so x² − 5x + 5 = 0 [1]. Then b² − 4ac = 25 − 20 = 5, which is positive, so there are two intersection points [1]. (c) Equate: x² − 4x + 3 = x − 4, so x² − 5x + 7 = 0 [1]. Then b² − 4ac = 25 − 28 = −3, which is negative, so there are no real solutions and the line does not meet the curve [1]. [9] Examiner insight: Parts (b) and (c) need only the discriminant, not the solutions. State the sign and what it means, rather than solving a quadratic you do not need.

Where marks are usually lost

• Forgetting to multiply the right-hand side when you scale an equation.
• Subtracting one equation from another but changing the sign of only one term.
• Substituting an expression without brackets, as in question 3.
• Stopping after finding x, then not finding y.
• Reading values from a graph without a ruler, or quoting more decimal places than the scale allows.
• Giving a graph reading when the question asks for an algebraic solution.
• Moving a term across an equals sign without changing its sign when rearranging a linear equation.
• Finding y from the quadratic rather than the line, and pairing values wrongly.
• Giving separate lists of x-values and y-values instead of matched coordinates.
• Giving a decimal for an answer that should stay as a surd, or the other way round.

Next steps

• Recap the methods: revision notes
• Full explanations: study guide
• Course hub: OxfordAQA International GCSE Mathematics
• Tick off what you can do: printable checklist
• Find your weak spots with all free 10-minute diagnostics
• Book a free trial class: /trial/

Official syllabus

OxfordAQA International GCSE Mathematics (9260), Version 5.1, for exams May/June 2018 onwards, published by OxfordAQA. This page follows section 3.2.3, statement A21.