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0580/32 May/Jun 2023, Question 8: Simultaneous Equations by Elimination

Cambridge IGCSE Mathematics 0580 · Paper 32 (Core) · Calculator · 15 marks

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The question, in short

Six-part algebra question: evaluate a formula, simplify, expand, solve an equation, rearrange a formula, and form and solve simultaneous equations from castle entry costs.

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

Part (a) · 2 marks

Substitute P = 6 and Q = 8 into T = 5P + 3Q and work out T.

  1. 1T = 5 × 6 + 3 × 8
  2. 2= 30 + 24 = 54

Answer54

Part (b) · 2 marks

Collect like terms to simplify 3a − 7b + 2a + 4b.

  1. 13a + 2a = 5a
  2. 2−7b + 4b = −3b
  3. 3= 5a − 3b

Answer5a − 3b

Part (c) · 1 mark

Expand 5(2x − 3y) by multiplying out the bracket.

  1. 15 × 2x = 10x and 5 × (−3y) = −15y
  2. 2= 10x − 15y

Answer10x − 15y

Part (d) · 2 marks

Solve the equation 5x − 1 = 3x + 19 for x.

  1. 15x − 3x = 19 + 1
  2. 22x = 20
  3. 3x = 10

Answer10

Part (e) · 2 marks

Rearrange p = 5t − 3 to make t the subject.

  1. 1p + 3 = 5t
  2. 2t = (p + 3)/5

Answer(p + 3)/5

Part (f) · 6 marks

Using $x per adult and $y per child, write simultaneous equations for 2 adults and 3 children costing $15.00 and 3 adults and 5 children costing $23.50, then solve for x and y showing all working.

  1. 12x + 3y = 15 and 3x + 5y = 23.5
  2. 2× 3: 6x + 9y = 45; × 2: 6x + 10y = 47
  3. 3Subtract: y = 2
  4. 42x + 3 × 2 = 15, so 2x = 9
  5. 5x = 4.5, y = 2

Answer2x + 3y = 15 and 3x + 5y = 23.5; x = 4.5, y = 2

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

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Also searched as: 0580/32/M/J/23 · 0580_s23_qp_32 · 0580 s23 32 · question 8.
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