If |x| - |y| = |x+y| and xy not equal zero , which of the following

there can be 4 instances:-
i)x +,y -....|x| - |y| = |x+y| => x-y=x-y
ii) both +... x-y=x+y.. not possible as xy not equal zero
iii) both -....x-y=-(x+y) .. not possible as xy not equal zero
iv) x-,y+...x-y=y-x......x=y..
therefore x and y are of different signs, when multiplied ,it should be -ive

If |x| - |y| = |x+y| and xy does not equal to 0, which of the following must be true?

A. x-y > 0
B. x-y < 0
C. x+y > 0
D. xy > 0
E. xy < 0

Also, if you're short on time, and since you know that xy is not 0, then you know it MUST be greater than or less than 0, so you can narrow your choices down to D or E.

the only way lxl - lyl can equal lx+yl is when one number is positive and one number is negative or both are zero. So then xy must be negative. I think I read if somewhere in Bunuel's post

mnpqxyzt has given a great solution above. I would like to add here that it is possible that it doesn't occur to you that you should square both sides. If you do get stuck with such a question, notice that it says 'which of the following MUST be true'. So as a back up, you can rely on plugging in numbers. If you get even one set of values for which the condition does not hold, the condition is not your answer.

|x| - |y| = |x+y|
First set of non-zero values that come to mind is x = 1, y = -1
This set satisfies only options (A) and (E).
Now, the set x = -1, y = 1 will also satisfy the given equation.
But this set will not satisfy option (A).
Hence answer (E).

Krishma can you pls explain how to eliminate answer choices precisely.How can xy<0 not be true when one is negative and the other positive.

Ok.
Given: |x| - |y| = |x+y|
There are infinite set of values for x and y that satisfy this equation. Let us try one of them. Say x = 1, y = -1
a) x-y> 0 .......... 1 - (-1) > 0; 2 > 0; True
b) x-y< 0........... 1 - (-1) < 0; 2 < 0; Eliminate
c) x+y> 0........... 1 + (-1) > 0; 0 > 0; Eliminate
d) xy>0.............. 1 *(-1) > 0; -1 > 0; Eliminate
e) xy<0.............. 1 *(-1) < 0; -1 < 0; True

So I have two options that satisfy the assumed values of x and y.
We need to eliminate one of them.
They are
a) x-y> 0
e) xy<0

We see that x = 1, y = -1 satisfies both these inequalities. But option (a) is not symmetric i.e. if you interchange the values of x and y, it will not hold. That is, if x = -1 and y = 1, our original equation |x| - |y| = |x+y| is still satisfied but
a) x-y> 0 .............. (-1) - (1) > 0; -2>0; False. Eliminate
e) xy<0................. (-1)(1) < 0; True
Since option (e) still holds, it is the answer.

xy<0 is certainly true when one of them is negative and the other is positive.

Takeaways:

|x| + |y| = |x+y|
when x and y have the same signs - either both are positive or both are negative (or one or both of them are 0)

|x| - |y| = |x+y|
when x and y have opposite signs - one is positive, the other negative (or y is 0 or both are 0)

Karishma, I think the 2nd takeaway has some extra condition missed out :

|x|-|y|=|x+y|
if and only if
1) Both have opposite signs and
2) |x| >= |y|

because
x, y, |x|-|y| , |x+y|
5 ,-6, -1,1
5 ,-5 , 0 , 0
5, -4, 1 , 1

so, |5| cannot be greater than |-6| for the condition to occur

The takeaway is that if x and y satisfy the condition |x|-|y|=|x+y|, then they must have opposite signs (or y is 0 or both are 0).

But, x and y having opposite signs is not sufficient to satisfy the condition |x|-|y|=|x+y|. As you said, in that case we will also need to check for their absolute values. (Good thinking, btw)

I plugged in numbers:

X = -6, Y =3 and X = 6, Y = -3. The answer is E.

When you see |x| or absolute values being tested in the GMAT, this means testing
(1) positive and negative signs
(2) zeroes
(3) nature of |x| and |y| against each other

Looking at the equation we know |x| - |y| = |x + y| where xy is not 0.
(1) We figure that |x| and |y| are non-zeroes
(2) We figure that |x| - |y| > 0. This means |x| > |y|

Now, we figure the signs of the two variables by lining up possibilities
(1) both negative ==> x=-5,y=-4 ==> |-5| - |-4| = |-9| FALSE
(2) both positive ==> x=5,y=4 ==> |5| - |4| = |5 + 4| FALSE
(3) x is positive, y is negative ==> |5| - |-4| = |5-4| TRUE
(4) y is positive, x is negative ==> |-5| - |4| = |-5+4| TRUE

Now, let's find the answer
A) x-y > 0 ==> -5-(-4) = -1 FALSE
E) xy <0 ==> This is exactly what we need. x and y have both different signs.

Answer: E

I think the best strategy is to square up the equation.

|x| - |y| = |x+y|

> (|x| - |y|)^2 = (|x+y|)^2

> x^2 + y^2 - 2|x|.|y| = x^2 + y^2 + 2xy

> |x|.|y| = - xy

'Cause xy#0 and |x|.|y|>=0 --> xy<0

this is specific to Eden,

For your rule. what if X = -1 and Y = 2 then it would be l - 1 l - l 2 l = l -1 + 2 l ... 1 - 2 = 1 incorrect. is this correct?

Hi Bunuel,
I thought an absolute value of a product can never be a negative. Could you please explain in your equation how did you progress after getting this negative?

Basically, remember in these exercises to do square root in both sides to simplify: (|x+y|)^2 = (x+y)^2 or (|x|)^2 = x^2
And also remember that: |x||y|=|xy|

Therefore, with: -|x||y| = xy I would say: xy is always equal to something negative. Solution: E.

Let us separate LHS and RHS :
1] |x| - |y| =0 It follows the curve x=y in 1st and 2nd quadrant

2] |x+y|=0. It follows he curve x=-y in 4rth and 2nd quadrant
The common set is 2nd quadrant which implies E with the exception of origin as xy<>0

Am I wrong in my interpretation ? But this is the way I visualize the problem.

Interpreting the given

the distance between x and 0 is larger than that between y and 0 and this difference in distance of each of x and y from zero is equal to the distance between x and -y

draw this on a number line

..................x..............-y...............0.............y

this means that x and y has to be on opposite sides from zero , i.e. different signs thus xy<0