Bunuel wrote:
What is \(y\) in terms of \(x\)?
(1) \(\frac{1}{\frac{x}{y} + \frac{y}{x}} = \frac{1}{2}\)
(2) \(\frac{x}{y} - \frac{y}{x} = 0\)
Statement 1: \(\frac{1}{\frac{x}{y} + \frac{y}{x}} = \frac{1}{2}\)
==> \(\frac{1}{((x^2 + y^2)/xy)} = \frac{1}{2}\)
\(\frac{xy}{(x^2 + y^2)} = \frac{1}{2}\)
2xy = \(x^2 + y^2\)
\(x^2 + y^2\) - 2xy=0 ==> \((x-y)^2 = 0\) ==> x-y = 0 ==> x=y
Sufficient.
Statement 2: \(\frac{x}{y} - \frac{y}{x} = 0\)
\(\frac{(x^2 - y^2)}{xy} = 0\)
Multiply both sides by xy
\(x^2 - y^2 = 0\)
\(x^2 = y^2\) ==> y= +x OR y= -x
Insufficient.
Answer: A
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