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If 1/x + 1/y = 1/3, then xy/(x + y) =

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Bunuel
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If 1/x + 1/y = 1/3, then xy/(x + y) = [#permalink] New post  03 May 2022, 02:30
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If \(\frac{1}{x} + \frac{1}{y} = \frac{1}{3}\), then \(\frac{xy}{x + y} =\)

(A) 1/5
(B) 1/3
(C) 1
(D) 3
(E) 5
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Re: If 1/x + 1/y = 1/3, then xy/(x + y) = [#permalink] New post  03 May 2022, 06:12
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Bunuel wrote:
If \(\frac{1}{x} + \frac{1}{y} = \frac{1}{3}\), then \(\frac{xy}{x + y} =\)

(A) 1/5
(B) 1/3
(C) 1
(D) 3
(E) 5


Given: \(\frac{1}{x} + \frac{1}{y} = \frac{1}{3}\)

Rewrite the left side with common denominators: \(\frac{y}{xy} + \frac{x}{xy} = \frac{1}{3}\)

Simplify the left side: \(\frac{x+y}{xy} = \frac{1}{3}\)

Useful property: If \(abcd ≠ 0\), and \(\frac{a}{b} = \frac{c}{d}\), then it's also true that \(\frac{b}{a} = \frac{d}{c}\)

So take the reciprocal of both sides: \(\frac{xy}{x+y} = \frac{3}{1} = 3\)

Answer: D
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Re: If 1/x + 1/y = 1/3, then xy/(x + y) = [#permalink]
03 May 2022, 06:12
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