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Re: I need a strategy for this one. [#permalink]
09 Sep 2009, 10:35

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

I want to know wether there is a rule involved that i am missing, or an effective strategy to tackle that kind of questions

Thank you

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). _________________

Re: I need a strategy for this one. [#permalink]
18 Jun 2011, 20:49

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. _________________

Re: I need a strategy for this one. [#permalink]
19 Jun 2011, 03:57

Expert's post

AnkitK wrote:

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) _________________

Re: I need a strategy for this one. [#permalink]
21 Jun 2011, 09:25

VeritasPrepKarishma wrote:

AnkitK wrote:

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 _________________

Re: I need a strategy for this one. [#permalink]
21 Jun 2011, 10:42

Expert's post

krishp84 wrote:

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) _________________