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

Question

If |x| - |y| = |x + y| and xy not equal zero, 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

Bunuel · Accepted Answer

If |x| - |y| = |x + y| and xy not equal zero, 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

\(|x|-|y|=|x+y|\) --> square both sides --> \((|x|-|y|)^2=(|x+y|)^2\) --> note that \((|x+y|)^2=(x+y)^2\) --> \((|x|-|y|)^2=(x+y)^2\) --> \(x^2-2|xy|+y^2=x^2+2xy+y^2\) --> \(|xy|=-xy\) --> \(xy\leq{0}\), but as given that \(xy\neq{0}\), then \(xy<0\).

Answer: E.

Another way:

Right hand side, \(|x+y|\), is an absolute value, which is always non-negative. So LHS must also be more than or equal to zero: \(|x|-|y|\geq 0\), or \(|x|\geq |y|\).

So we can have following 4 scenarios:

1. ------0--y----x--: \(0<y<x\) --> \(|x|-|y|=x-y\) and \(|x+y|=x+y\) --> \(x-y\neq{x+y}\). Not correct.

2. ----y--0------x--: \(y<0<x\) --> \(|x|-|y|=x+y\) and \(|x+y|=x+y\) --> \(x+y={x+y}\). Correct.

3. --x------0--y----: \(x<0<y\) --> \(|x|-|y|=-x-y\) and \(|x+y|=-x-y\) --> \(-x-y={-x-y}\). Correct.

4. --x----y--0------: \(x \geq y<0\) --> \(|x|-|y|=-x+y\) and \(|x+y|=-x-y\) --> \(-x+y\neq{-x-y}\). Not correct.

So we have that either \(y<0<x\) (case 2) or \(x<0<y\) (case 3) --> \(x\) and \(y\) have opposite signs --> \(xy<0\).

Answer: E.

Hope it helps.

mnpqxyzt · Answer

|x| - |y| = |x+y|
=> (|x| - |y|)^2 = |x+y|^2 = (x+y)^2
=> x^2 - 2|x||y| + y^2 = x^2 + 2xy + y^2 
=> |xy| = -xy
because xy != 0 so xy <0
Ans: E

chetan2u · Answer

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

SnehaC · Answer

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.

Eden · Answer

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

KarishmaB · Answer

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).

AnkitK · Answer

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.

KarishmaB · Answer

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)

krishp84 · Answer

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

KarishmaB · Answer

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)

mbaiseasy · Answer

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

lunar255 · Answer

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

WholeLottaLove · Answer

How does x^2-2|xy|+y^2 become x^2+2xy+y^2? What happened to the -2|xy|? Even if xy were negative |xy| will be positive and pos. * neg (In this case, the -2) = negative, right?

Also, how do we know that xy is negative?

walker · Answer

Another 15 sec approach:

1. The equation isn't sensitive to changing signs of x and y simultaneously. In other words, the equation is the same for (-x,-y). So, A, B, C are out.
2. If x and y had the same sign, |x+y| would be always greater than |x| (and |x| - |y|). So, x and y have different signs and only E remains.

Bunuel · Answer

Step by step:

Step 1:  square  |x|-|y|=|x+y|  -->  (|x|-|y|)^2=(|x+y|)^2  (note that  (|x+y|)^2=(x+y)^2 ) -->  (|x|-|y|)^2=(x+y)^2  -->  x^2-2|xy|+y^2=x^2+2xy+y^2

Step 2:  cancel x^2+y^2 in both sides to get  -2|xy|=2xy

Step 3:  reduce by -2 to get  |xy|=-xy

Step 4:  apply absolute value property, which says that the absolute value is always non-negative -->  LHS=|xy|  is always non negative, thus  |xy|\geq{0} , therefore RHS is also non-negative:  |xy|=-xy\geq{0}  -->  -xy\geq{0} .

Step 5:  multiply by -1 and flip the sign of the inequality -->  xy\leq{0}

Step 6:  apply info given in the stem --> since given that  xy
eq{0} , then  xy<0 .

Hope it's clear.

PerfectScores · Answer

Let us plug in and eliminate: x = -2 and y = 2 OR x = 2 and y = -2 then |2| - |-2| = |2-2|, let us check the options A. ELIMINATED (-2 - 2>0) B. ELIMINATED (2 - (-2) <0) C. ELIMINATED (2 - 2 > 0) D. ELIMINATED (-2(2) < 0 ) E. The only one which stays Hence answer is E.

Natansha · Answer

In the second approach, how is |x + y | > 0 , we can have |-5 + 5| =0, while still satisfying xy = 0
and |-5| + |5| = 0 as well
how do we have |x + y | > 0 and |x| + |y|>0

Bunuel · Answer

We'd have  |x + y | \geq 0  and  |x| + |y|\geq 0 . Edited.

Manbhav7 · Answer

square both sides  
  (|x|−|y|)2=(|x+y|)2(|x|−|y|)2=(|x+y|)2

x2−2|xy|+y2=x2+2xy+y2x2−2|xy|+y2=x2+2xy+y2  
 
  |xy|=−xy|xy|=−xy

xy≤0xy≤0, because xy not equal to zero

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

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