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Positive integer solutions of a linear equation under an inequality

Positive integer solutions of a linear equation under an inequality

IP 2
7 marks

The positive integers \(x\) and \(y\) satisfy \(3x+5y=100\) and \(x>2y\).

  1. Show that \(y<\dfrac{100}{11}\).[2]
  2. Explain why \(y\) must leave a remainder of \(2\) when divided by \(3\).[2]
  3. Hence find all possible pairs \((x,y)\).[3]

Solution:

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Answer:(a) Shown (b) \(1+y\) must be a multiple of \(3\) (c) \((30,2)\), \((25,5)\), \((20,8)\)

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