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Is |x - y| > |x + y|? (1) x^2 - y^2 = 9 (2) x - y = 2
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Bunuel
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Is |x - y| > |x + y|? (1) x^2 - y^2 = 9 (2) x - y = 2 [#permalink] New post  02 Dec 2017, 23:26
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Is |x - y| > |x + y|?

(1) x^2 - y^2 = 9
(2) x - y = 2


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Re: Is |x - y| > |x + y|? (1) x^2 - y^2 = 9 (2) x - y = 2 [#permalink] New post  03 Dec 2017, 02:38
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Bunuel wrote:
Is |x - y| > |x + y|?

(1) x^2 - y^2 = 9
(2) x - y = 2


\(|x-y|>|x+y|\), square both sides to get

\(x^2+y^2-2xy>x^2+y^2+2xy => 4xy<0\)

so the question Is \(xy<0\). Hence we need to know the value of \(x\) & \(y\)

Statement 1: implies \((x-y)(x+y)=9\). two variable one equation not possible to solve. Insufficient

Statement 2: \(x-y=2\). again two variable one equation not possible to solve. Insufficient

Combining 1 & 2 we get \(2(x+y)=9\) so \(x+y=4.5\) and \(x-y=2\). Two variables and two equations we can calculate the value of \(x\) & \(y\). Sufficient

Option C
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Re: Is |x - y| > |x + y|? (1) x^2 - y^2 = 9 (2) x - y = 2 [#permalink] New post  03 Dec 2017, 05:53
antztheman wrote:
Hmm.. why not B?


Hi antztheman

if x=1 & y =-1, then x-y=2 and x+y=0 so |x-y|>|x+y|

but if x=4 y=2, then x-y=2 and x+y=6 but |x-y|<|x+y|

Hence we have a Yes and a NO answer for B.
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Re: Is |x - y| > |x + y|? (1) x^2 - y^2 = 9 (2) x - y = 2 [#permalink] New post  08 Jul 2019, 17:50
How do we know we can safely square each side of the inequality? is it because they are both absolute values? I know that we can safely square each sides of inequalities if we know they are both positive values.
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Re: Is |x - y| > |x + y|? (1) x^2 - y^2 = 9 (2) x - y = 2 [#permalink] New post  09 Jul 2019, 05:46
pcho409 wrote:
How do we know we can safely square each side of the inequality? is it because they are both absolute values? I know that we can safely square each sides of inequalities if we know they are both positive values.

Square is always positive. Hence you can square without worrying about the sign

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Re: Is |x - y| > |x + y|? (1) x^2 - y^2 = 9 (2) x - y = 2 [#permalink] New post  09 Jul 2019, 05:59
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pcho409 wrote:
How do we know we can safely square each side of the inequality? is it because they are both absolute values? I know that we can safely square each sides of inequalities if we know they are both positive values.


We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).

In the question at hand, we have an absolute value on both sides, which is always non-negative, so we can safely square.

How to manipulate inequalities (adding, subtracting, squaring etc.).
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Re: Is |x - y| > |x + y|? (1) x^2 - y^2 = 9 (2) x - y = 2 [#permalink] New post  09 Jul 2019, 06:27
Bunuel wrote:
Is |x - y| > |x + y|?

(1) x^2 - y^2 = 9
(2) x - y = 2


Statement 1
(x+y)(x-y)=9
9 = 1*9 =9*1 = (-1)(-9) = (-9)(-1) = 3*3 = (-3)*(-3)
x & y can take multiple values
NOT SUFFICIENT

Statement 2
x-y=2
x & y can take multiple values
NOT SUFFICIENT

Statement 1 and statement 2 combines
\(x=\frac{13}{4}\)
\(y=\frac{5}{4}\)
|x-y| = 2 & |x+y| = 4.5
|x-y| is NOT > |x+y|
SUFFICIENT

IMO C
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Re: Is |x - y| > |x + y|? (1) x^2 - y^2 = 9 (2) x - y = 2 [#permalink] New post  09 Jul 2019, 06:36
Bunuel wrote:
Is |x - y| > |x + y|?

(1) x^2 - y^2 = 9
(2) x - y = 2



Analyzing the main equation |x-y|>|x + y|. It is possible when both x and y have opposite signs.

Since this (that is when x and y are of opp. signs) leads to reduction in RHS value and combined effect in LHS will be greater .

Now we need to know x and y are of opposite signs. That is is xy<0 or xy>0

STAT1:

x^2 - y^2 = 9
implies (x+y)(x-y) = 9

Clearly Insufficient since we dont know values of x and y

STAT2:

x - y = 2

STAT2 is insufficient since we dont know x and y .

Combining STAT1 and STAT2 ,

Putting stat2 in stat1
(x+y)(x-y) = 9
we get (x+y)(2)=9
x+y = 4.5 .............1

Now we know x-y=2 .............2

From 1 and 2 we can find exact values..We dont need to calculate even till this much..

Two variables two equations is enough. So OA C.


Hope it helps!!
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Re: Is |x - y| > |x + y|? (1) x^2 - y^2 = 9 (2) x - y = 2 [#permalink] New post  09 Jul 2019, 09:51
Bunuel wrote:
Is |x - y| > |x + y|?

(1) x^2 - y^2 = 9
(2) x - y = 2


solve for |x - y| > |x + y|
we get -4xy>0
#1
(x+y)*(x-y)=9
insufficient
#
x-y=2
x=y+2
insufficient
from 1&2
x+y=9/2
sufficient to say -4xy>0
IMO C
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