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10 May 2008, 19:06
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N
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Any help would be greatly appreciated.
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10 May 2008, 19:23
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d
factor out the original equation
((x^2) +2xy + (y^2)) / ((x^2) -2xy + (y^2))
or rewrite ((x^2) + (y^2) + 2xy) / ((x^2) + (y^2) - 2xy)
xy = 3 ---> if x =1, y =2 or x = 6, y = .5
substitute in the x & y numbers, and the value will always be the same
factor out the original equation
((x^2) +2xy + (y^2)) / ((x^2) -2xy + (y^2))
or rewrite ((x^2) + (y^2) + 2xy) / ((x^2) + (y^2) - 2xy)
xy = 3 ---> if x =1, y =2 or x = 6, y = .5
substitute in the x & y numbers, and the value will always be the same
10 May 2008, 20:02
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Hi , I would agree with A
Stat 1 sufficient
stat 2
xy=3
x=1, y=3
(x+y)^2/(x-y)^2
=16/4=4
OR
x=1/2, y=6
=(169/121) not equal to 4 or from stat 1. 5
Insufficient
Stat 1 sufficient
stat 2
xy=3
x=1, y=3
(x+y)^2/(x-y)^2
=16/4=4
OR
x=1/2, y=6
=(169/121) not equal to 4 or from stat 1. 5
Insufficient
