Edexcel International GCSE Maths Simultaneous Equations Past Paper Questions and Solutions (4MA1 Higher)

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Edexcel International GCSE Maths A (4MA1) · Higher · Equations, Formulae and Identities

Simultaneous Equations: Edexcel 4MA1 Past Paper Questions and Worked Solutions

This page collects every simultaneous equations question from the Pearson Edexcel International GCSE Mathematics A (4MA1) Higher tier Paper 1H and Paper 1HR, January 2020 to June 2025: 14 questions worth 62 marks. Questions run from easier to harder. Try each one first, then click “Show solution” for a step-by-step method with the reason for every step. All answers are checked against the official Pearson mark schemes.

15 marksHigherNovember 2021 · Paper 1H (4MA1/1H) · Q5

Yasmin has some identical rectangular tiles. Each tile is L cm by W cm. Using 9 of her tiles, Yasmin makes rectangle ABCD: the top row has 5 tiles standing upright and the bottom row has 4 tiles lying on their sides, filling the same width. The area of ABCD is 1620 cm2. Work out the value of L and the value of W.

ABCD
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  1. The top row is 5 tile-widths across and the bottom row is 4 tile-lengths across. They are the same width, so 5W = 4L, which gives L = 5⁄4W.
  2. The rectangle is made of 9 tiles, each with area LW: 9LW = 1620, so LW = 180
  3. Substitute: 5⁄4W × W = 180, so W2 = 144 and W = 12
  4. L = 5⁄4 × 12 = 15

Answer: L = 15, W = 12

25 marksHigherNovember 2021 · Paper 1H (4MA1/1H) · Q6

Alison buys 5 apples and 3 pears for a total cost of 1.96. Greg buys 3 apples and 2 pears for a total cost of 1.22. Michael buys 10 apples and 10 pears. Work out how much Michael pays for his 10 apples and 10 pears. Show your working clearly.

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  1. Let an apple cost a cents and a pear p cents: 5a + 3p = 196 and 3a + 2p = 122
  2. Make the p terms equal: multiply the first by 2 and the second by 3: 10a + 6p = 392 and 9a + 6p = 366
  3. Subtract: a = 26
  4. Substitute: 5(26) + 3p = 196, so 3p = 66 and p = 22
  5. 10 apples and 10 pears = 10 × 26 + 10 × 22 = 260 + 220 = 480 cents

Answer: $4.80

33 marksHigherJanuary 2023 · Paper 1HR (4MA1/1HR) · Q6

Solve the simultaneous equations x + 2y = 15 and 4x − 6y = 4. Show clear algebraic working.

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  1. Multiply the first equation by 4: 4x + 8y = 60
  2. Subtract the second equation: 8y − (−6y) = 60 − 4, so 14y = 56 and y = 4
  3. Substitute: x + 8 = 15, so x = 7
Tip: Check in the second equation: 4 × 7 − 6 × 4 = 28 − 24 = 4 ✓

Answer: x = 7, y = 4

43 marksHigherJune 2023 · Paper 1H (4MA1/1H) · Q9

Solve the simultaneous equations 2x + 9y = 14.5 and 7x + 3y = 8. Show clear algebraic working.

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  1. Make the y coefficients equal: multiply the second equation by 3, giving 21x + 9y = 24
  2. Subtract the first equation: (21x + 9y) − (2x + 9y) = 24 − 14.5, so 19x = 9.5 and x = 0.5
  3. Substitute into 7x + 3y = 8: 3.5 + 3y = 8, so 3y = 4.5 and y = 1.5

Answer: x = 0.5, y = 1.5

54 marksHigherJune 2020 · Paper 1H (4MA1/1H) · Q12

Solve the simultaneous equations 7x − 2y = 34 and 3x + 5y = −3. Show clear algebraic working.

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  1. Make the y coefficients opposites: multiply the first by 5 and the second by 2: 35x − 10y = 170 and 6x + 10y = −6
  2. Add: 41x = 164, so x = 4
  3. Substitute: 7(4) − 2y = 34, so −2y = 6 and y = −3

Answer: x = 4, y = −3

64 marksHigherJune 2020 · Paper 1HR (4MA1/1HR) · Q12

Solve the simultaneous equations 7x + 2y = 5.5 and 3x − 5y = 17. Show clear algebraic working.

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  1. Multiply the first by 5 and the second by 2 so the y terms are opposites: 35x + 10y = 27.5 and 6x − 10y = 34
  2. Add: 41x = 61.5, so x = 1.5
  3. Substitute: 7(1.5) + 2y = 5.5, so 2y = −5 and y = −2.5

Answer: x = 1.5, y = −2.5

75 marksHigherJune 2023 · Paper 1HR (4MA1/1HR) · Q12

Moeen has a biased 6-sided dice. The table gives the probability that, when the dice is thrown, it will land on each number.

Number123456
Probabilityx0.150.5y0.130.03

(a) Show that x + y = 0.19 (2)

Given that 3x − y = 0.09 and x + y = 0.19

(b) work out the value of x and the value of y. Show clear algebraic working. (3)

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Part (a)

  1. The probabilities add up to 1: x + 0.15 + 0.5 + y + 0.13 + 0.03 = 1
  2. 0.15 + 0.5 + 0.13 + 0.03 = 0.81, so x + y = 1 − 0.81

Answer: x + y = 0.19

Part (b)

  1. Add the two equations so that y cancels: (3x − y) + (x + y) = 0.09 + 0.19, so 4x = 0.28
  2. x = 0.07
  3. Substitute into x + y = 0.19: y = 0.19 − 0.07 = 0.12

Answer: x = 0.07, y = 0.12

84 marksHigherJune 2024 · Paper 1HR (4MA1/1HR) · Q12

Solve the simultaneous equations 4x + 3y = 9.6 and 6x + 5y = 16.8. Show clear algebraic working.

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  1. Make the x coefficients the same. Multiply the first equation by 3 and the second by 2: 12x + 9y = 28.8 and 12x + 10y = 33.6
  2. Subtract the first new equation from the second to eliminate x: y = 4.8
  3. Substitute y = 4.8 into 4x + 3y = 9.6: 4x + 14.4 = 9.6, so 4x = −4.8
  4. x = −1.2
  5. Check in the second equation: 6(−1.2) + 5(4.8) = −7.2 + 24 = 16.8 ✓

Answer: x = −1.2, y = 4.8

94 marksHigherJune 2022 · Paper 1HR (4MA1/1HR) · Q14

Solve the simultaneous equations 3x − 5y = 25 and 4x + 3y = 14. Show clear algebraic working.

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  1. Make the y coefficients equal and opposite: multiply the first equation by 3 and the second by 5. 9x − 15y = 75 and 20x + 15y = 70
  2. Add to eliminate y: 29x = 145, so x = 5
  3. Substitute into 4x + 3y = 14: 20 + 3y = 14, so 3y = −6 and y = −2
  4. Check: 3(5) − 5(−2) = 15 + 10 = 25 ✓

Answer: x = 5, y = −2

105 marksHigherJanuary 2021 · Paper 1HR (4MA1/1HR) · Q17

Solve the simultaneous equations x − 6y = 5 and xy − 2y2 = 6. Show clear algebraic working.

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  1. From the linear equation: x = 5 + 6y
  2. Substitute: (5 + 6y)y − 2y2 = 6, so 5y + 6y2 − 2y2 = 6
  3. 4y2 + 5y − 6 = 0, which factorises as (4y − 3)(y + 2) = 0
  4. y = 0.75 gives x = 5 + 4.5 = 9.5. y = −2 gives x = 5 − 12 = −7

Answer: x = 9.5, y = 0.75 and x = −7, y = −2

115 marksHigherJanuary 2021 · Paper 1H (4MA1/1H) · Q19

Solve the simultaneous equations x2 − 9y − x = 2y2 − 12 and x + 2y − 1 = 0. Show clear algebraic working.

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  1. From the linear equation: x = 1 − 2y
  2. Substitute: (1 − 2y)2 − 9y − (1 − 2y) = 2y2 − 12
  3. Expand: 1 − 4y + 4y2 − 9y − 1 + 2y = 2y2 − 12, so 4y2 − 11y = 2y2 − 12
  4. 2y2 − 11y + 12 = 0, which factorises as (2y − 3)(y − 4) = 0
  5. y = 1.5 gives x = 1 − 3 = −2. y = 4 gives x = 1 − 8 = −7

Answer: x = −2, y = 1.5 and x = −7, y = 4

125 marksHigherJanuary 2022 · Paper 1H (4MA1/1H) · Q19

Solve the simultaneous equations 3x2 + y2 − xy = 5 and y = 2x − 3. Show clear algebraic working.

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  1. Substitute: 3x2 + (2x − 3)2 − x(2x − 3) = 5
  2. Expand: 3x2 + 4x2 − 12x + 9 − 2x2 + 3x = 5
  3. Simplify: 5x2 − 9x + 4 = 0, which factorises as (5x − 4)(x − 1) = 0
  4. x = 1 gives y = −1. x = 0.8 gives y = 1.6 − 3 = −1.4

Answer: x = 1, y = −1 and x = 0.8, y = −1.4

135 marksHigherJune 2025 · Paper 1H (4MA1/1H) · Q22

Solve the simultaneous equations x2 + 3y + y2 = 7 and y = x + 2. Show clear algebraic working.

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  1. Substitute y = x + 2 into the first equation: x2 + 3(x + 2) + (x + 2)2 = 7
  2. Expand: x2 + 3x + 6 + x2 + 4x + 4 = 7
  3. Simplify: 2x2 + 7x + 10 = 7, so 2x2 + 7x + 3 = 0
  4. Factorise: (2x + 1)(x + 3) = 0, so x = −0.5 or x = −3
  5. Find each y from y = x + 2: x = −0.5 gives y = 1.5; x = −3 gives y = −1

Answer: x = −0.5, y = 1.5 and x = −3, y = −1

145 marksHigherJune 2025 · Paper 1HR (4MA1/1HR) · Q22

Solve the simultaneous equations x2 + y2 = 41 and 2x + y = 3. Show clear algebraic working.

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  1. Rearrange the linear equation: y = 3 − 2x
  2. Substitute into the circle equation: x2 + (3 − 2x)2 = 41
  3. Expand: x2 + 9 − 12x + 4x2 = 41, so 5x2 − 12x − 32 = 0
  4. Factorise: (5x + 8)(x − 4) = 0, so x = 4 or x = −1.6
  5. x = 4: y = 3 − 8 = −5. x = −1.6: y = 3 + 3.2 = 6.2

Answer: x = 4, y = −5 and x = −1.6, y = 6.2

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