If $x>y$, is $|x+y|$ or $|x-y|$ bigger, or neither?

Question

Question: If $x>y$, is $|x+y|$ or $|x-y|$ bigger, or neither?

I got this question wrong on a GRE practice test, and now I know the correct answer, but I am curious what your thought process is for solving it efficiently.

Edit: This is the full problem as given in the GRE. Problems like this as given in the GRE always have 4 possible solutions: 1) |x+y| is bigger, 2) |x-y| is bigger, 3) |x+y|=|x-y|, or 4) it is not possible to determine.

Answer 1

Depends.

If $x=2$ and $y=1$, then $|x+y|=3 > 1=|x-y|$.

If $x=2$ and $y=-5$, then $|x+y|=3 <7= |x-y|$.

Addendum: And of course if $x=2$ and $y=0$, then $|x+y|=|x-y|$ (as Love invariants noted in a comment)

Answer 2

For $x=1$ and $y=-1$ we have $$|x+y|<|x-y|.$$

For $x=2$ and $y=1$ we have $$|x+y|>|x-y|.$$

Answer 3

Both quantities are non-negative, so $>$ orders them the same way it does their squares, which are $x^2+y^2+2xy,\,x^2+y^2-2xy$. Therefore, $|x+y|>|x-y|$ iff $xy>0$, i.e. $x,\,y$ are both positive or both negative.

Answer 4

The diagram below shows the plots of $y = x$ and $y = -x$.

Then $|x+y|$ is the taxicab ($L_1$) distance of a point $(x, y)$ from the red line, while $|x-y|$ is the taxicab distance of that same point from the blue line.

(Note that those taxicab distances are equal to $\sqrt{2}$ times the Euclidean ($L_2$) distances.)

Therefore, either quantity can be greater. They are equal along the axes.

Answer 5

Let $$y=x-t, t>0$$ Then $$|x+y|=|2x-t|$$ $$|x-y|=|t|$$

Observe what happens as we vary our variables.

If $x<0$, one can immediately see that $$|x+y|>|x-y|$$

On the contrary, if $t>x$ while $x>0$, we can establish that: $$0<|2x-t|<|t|$$ and hence $$|x+y|<|x+y|$$ This occurs when $y<0$.