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If x is not equal to -y, is (x-y)/(x+y) > 1? 1) x>0 2)

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If x is not equal to -y, is (x-y)/(x+y) > 1? 1) x>0 2) [#permalink] New post 07 Sep 2007, 07:32
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If x is not equal to -y, is (x-y)/(x+y) > 1?
1) x>0
2) y<0> (x+y) ?

SO, multiplying the inequality by (x+y) both sides, the question becomes is (x-y) > (x+y)?

From 1) if x>0, x is +ve
From 2) if y<0, y is -ve


So From 1 and 2, (x-y) is postive and (x+y) could be positive or negative, but regardless of (x+y) being postive or negative, (x-y) will always be greater than (x+y). I thought the official answer to be C

but the OE is E, can some one pls explain what is wrong with my analysis.
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 [#permalink] New post 07 Sep 2007, 07:53
I think you just explaned the answer yourself.

Quote:
So From 1 and 2, (x-y) is postive and (x+y) could be positive or negative, but regardless of (x+y) being postive or negative, (x-y) will always be greater than (x+y).


If (x-y) is positive and (x+y) is negative the whole expression will be negative, therefore <1>1. Therefore, insufficient. Answer E.
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 [#permalink] New post 07 Sep 2007, 08:13
eileen1017 wrote:
I think you just explaned the answer yourself.

Quote:
So From 1 and 2, (x-y) is postive and (x+y) could be positive or negative, but regardless of (x+y) being postive or negative, (x-y) will always be greater than (x+y).


If (x-y) is positive and (x+y) is negative the whole expression will be negative, therefore <1>1. Therefore, insufficient. Answer E.


I could see that if x-y is positive and x+y negative then less than 1 , and x-y) is positive and x+y positive then greater than 1.

But if you the change the inequality from is (x-y)/(x+y) > 1 to (x-y) > (x+y) and ask yourself is (x-y) > (x+y), you will end up with answer YES.

So probably i should not be altering the inequlaity by multiplying (x+y), which changes the whole picture.
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 [#permalink] New post 07 Sep 2007, 08:25
zion wrote:
eileen1017 wrote:
I think you just explaned the answer yourself.

Quote:
So From 1 and 2, (x-y) is postive and (x+y) could be positive or negative, but regardless of (x+y) being postive or negative, (x-y) will always be greater than (x+y).


If (x-y) is positive and (x+y) is negative the whole expression will be negative, therefore <1>1. Therefore, insufficient. Answer E.


I could see that if (x-y) is negative , and (x-y) is positive then (x-y)/(x+y) <1>1, hence E.

But if you the change the inequality from is (x-y)/(x+y) > 1 to (x-y) > (x+y) and ask yourself is (x-y) > (x+y), you will end up with answer YES.

So probably i should not be altering the inequlaity by multiplying (x+y), which changes the whole picture.


In these type of equations we can not change the equality, the way u did.
Cuz (x+y) can be both posotive or negative, if (x+y) is positive then the way u changed the equation is correct, but if it is negative, the sign <> will change to opposite.

I keep getting C,

Plz, can you check if I understand what is given, the HTML sometimes changes the inequality sings, so let me write in words:

X is not equal to minus Y. Is X-Y over X+Y greater than 1?

1) X is more than zero
2) X+Y is more than zero, Y is less than zero, and Y is less than (X+Y)

I see two inequality sings in the 2nd stem, is it right?
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 [#permalink] New post 07 Sep 2007, 08:36
zion wrote:
eileen1017 wrote:
I think you just explaned the answer yourself.

Quote:
So From 1 and 2, (x-y) is postive and (x+y) could be positive or negative, but regardless of (x+y) being postive or negative, (x-y) will always be greater than (x+y).


If (x-y) is positive and (x+y) is negative the whole expression will be negative, therefore <1>1. Therefore, insufficient. Answer E.


I could see that if x-y is positive and x+y negative then less than 1 , and x-y) is positive and x+y positive then greater than 1.

But if you the change the inequality from is (x-y)/(x+y) > 1 to (x-y) > (x+y) and ask yourself is (x-y) > (x+y), you will end up with answer YES.

So probably i should not be altering the inequlaity by multiplying (x+y), which changes the whole picture.


While solving inequlaity problems, multipling denominator (bottom part of fraction) to the other part of inequality sign is INCORRECT, it is only possible in equations.

Instead take 1 to left side and simplify the expression.

you will have 2y/(x+y)<0>0 and y>0 will give postive value (to satisfy inequlaity it should be negative since 2y/(x+y)<0)
2) is insufficient because y<0 and x<0 will give positive value (to satisfy inequlaity it should be negative since 2y/(x+y)<0)

both together will not sufficient since we don't know the values, if x is more than two times higher than y than it will give postive value and in other cases negative so Insufficient

Ans: E

Last edited by Ferihere on 07 Sep 2007, 08:39, edited 1 time in total.
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 [#permalink] New post 07 Sep 2007, 08:36
IrinaOK wrote:
zion wrote:
eileen1017 wrote:
I think you just explaned the answer yourself.

Quote:
So From 1 and 2, (x-y) is postive and (x+y) could be positive or negative, but regardless of (x+y) being postive or negative, (x-y) will always be greater than (x+y).


If (x-y) is positive and (x+y) is negative the whole expression will be negative, therefore <1>1. Therefore, insufficient. Answer E.


I could see that if (x-y) is negative , and (x-y) is positive then (x-y)/(x+y) <1>1, hence E.

But if you the change the inequality from is (x-y)/(x+y) > 1 to (x-y) > (x+y) and ask yourself is (x-y) > (x+y), you will end up with answer YES.

So probably i should not be altering the inequlaity by multiplying (x+y), which changes the whole picture.


In these type of equations we can not change the equality, the way u did.
Cuz (x+y) can be both posotive or negative, if (x+y) is positive then the way u changed the equation is correct, but if it is negative, the sign <> will change to opposite.

I keep getting C,

Plz, can you check if I understand what is given, the HTML sometimes changes the inequality sings, so let me write in words:

X is not equal to minus Y. Is X-Y over X+Y greater than 1?

1) X is more than zero
2) X+Y is more than zero, Y is less than zero, and Y is less than (X+Y)

I see two inequality sings in the 2nd stem, is it right?



Thanks for the explanation.. Here is the question agian

If X is not equal to minus Y. Is X-Y over X+Y greater than 1?

1) X is more than zero
2) Y is less than zero
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 [#permalink] New post 07 Sep 2007, 08:40
zion wrote:
IrinaOK wrote:
zion wrote:
eileen1017 wrote:
I think you just explaned the answer yourself.

Quote:
So From 1 and 2, (x-y) is postive and (x+y) could be positive or negative, but regardless of (x+y) being postive or negative, (x-y) will always be greater than (x+y).


If (x-y) is positive and (x+y) is negative the whole expression will be negative, therefore <1>1. Therefore, insufficient. Answer E.


I could see that if (x-y) is negative , and (x-y) is positive then (x-y)/(x+y) <1>1, hence E.

But if you the change the inequality from is (x-y)/(x+y) > 1 to (x-y) > (x+y) and ask yourself is (x-y) > (x+y), you will end up with answer YES.

So probably i should not be altering the inequlaity by multiplying (x+y), which changes the whole picture.


In these type of equations we can not change the equality, the way u did.
Cuz (x+y) can be both posotive or negative, if (x+y) is positive then the way u changed the equation is correct, but if it is negative, the sign <> will change to opposite.

I keep getting C,

Plz, can you check if I understand what is given, the HTML sometimes changes the inequality sings, so let me write in words:

X is not equal to minus Y. Is X-Y over X+Y greater than 1?

1) X is more than zero
2) X+Y is more than zero, Y is less than zero, and Y is less than (X+Y)

I see two inequality sings in the 2nd stem, is it right?



Thanks for the explanation.. Here is the question agian

If X is not equal to minus Y. Is X-Y over X+Y greater than 1?

1) X is more than zero
2) Y is less than zero


Thank you for clarification,

I agree with E. =)
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 [#permalink] New post 08 Sep 2007, 04:09
Answer is 'E'.
I find easy to solve such questions by picking up numbers rathen than solving the equations.

x=6, y =-16 x-y/x+y <1> 1

So Insuff.
  [#permalink] 08 Sep 2007, 04:09
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