John Corn wrote:x not equal to -y

is (x-y)/(x+y)>1?

1) x>0
2) y<0

Please help [/u]

x-y
---- -1 > 0
x+y

x-y-x-y
-------- >0
x+y

y/(x+y) < 0

If x>0 we do not know anything about y insuff
If y<0 we do not know anything abt x insuff

If x>0 and y<0 depends on its value for the sign of x+y, not since y<0 the sum y/(x+y) < 0 means 1/(x+y) > 0 but that is not always true, if y=-2 x =1 we have x+y = -1 while y=-2 x=3 means x+y =1

so E.

Thanks maihuna for your response.
I have a question

Why can't you solve it with this equality
(x-y)>(x+y)?

With this I am getting that B is sufficient.

Thanks in advance.

John Corn wrote:Thanks maihuna for your response.
I have a question

Why can't you solve it with this equality
(x-y)>(x+y)?

With this I am getting that B is sufficient.

Thanks in advance.

in inequalities dividing or multipying any unknown variable(+or-ve) always leads to two outcomes a positive or a negative

since x-y is + or -ve it is unknown cross multiplying may or may not req change of sign....

therefore whn sign is unknown its always better to limit the operation upto addition or sub...

HTH

Thanks so much.

maihuna wrote: x-y-x-y
-------- >0
x+y

y/(x+y) < 0

maihuna - is the second formula supposed to read 2y/x+y?

I dont think it makes a difference to the ans but just wanted to check i was reducing it correctly when i worked it out.

thks

x-y
---- -1 > 0
x+y

How did you get -x-y in the numerator from -1?

x-y-x-y
-------- >0
x+y

x-y
---- -1 > 0
x+y

x-y-x-y
-------- >0
x+y

till here i got it but then hoe did u got the below step..

y/(x+y) < 0

i got stucked at
-2y
---- > 0
x+y

pls help!!!