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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 25 Mar 2013, 05:29
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kancharana
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Is |x-y|>|x|-|y| ?

(1) y < x
(2) xy < 0


How it is B? Did they mention that X and Y are integers? No right, the answer should be E. If they provide details about X and Y as integers then it will be B otherwise it will be E.

can anyone help me about the scenario whether we consider fractions or not in this case?

Scenario:

x=1/2, y=1/3 ==> |1/2-1/3|=1/6 and |1/2|-|1/3|=1/6
Show more

It is B because if you use the data of statement 2, you can say, "Yes, |x-y| is greater than |x|-|y|"

(2) xy < 0
This means that one of x and y is positive and the other is negative. You cannot take x = 1/2 and y = 1/3.
It is not about fractions/integers. It is about positive/negative numbers (most mod questions are about positive/negative numbers)
When xy < 0, |x-y|>|x|-|y| always holds.
Only when x and y both are positive or both are negative and |x|>|y|, then |x-y|=|x|-|y|
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 06 Aug 2013, 22:25
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kancharana
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Is |x-y|>|x|-|y| ?

(1) y < x
(2) xy < 0


How it is B? Did they mention that X and Y are integers? No right, the answer should be E. If they provide details about X and Y as integers then it will be B otherwise it will be E.

can anyone help me about the scenario whether we consider fractions or not in this case?

Scenario:

x=1/2, y=1/3 ==> |1/2-1/3|=1/6 and |1/2|-|1/3|=1/6
Show more

Fractions and integers have no role to play here. Check Bunuel's post above.

Whenever xy < 0, i.e. x is negative but y is positive OR x is positive but y is negative, |x-y| is greater than |x|-|y|.

e.g. x = -1/2, y = 1/3
|x-y| = |-1/2-1/3| = 5/6
|x|-|y| = 1/2 - 1/3 = 1/6

So |x - y| > |x|-|y|

Do you see the logic here? If one of x and y is positive and the other is negative, in |x - y|, absolute values of x and y get added and the sum is positive. While in |x|-|y|, the absolute values are subtracted.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 02 Jan 2014, 23:36
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Is |x - y| > |x| - |y|?

(1) y < x
(2) xy < 0
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Statement I is insufficient

x....y.....(|x|-|y|)....|x-y|.....YES/NO
3....2.......1..............1........NO
3..-3........0..............6........YES

Statement II is sufficient
xy<0 means that either one of the numbers is negative and other is positive

|x-y| will always result to be a greater value as negative - positive or positive - negative will result in addition of the numbers and |x| - |y| will always result in subtraction.

Hence the answer is B
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 19 Oct 2015, 23:55
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is |x - y| > |x| - |y|?

(1) y < x
(2) xy < 0

If we modify the original condition,
if |x|>=|y|,
|x-y|>|x|-|y|>=0, so we can square both sides, which gives us (|x-y|)^2>(|x|-|y|)^2, and (x-y)^2>(|x|-|y|)^2
This becomes x^2+y^2-2xy>x^2+y^2-2|xy|, and if we simplify the inequality, -2xy>-2|xy|, and we ultimately want to know whether xy<0. This makes condition 2 sufficient.
if |x|<|y|, then |x|-|y|<0,
so |x-y|>|x|-|y| always work. Hence we do not need to deal with this, so the answer becomes (B).

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 12 Feb 2017, 17:12
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Bunuel
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Please tell the quick approach.... it took me loner than I should have taken....

Algebraic approach is given in my first post. Below is another approach:

\(|x-y|>|x|-|y|\)?

(1) \(y<x\)

Try two positive number \(x=3>y=1\) --> is \(|3-1|>|3|-|1|\)? --> is \(2>2\)? Answer NO.

Try ANY other case but both positive: \(x=-5>y=-7\) --> is \(|-5-(-7)|>|-5|-|-7|\)? --> is \(2>-2\)? Answer YES.

Two different answers. Not sufficient.

(2) \(xy<0\), means \(x\) and \(y\) have different signs.

Now we can spot here that when \(x\) and \(y\) have different signs \(x-y\) always "contribute" to each other so that its absolute value will increase: \(x=3\), \(y=-1\) --> \(|x-y|=|3+1|=|4|=4\) or \(x=-3\), \(y=1\) --> \(|x-y|=|-3-1|=|4|=4\).

But \(|x|-|y|\) is difference (thus not "contributing") of two positive values (as neither equals to zero). \(x=3\), \(y=-1\) --> \(|x|-|y|=|3|-|-1|=2\) or \(x=-3\), \(y=1\) --> \(|x|-|y|=|-3|-|1|=2\).

So \(xy<0\) means \(|x-y|>|x|-|y|\) is always true.

Sufficient.

Answer: B.

Hope it helps.
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This graphical approach will help to verify those value.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 12 Mar 2017, 20:53
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Is \(|x - y| > |x| - |y|\)?

(1) \(y < x\)
(2) \(xy < 0\)
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Explanation from RON (Manhattan)

Statement (1)
let's try x = 2, y = 1
is |2 - 1| > |2| - |1| ?
no.
let's try x = -1, y = -2
is |-1 - (-2)| > |-1| - |-2| ?
yes.
Insufficient.

Statement (2)

this means we have to pick OPPOSITE SIGNS. therefore, there are basically 6 cases to try:
* x is negative "bigger", y is positive "smaller"
* x is negative, y is positive, same magnitude
* x is negative "smaller", y is positive "bigger"
* x is positive "bigger", y is negative "smaller"
* x is positive, y is negative, same magnitude
* x is positive "smaller", y is negative "bigger"

If you try all of these --

-2, 1 --> is |-3| > 2 - 1? YES
-1, 1 --> is |-2| > 0? YES
-1, 2 --> is |-3| > 1 - 2? YES
2, -1 --> is |3| > 2 - 1? YES
1, -1 --> is |2| > 0? YES
1, -2 --> is |3| > 1 - 2? YES

the pattern is pretty clear: in each of these cases, the two numbers' magnitudes are working together on the left, but against each other on the right. therefore, the left-hand side is always going to be bigger.
Sufficient.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 20 Aug 2017, 10:44
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Answer: B
IS |X-Y| > |X| - |Y| ?
a) Y<X
b) XY<0

Statement A:
If, Y=-1 X=1 then LHS=2 and RHS=0 -->Correct
If, Y=1 X=3 then LHS=2 and RHS=2 -->Incorrect
Thus INSUFFICIENT

Statement B:
Either of X or Y is -ve.
If, Y=-1 X=1 then LHS=2 and RHS=0 -->Correct
If, Y=1 X=-3 then LHS=4 and RHS=2 -->Correct
Thus SUFFICIENT
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 15 Dec 2017, 22:52
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Bunuel
Basically the question asks whether the distance between the two points x and y on the line is greater than the difference between the individual distances of x and y from 0.

\(|x-y|>|x|-|y|\)?

(1) \(y<x\), 3 possible cases for \(|x-y|>|x|-|y|\):

A. ---------------\(0\)---\(y\)---\(x\)---, \(0<y<x\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>x-y\) --> \(0>0\). Which is wrong;
B. ---------\(y\)---\(0\)---------\(x\)---, \(y<0<x\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>x+y\) --> \(y<0\). Which is right, as we consider the range \(y<0<x\);
C. ---\(y\)---\(x\)---\(0\)--------------, \(y<x<0\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>-x+y\) --> \(x>y\). Which is right, as we consider the range \(y<x<0\).

Two different answers. Not sufficient.

(2) \(xy<0\), means \(x\) and \(y\) have different signs, hence 2 cases for \(|x-y|>|x|-|y|\):

A. ----\(y\)-----\(0\)-------\(x\)---, \(y<0<x\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>x+y\) --> \(y<0\). Which is right, as we consider the range \(y<0<x\);
B. ----\(x\)-----\(0\)-------\(y\)---, \(x<0<y\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(-x+y>-x-y\) --> \(y>0\). Which is right, as we consider the range \(x<0<y\).

In both cases inequality holds true. Sufficient.

Answer: B.
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Hi Bunuel,

Tried hard to understand the following related to the red part above but at the end no success. Would you please clear my doubt?

In 1 B & C, you have changed the sign of x and y in RHS, and not modified the LHS(red part) according to the sign of x and y.---- Understood the explanation of st. 1
But in 2 B, You have modified also the LHS part according to the sign of x and y. (red part) ------- Why have you not followed the process that you have followed in st 1 ?

What did this difference in procedure depend on?
Thank You in advance.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 16 Dec 2017, 00:05
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TaN1213
Bunuel
Basically the question asks whether the distance between the two points x and y on the line is greater than the difference between the individual distances of x and y from 0.

\(|x-y|>|x|-|y|\)?

(1) \(y<x\), 3 possible cases for \(|x-y|>|x|-|y|\):

A. ---------------\(0\)---\(y\)---\(x\)---, \(0<y<x\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>x-y\) --> \(0>0\). Which is wrong;
B. ---------\(y\)---\(0\)---------\(x\)---, \(y<0<x\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>x+y\) --> \(y<0\). Which is right, as we consider the range \(y<0<x\);
C. ---\(y\)---\(x\)---\(0\)--------------, \(y<x<0\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>-x+y\) --> \(x>y\). Which is right, as we consider the range \(y<x<0\).

Two different answers. Not sufficient.

(2) \(xy<0\), means \(x\) and \(y\) have different signs, hence 2 cases for \(|x-y|>|x|-|y|\):

A. ----\(y\)-----\(0\)-------\(x\)---, \(y<0<x\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>x+y\) --> \(y<0\). Which is right, as we consider the range \(y<0<x\);
B. ----\(x\)-----\(0\)-------\(y\)---, \(x<0<y\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(-x+y>-x-y\) --> \(y>0\). Which is right, as we consider the range \(x<0<y\).

In both cases inequality holds true. Sufficient.

Answer: B.
Hi Bunuel,

Tried hard to understand the following related to the red part above but at the end no success. Would you please clear my doubt?

In 1 B & C, you have changed the sign of x and y in RHS, and not modified the LHS(red part) according to the sign of x and y.---- Understood the explanation of st. 1
But in 2 B, You have modified also the LHS part according to the sign of x and y. (red part) ------- Why have you not followed the process that you have followed in st 1 ?

What did this difference in procedure depend on?
Thank You in advance.
Show more

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\)


(2) \(xy<0\), means \(x\) and \(y\) have different signs, hence 2 cases for \(|x-y|>|x|-|y|\):

B. ----\(x\)-----\(0\)-------\(y\)---.

This means that \(x<0<y\). So, \(x - y < 0\), \(x<0\) and \(y>0\). According to the properties above if \(x - y < 0\), then \(|x-y|=-(x-y)\), if \(x<0\), then \(|x|=-x\) and if \(y>0\), then \(|y|=y\).

Hope it's clear.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 16 Dec 2017, 10:56
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Bunuel

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\)


(2) \(xy<0\), means \(x\) and \(y\) have different signs, hence 2 cases for \(|x-y|>|x|-|y|\):

B. ----\(x\)-----\(0\)-------\(y\)---.

This means that \(x<0<y\). So, \(x - y < 0\), \(x<0\) and \(y>0\). According to the properties above if \(x - y < 0\), then \(|x-y|=-(x-y)\), if \(x<0\), then \(|x|=-x\) and if \(y>0\), then \(|y|=y\).

Hope it's clear.
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Following this same approach, shouldn't the following part become (-x-y)?
Quote:

1C. ---\(y\)---\(x\)---\(0\)--------------, \(y<x<0\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>-x+y\) --> \(x>y\). Which is right, as we consider the range \(y<x<0\).
Show more

in 2 B you have modified the \(x-y\) to -(x-y) because x is negative and y is positive. So in the same way the 1C should take (-x-y) as per the sign of x.
What am I missing?
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 16 Dec 2017, 11:23
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TaN1213
Bunuel

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\)


(2) \(xy<0\), means \(x\) and \(y\) have different signs, hence 2 cases for \(|x-y|>|x|-|y|\):

B. ----\(x\)-----\(0\)-------\(y\)---.

This means that \(x<0<y\). So, \(x - y < 0\), \(x<0\) and \(y>0\). According to the properties above if \(x - y < 0\), then \(|x-y|=-(x-y)\), if \(x<0\), then \(|x|=-x\) and if \(y>0\), then \(|y|=y\).

Hope it's clear.
Following this same approach, shouldn't the following part become (-x-y)?
Quote:

1C. ---\(y\)---\(x\)---\(0\)--------------, \(y<x<0\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>-x+y\) --> \(x>y\). Which is right, as we consider the range \(y<x<0\).

in 2 B you have modified the \(x-y\) to -(x-y) because x is negative and y is positive. So in the same way the 1C should take (-x-y) as per the sign of x.
What am I missing?
Show more

C. ---\(y\)---\(x\)---\(0\)--------------, \(y<x<0\):

x - y > 0, so |x - y| = x - y;
y < 0, so |y| = y;
x < 0, so |x| = x.

\(|x-y|>|x|-|y|\) --> x - y > -x - (-y) --> x - y > -x + y
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I am a little bit concerned here.
If the question states x and y, can we make an assumption that x ≠ y?
For example here, without this assumption, i can let x = y and (2) is NS?
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khiemchii
I am a little bit concerned here.
If the question states x and y, can we make an assumption that x ≠ y?
For example here, without this assumption, i can let x = y and (2) is NS?
Show more

Unless it is explicitly stated otherwise, different variables CAN represent the same number.

For (2) though x = y is not possible because in this case we'd get x^2 < 0, which is not true for any real number, so x = y does not satisfy xy < 0 and therefore should not be considered for the second statement. Similarly, for (1) it's give that y < x, so x = y cannot be true.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 20 Jan 2018, 21:31
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Can anyone tell me the flaw in my logic

|x - y| > |x| - |y|
Squaring both sides
x^2 +y^2-2xy> x^2 +y^2-2|x||y|
simplifying, xy<|x||y|,

It will only be possible when product of xy is -ve. or xy<0, Hence B.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 22 Jan 2018, 03:27
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ManishKM1
Can anyone tell me the flaw in my logic

|x - y| > |x| - |y|
Squaring both sides
x^2 +y^2-2xy> x^2 +y^2-2|x||y|
simplifying, xy<|x||y|,

It will only be possible when product of xy is -ve. or xy<0, Hence B.
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Given:
|x - y| > |x| - |y|, you cannot square it. You can square only when you know that both sides are positive. Here, the right hand side may not be positive. For example, if x is 2 and y is 5.
3 > -3
Squaring does not work here since you get 9 > 9 which doesn't hold.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 04 Mar 2018, 00:17
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VeritasPrepKarishma

Is it incorrect to say that question is asking is |x|>|y| ?

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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 05 Mar 2018, 05:23
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Mudit27021988
VeritasPrepKarishma

Is it incorrect to say that question is asking is |x|>|y| ?

Posted from my mobile device
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For all real x and y, |x – y| >= |x| – |y|

|x – y| = |x| – |y| when
(1) x and y have the same sign and x has greater (or equal) absolute value than y
(2) y is 0

|x – y| > |x| – |y| in all other cases (i.e. y is not 0 and |x| < |y| or x and y have opposite signs)

So |x – y| > |x| – |y| is not the same as |x| > |y|

Check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/0 ... -the-gmat/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/0 ... t-part-ii/
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 05 Feb 2019, 03:53
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Dear GMATGuruNY

In this question stem, can I square both sides? or is it wrong?

Thanks
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 05 Feb 2019, 04:40
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Mo2men
Dear GMATGuruNY

In this question stem, can I square both sides? or is it wrong?

Thanks
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Squaring both sides of an inequality is a good approach when both sides are guaranteed to be nonnegative.
Here, the right side of the question stem is NOT guaranteed to be nonnegative, so squaring the inequality is not recommended:
If x=2 and y=1, then |x| - |y| > 0
If x=1 and y=2, then |x| - |y| < 0
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 13 Mar 2020, 14:57
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|x - y| > |x| - |y|

Above inequality will be true when xy<0. As in LHS, absolute values of x and y will add up, while in RHS absolute values of x and y will get subtracted.

Also,inequality will be true, when xy>0 and |x|<|y|, As LHS will be positive, but RHS will be negative.

Statement 1- y<x

if x<0 and y<0, |y|>|x| (inequality holds)
if x>0 and y>0, |y|<|x| (inequality won't hold)

Statement 2- xy<0
Inequality holds (As explained above)


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Is |x - y| > |x| - |y|?

(1) y < x
(2) xy < 0


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3_DS_Absolute_B.JPG
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