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4024/22 May/Jun 2020, Question 5: Simultaneous Equations by Elimination

Cambridge O Level Mathematics 4024 · Paper 22 (Calculator) · Calculator · 12 marks

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The question, in short

Solve a pair of linear simultaneous equations, a quadratic equation to 2 decimal places, and work with inverse proportionality involving a cube, finding a formula and a value.

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

Part (a) · 4 marks

Find x and y satisfying 2x − 4y = 11 and 3x + 3y = −6, showing working.

  1. 13x + 3y = −6, so x + y = −2 and 4x + 4y = −8
  2. 2Add to 2x − 4y = 11: 6x = 3
  3. 3x = 0.5
  4. 40.5 + y = −2, so y = −2.5
  5. 5x = 0.5, y = −2.5

Answerx = 0.5, y = −2.5

Part (b) · 4 marks

Solve 2x² = 3(8 − x), showing working, giving both answers to 2 decimal places.

  1. 12x² = 24 − 3x, so 2x² + 3x − 24 = 0
  2. 2x = (−3 ± √(3² − 4 × 2 × (−24)))/(2 × 2) = (−3 ± √201)/4
  3. 3x = (−3 + 14.177…)/4 = 2.794… or x = (−3 − 14.177…)/4 = −4.294…
  4. 4x = 2.79 or x = −4.29

Answer2.79, −4.29

Part (ci) · 2 marks

Given h inversely proportional to g³ with h = 4.5 when g = 2, derive the equation linking h and g.

  1. 1h = k/g³
  2. 24.5 = k/2³, so k = 4.5 × 8 = 36
  3. 3h = 36/g³

Answer36/g³

Part (cii) · 2 marks

Using the relationship, determine g when h = 32/3.

  1. 132/3 = 36/g³
  2. 2g³ = 36 × 3/32 = 3.375
  3. 3g = ∛3.375 = 1.5

Answer1.5

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

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Also searched as: 4024/22/M/J/20 · 4024_s20_qp_22 · 4024 s20 22 · question 5.
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