Is x>y? 1) x+y>1/x-y 2) 1/x<1/y

Statement 1
\(x+y>\frac{1}{x-y}\)
\(\frac{x^2-y^2-1}{x-y}\)>0
When x^2-y^2>1, x>y
and when x^2-y^2<1, x<y
Insufficient
Statement 2
\(\frac{1}{x}<\frac{1}{y}\)
\(\frac{x-y}{xy}>0\)

When x and y has same signs, x>y
When x and y have opposite signs, x<y
Insufficient

Combining both equations
From equation 1
\(x+y>\frac{1}{x-y}\)
x+y>\(\frac{xy}{x-y}\)*\(\frac{1}{xy}\)
From statement 2, we can say that highlighted portion is positive. Hence we can cross-multiply it without the change of inequality sign.

We get, \(\frac{x^2-y^2-1}{xy}\)>0

When x and y has same sign x^2-y^2-1 is positive. Hence when x and y are both positive, then x>y. But when x and y are both negative, x can be greater or smaller than y. We don't need to analyse further, as we get our answer.

Option E

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