This question is also very easily solved algebraically.
1) x^2 - y^2 <0. (x+y)(x-y) <0 . One of these is negative, soeitherx+y<0 , which means thatx<-y , orx<y . Two options; not sufficient.
2)-x-y<0 . Not sufficient.
Using number 2), we can multiply by a negative to see thatx+y>0 . Now using option 1, this means that x-y must be less than 0. Therefore,x-y<0 andx<y . Sufficient.
Answer: C
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1) x^2 - y^2 <0. (x+y)(x-y) <0 . One of these is negative, soeitherx+y<0 , which means thatx<-y , orx<y . Two options; not sufficient.
2)-x-y<0 . Not sufficient.
Using number 2), we can multiply by a negative to see thatx+y>0 . Now using option 1, this means that x-y must be less than 0. Therefore,x-y<0 andx<y . Sufficient.
Answer: C
...
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