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

\(|x| - |y| = |x+y|\) , square both sides \((|x| - |y|)^2 = (|x+y|)^2\), \(x^2 - 2|xy|=y^2 = x^2+2xy+y^2\), eliminate similar terms and divide by 2
\(-|xy|=xy\) from here multiply by -1 just for clarity and obtain \(|xy|=-xy\)
remember that \(|abs|\geq{0}\) to the equation translates into \(-xy\geq{0}\) or \(xy\leq{0}\)

The text says that \(xy\neq{0}\) so \(xy\leq{0}\)=>\(xy<0\)

but I don't understand how the -2|xy| simplifies to + 2xy.

And why do we multiply by -1?!

You arrive at \(-2|xy|=2xy\) right? Now divide by 2: \(-|xy|=xy\).
From here the -1 multiplication is not necessary to arrive at the result, but IMO it helps.

Keeping in mind that \(|abs|\geq{0}\), we can write \(-|abs|\leq{0}\), right?
The abs expression stands for any value or expression we find inside "| |", xy included.

Now we can merge the equations \(xy=-|xy|\) with \(-|abs|\leq{0}\) into \(xy=-|xy|\leq{0}\) => \(xy\leq{0}\)

Yes, but why does -2|xy| = 2xy?

and is this example posted by Bunuel...
2. ----y--0------x--: y<0<x --> |x|-|y|=x+y and |x+y|=x+y --> x+y={x+y}. Correct.

How does |x| - |y| = x+y? wouldn't it be x-y?

I am sorry about the stupid questions. This is an extremely difficult topic for me and I am having a tough time picking it up.

Yes, but why does -2|xy| = 2xy?

and is this example posted by Bunuel...
2. ----y--0------x--: y<0<x --> |x|-|y|=x+y and |x+y|=x+y --> x+y={x+y}. Correct.

How does |x| - |y| = x+y? wouldn't it be x-y?

I get the mechanics of flipping the signs when y is negative, but I guess I don't understand the logic.

If I take an absolute value of a number (always positive) then subtract from it an absolute value of a smaller number, which is negative how does that end up being X+Y? Are we looking just for the values of x and y, as opposed to the values of |x|-|y|?

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?

By the way, I would be extremely careful with using the "squaring" approach to solve absolute value problems.

For example, if |x|=-|x|, what is x?

The answer is obvious. x=0

But,

(|x|)^2 = (-|x|)^2 --> x^2 = x^2 --> x can be any numbers (incorrect!).

\(|x|-|y|=|x+y|\) --> 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.

Hope it helps.

Could you please explain following line as above defined in solution

note that (|x+y|)^2=(x+y)^2

Thanks.

\(|x|-|y|=|x+y|\) --> 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.

Hope it helps.