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11 Nov 2017, 09:39
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D
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Difficulty:
35%
(medium)
Question Stats:
68% (01:29) correct
32%
(01:53)
wrong
based on 141
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If x ≠ y, is x + y = xy?
(1) \((1 - x)(1 - y) = 1\)
(2) \(x^2 - y^2 = x^2y - xy^2\)
(1) \((1 - x)(1 - y) = 1\)
(2) \(x^2 - y^2 = x^2y - xy^2\)
11 Nov 2017, 09:59
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Bunuel
(1) \((1 - x)(1 - y) = 1\).
\(-x-y+1+xy=1...........xy=x+y\)
suff
(2) \(x^2 - y^2 = x^2y - xy^2\).
\((x-y)(x+y)=xy(x-y)..................(x-y)(x+y-xy)=0\)
so either x=y or xy=x+y
since x ≠ y, xy=x+y
suff
D
11 Nov 2017, 10:19
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1) 1-y-x+xy=1
xy=1-1+x+y
=> xy=x+y -> sufficient
2)(x-y)(x+y)=xy(x-y)
=> x+y=xy -> sufficient
=> D
xy=1-1+x+y
=> xy=x+y -> sufficient
2)(x-y)(x+y)=xy(x-y)
=> x+y=xy -> sufficient
=> D
