If |x|-|y|=|x+y|, then which of the following must be true?

Thanx for the reply Macfauz. Agree to your illustration but it will be great if you can go with the algebraic method.
Such modulus questions are painful if one doesn't the knows the correct approach.

The correct answer is E, but should be \(xy\leq{0}\) and not \(xy<0\). Otherwise, none of the answers is correct.
The given equality holds for \(x=y=0\), for which none of the given answers is correct.

The given equality can be rewritten as \(|x| = |y| + |x + y|\).
If \(y=0\), the equality becomes \(|x|=|x|\), obviously true.
From the given answers, D cannot hold, and A,B or C holds, depending on the value of \(x\). Corrected E holds.
If \(y>0\), then necessarily \(x\) must be negative, because if \(x>0\), then \(|x+y|>|x|\) (\(x+y>x\)), and the given equality cannot hold.
If \(y<0\), then necessarily \(x\) must be positive, because if \(x<0\), then again \(|x+y|>|x|\) (\(-x-y>-x\)) and the given equality cannot hold.
It follows that \(x\) and \(y\) must have opposite signs or \(y=0\).

Answer corrected version of E \(\,\,xy\leq{0}\).

Macfauz , the solution was perfect. Thanks.

Many thanks for the explanation.
It will be great if you elaborate on how to solve split modulus questions such as given above.

OR 0, that's why the correct answer should be \(xy\leq{0}\).
Otherwise, very nice solution.

I was solving on the basis of my previous comment where I had just added the phrase "where x and y are non zero" to the question.

But seeing as how it is much more probable to leave out a <= sign than an entire sentence, I guess the question frame is right and the answer should be xy <= 0

Squaring both sides of equation:-

x^2+y^2 - 2 lxl lyl= x^2 +y^2 +2xy

-lxl lyl= xy

for this to hold true xy should be -ve i.e <0

E is the answer

I think xy=0 wouldn't be the answer because when xy = 0 , either only x or only y can also be zero. Then if x=0 but y is not, then it won't hold true for xy=0. Let me know if I am wrong

say x = 4 and y = 2. then x+y = 6. |x|-|y| = 2. Not equal. Hence D is out. Also C is out.

Now say x = 4 and y = -2. then x+y = 2. |x|-|y| = 2. Equal. Hence E could be correct. Also B is out.

take Now say x = -4 and y = 2. then |x+y| = 2. |x|-|y| = 2. Equal. Hence A is out.

E is correct option.

If lxl - lyl = lx+yl and xy does not equal to o, 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, but as \(xy\neq{0}\), then in this case it's positive --> \(RHS=|x+y|>0\). So LHS must also be more than zero \(|x|-|y|>0\), or \(|x|>|y|\).

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------: \(x<y<0\) --> \(|x|-|y|=-x+y\) 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.

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