Question

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Answer 1

Answer:

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Answer 2

$x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } = \frac{ (\sqrt{3} + \sqrt{2}) ^{2} }{3 - 2} = 5 + 2 \sqrt{6} \\hence \: y = 1 \\ \frac{x - y}{x - 3y} = \frac{4 + 2 \sqrt{6} }{2 + 2 \sqrt{6} } = \frac{2 + 2 \sqrt{6} }{1 + \sqrt{6} } = 2$