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0580/43 Oct/Nov 2021, Question 4: Simultaneous Equations by Elimination

Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 15 marks

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The question, in short

Solve a set of algebra problems: simultaneous equations, a fractional equation, a double inequality, factorising, and simplifying fractions.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 3 marks

Solve 2p − q = 7 and 3p + 2q = 7 simultaneously, showing all working.

  1. 1Double the first: 4p − 2q = 14
  2. 2Add to 3p + 2q = 7: 7p = 21, so p = 3
  3. 32 × 3 − q = 7, so q = −1
  4. 4p = 3, q = −1

Answerp = 3, q = −1

Part (b) · 2 marks

Solve x/4 + 2x/3 = 1 for x.

  1. 13x/12 + 8x/12 = 1
  2. 211x = 12
  3. 3x = 12/11 = 1 1/11

Answer1 1/11 or 12/11 or 1.09 or 1.090 to 1.091

Part (ci) · 3 marks

Solve the double inequality −8 < 3x − 2 ≤ 7 for x.

  1. 1−8 + 2 < 3x ≤ 7 + 2
  2. 2−6 < 3x ≤ 9
  3. 3−2 < x ≤ 3

Answer−2 < x ≤ 3

Part (cii) · 1 mark

List the integer values of x satisfying −8 < 3x − 2 ≤ 7.

  1. 1Integers with −2 < x ≤ 3: −1, 0, 1, 2, 3

Answer−1, 0, 1, 2, 3

Part (d) · 2 marks

Factorise 16a − 4a² completely.

  1. 1Both terms share 4 and a, so the HCF is 4a.
  2. 216a ÷ 4a = 4 and 4a² ÷ 4a = a
  3. 316a − 4a² = 4a(4 − a)

Answer4a(4 − a)

Part (ei) · 2 marks

Write 1/(2a) ÷ 3/(4b) as a single simplified fraction.

  1. 11/(2a) × 4b/3
  2. 2= 4b/(6a) = (2b)/(3a)

Answer(2b)/(3a)

Part (eii) · 2 marks

Write 2 − x/(x − 1) as a single simplified fraction.

  1. 12 − x/(x − 1) = (2(x − 1) − x)/(x − 1)
  2. 2= (2x − 2 − x)/(x − 1) = (x − 2)/(x − 1)

Answer(x − 2)/(x − 1)

Stuck on a step? Read the free lesson: Simultaneous Equations by Elimination

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Also searched as: 0580/43/O/N/21 · 0580_w21_qp_43 · 0580 w21 43 · question 4.
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