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If x ≠ y and x and y are non-zero integers, is (x+y)/(x−y)>0?

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If x ≠ y and x and y are non-zero integers, is (x+y)/(x−y)>0?  [#permalink]

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New post 29 Apr 2016, 03:57
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If x ≠ y and x and y are non-zero integers, is (x+y)/(x−y) > 0?

(1) x > y
(2) |x| > |y|
Spoiler: OA

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Re: If x ≠ y and x and y are non-zero integers, is (x+y)/(x−y)>0?  [#permalink]

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New post 29 Apr 2016, 08:30
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1
The question basically asks are both x+y and x-y positive or negative?

(1) x > y
(2) |x| > |y|

Stmnt 1:
x-y> 0 => x-y is positive
says nothing about x+y which can either be positive or negative
Insufficient

Stmnt 2
\(\sqrt{x^2}>\sqrt{y^2}\)
x^2-y^2 > 0
(x+y)(x-y) > 0
so both x+y and x-y can be positive or
both x+y and x-y can be negative
This means that x+y /x-y will always be positive and greater than 0
Sufficient

Ans B
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Re: If x ≠ y and x and y are non-zero integers, is (x+y)/(x−y)>0?  [#permalink]

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New post 22 Aug 2018, 23:26
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(x + y)/(x - y) = (x + y)(x - y) / (x - y) (x - y) = (x^2 -y^2)/(x - y)^2
we know x is not equal to y , so (x - y)^2 is always > 0
Therefore x^2 - y^2 > 0
So , x^2 > y^2
Hence, |x| > |y|.

Ans- B
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If x ≠ y and x and y are non-zero integers, is (x+y)/(x−y)>0?  [#permalink]

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New post 26 Aug 2018, 21:45
\(\frac{x+y}{x-y} > 0\)

Multiplying numerator and denominator by x-y, we get,

\(x^2-y^2 >0\) i.e. \(x^2>y^2\) and hence |x| > |y|

1. x>y is clearly not sufficient
2|x|>|y| , this what our deduced expressions ask for!

Hence, B
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Re: If x ≠ y and x and y are non-zero integers, is (x+y)/(x−y)>0?  [#permalink]

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New post 27 Aug 2018, 07:48
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Bunuel wrote:
If x ≠ y and x and y are non-zero integers, is (x+y)/(x−y) > 0?

(1) x > y
(2) |x| > |y|


(1) x > y

If x=2, y=1 then (x+y)/(x−y) = 3/1 = 3 > 0 --> Yes.
If x=1, y=-1 then (x+y)/(x−y) = 0/2 = 0 --> No.
--> Not sufficient.

(2) |x| > |y| --> \(x^2 > y^2\) --> \(x^2-y^2 > 0\) --> \((x+y)(x-y) > 0\) --> \((x+y)\)/\((x−y)\) \(> 0\) --> Yes --> Sufficient.

Answer B.
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Re: If x ≠ y and x and y are non-zero integers, is (x+y)/(x−y)>0? &nbs [#permalink] 27 Aug 2018, 07:48
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