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

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

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

This page collects every linear equations question from the Pearson Edexcel International GCSE Mathematics A (4MA1) Higher tier Paper 1H and Paper 1HR, January 2020 to June 2025: 10 questions worth 58 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.

13 marksHigherJanuary 2023 · Paper 1H (4MA1/1H) · Q2

Solve 3(2 − 4x) = 5 − 8x. Show clear algebraic working.

Show solution
  1. Expand: 6 − 12x = 5 − 8x
  2. Add 12x to both sides and subtract 5: 1 = 4x

Answer: x = 0.25

26 marksHigherJune 2020 · Paper 1HR (4MA1/1HR) · Q3

(a) Simplify h7 × h2 (1)

G = c2 − 4c

(b) Find the value of G when c = −5 (2)

(c) Solve 5x − 3⁄4 = 2x + 3. Show clear algebraic working. (3)

Show solution

Part (a)

  1. Add the indices: 7 + 2 = 9

Answer: h9

Part (b)

  1. Substitute c = −5, using brackets: (−5)2 − 4 × (−5) = 25 + 20

Answer: G = 45

Part (c)

  1. Multiply both sides by 4: 5x − 3 = 4(2x + 3) = 8x + 12
  2. Subtract 5x and 12 from both sides: −15 = 3x

Answer: x = −5

35 marksHigherJanuary 2021 · Paper 1H (4MA1/1H) · Q5

(a) Factorise fully 15y4 + 20uy3 (2)

(b) Solve 4 − 3x = 5 − 8x⁄4. Show clear algebraic working. (3)

Show solution

Part (a)

  1. HCF of 15 and 20 is 5; lowest power of y is y3. (u is only in one term.)

Answer: 5y3(3y + 4u)

Part (b)

  1. Multiply both sides by 4: 16 − 12x = 5 − 8x
  2. Add 12x and subtract 5: 11 = 4x

Answer: x = 2.75

45 marksHigherJune 2022 · Paper 1H (4MA1/1H) · Q5

(a) Expand and simplify (n − 6)(n + 4) (2)

(b) Solve 2x − 3 = 3x − 5⁄4. Show clear algebraic working. (3)

Show solution

Part (a)

  1. n × n = n2, n × 4 = 4n, −6 × n = −6n, −6 × 4 = −24
  2. 4n − 6n = −2n

Answer: n2 − 2n − 24

Part (b)

  1. Multiply both sides by 4: 4(2x − 3) = 3x − 5, so 8x − 12 = 3x − 5
  2. Subtract 3x and add 12: 5x = 7

Answer: x = 1.4

55 marksHigherJune 2024 · Paper 1H (4MA1/1H) · Q6

(a) Expand and simplify (m + 5)(m − 8) (2)

(b) Solve 3n − 4 = 5n + 6⁄3. Show clear algebraic working. (3)

Show solution

Part (a)

  1. Multiply every term in the first bracket by every term in the second: m × m = m2, m × (−8) = −8m, 5 × m = 5m, 5 × (−8) = −40
  2. m2 − 8m + 5m − 40
  3. Collect the like terms: −8m + 5m = −3m

Answer: m2 − 3m − 40

Part (b)

  1. Remove the fraction by multiplying both sides by 3: 3(3n − 4) = 5n + 6
  2. Expand the bracket: 9n − 12 = 5n + 6
  3. Subtract 5n from both sides: 4n − 12 = 6
  4. Add 12 to both sides: 4n = 18
  5. Divide by 4: n = 18⁄4

Answer: n = 4.5

66 marksHigherJune 2025 · Paper 1H (4MA1/1H) · Q6

(a) Solve x − 4 = 3 + 2x⁄6. Show clear algebraic working. (3)

(b)(i) Factorise y2 − 11y + 30 (2)

(ii) Hence solve y2 − 11y + 30 = 0 (1)

Show solution

Part (a)

  1. Multiply both sides by 6 to clear the fraction: 6(x − 4) = 3 + 2x
  2. Expand: 6x − 24 = 3 + 2x
  3. Subtract 2x from both sides and add 24: 4x = 27

Answer: x = 6.75

Part (b)(i)

  1. Find two numbers that multiply to +30 and add to −11: −5 and −6.

Answer: (y − 5)(y − 6)

Part (b)(ii)

  1. One of the brackets must equal zero.

Answer: y = 5 or y = 6

79 marksHigherJanuary 2020 · Paper 1H (4MA1/1H) · Q7

(a) Expand and simplify (m − 8)(m + 5) (2)

(b) Factorise fully 5y + 20y2 (2)

(c) Simplify (p2 + 3)0 (1)

(d) Solve 3(2x − 5) = 9 − x⁄2. Show clear algebraic working. (4)

Show solution

Part (a)

  1. m2 + 5m − 8m − 40

Answer: m2 − 3m − 40

Part (b)

  1. Both terms share 5 and y.

Answer: 5y(1 + 4y)

Part (c)

  1. Anything (non-zero) to the power 0 is 1. p2 + 3 is always at least 3, so it is never 0.

Answer: 1

Part (d)

  1. Expand: 6x − 15 = 9 − x⁄2
  2. Multiply both sides by 2: 12x − 30 = 9 − x
  3. Add x and 30 to both sides: 13x = 39

Answer: x = 3

88 marksHigherJune 2020 · Paper 1H (4MA1/1H) · Q7

(a) Solve 5(4 − x) = 7 − 3x. Show clear algebraic working. (3)

(b) Factorise fully 16m3g3 + 24m2g5 (2)

(c)(i) Factorise y2 − 2y − 48 (2)

(ii) Hence, solve y2 − 2y − 48 = 0 (1)

Show solution

Part (a)

  1. Expand: 20 − 5x = 7 − 3x
  2. Add 5x to both sides and subtract 7: 13 = 2x

Answer: x = 6.5

Part (b)

  1. HCF of 16 and 24 is 8; lowest powers are m2 and g3.
  2. 16m3g3 ÷ 8m2g3 = 2m and 24m2g5 ÷ 8m2g3 = 3g2

Answer: 8m2g3(2m + 3g2)

Part (c)(i)

  1. Two numbers that multiply to −48 and add to −2: −8 and +6.

Answer: (y − 8)(y + 6)

Part (c)(ii)

  1. Set each bracket to zero.

Answer: y = 8 or y = −6

95 marksHigherJanuary 2020 · Paper 1HR (4MA1/1HR) · Q10

The diagram shows a regular hexagon ABCDEF with sides (x − 1) cm, and an isosceles triangle GHJ with GJ = HJ = (x + 5) cm and GH = (2x − 3) cm. The perimeter of the hexagon is equal to the perimeter of the triangle. Find the length of each side of the hexagon. Show clear algebraic working.

Show solution
  1. Perimeter of the hexagon = 6(x − 1) = 6x − 6
  2. Perimeter of the triangle = (x + 5) + (x + 5) + (2x − 3) = 4x + 7
  3. 6x − 6 = 4x + 7, so 2x = 13 and x = 6.5
  4. Side of the hexagon = x − 1 = 6.5 − 1

Answer: 5.5 cm

106 marksHigherJune 2025 · Paper 1HR (4MA1/1HR) · Q15

(a) Solve 5a + 8⁄3 − 2a + 5⁄4 = 23. Show clear algebraic working. (4)

(b) Express (√y⁄3)−1 in the form cyn where c and n are numbers to be found. (2)

Show solution

Part (a)

  1. The lowest common multiple of 3 and 4 is 12, so multiply every term by 12: 4(5a + 8) − 3(2a + 5) = 276
  2. Expand, taking care with the minus sign: 20a + 32 − 6a − 15 = 276
  3. Simplify: 14a + 17 = 276, so 14a = 259

Answer: a = 18.5

Part (b)

  1. A power of −1 flips the fraction: (√y⁄3)−1 = 3⁄√y
  2. √y = y½, and dividing by y½ is the same as multiplying by y−½.

Answer: 3y−½, so c = 3 and n = −½

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