MHB Graph Equation: x-|x|=y-|y| - Seeking Help

[NoWay1]

The equation is x-|x|=y-|y| and I need to make a graph for it.
So I thought I should solve it by breaking it down to 4 different equations, which would be -

x - x = y - y => 0=0
x - x = y + y => 0=2y => y = 0
x + x = y - y => 2x=0 => x = 0
x + x = y + y => 2x=2y => x = y

But this isn't correct, riight? I don't know what else to do, kinda lost here.
Would love some help, thanks.

 

[Greg]

Hi NoWay and welcome to MHB! :D

What if:

y = x

y $\ne$ x

?

 

[I like Serena]

The equation is x-|x|=y-|y| and I need to make a graph for it.
So I thought I should solve it by breaking it down to 4 different equations, which would be -

x - x = y - y => 0=0
x - x = y + y => 0=2y => y = 0
x + x = y - y => 2x=0 => x = 0
x + x = y + y => 2x=2y => x = y

But this isn't correct, riight? I don't know what else to do, kinda lost here.
Would love some help, thanks.

Hi NoWay! Welcome to MHB! ;)

It's correct all right, but you've left out the conditions.
The first equation is for the first quadrant ($x,y\ge 0$), so that $|x|=x,|y|=y$.
The second equation is for the 4th quadrant ($x\ge 0, y<0$).
The third equation is for the 2nd quadrant ($x<0, y\ge 0$).
And the fourth equation is for the 3rd quadrant ($x,y<0$).

Since the equation is always true in the first quadrant, all points in the first quadrant belong to the graph!
Your 2nd and 3rd equations show that only the boundary with the first quadrant belongs to the graph in those respective quadrants.
And the 4th equation shows that the line $y=x$ belongs to the graph with negative x and y.

In other words, the graph is:
View attachment 5967

 

[NoWay1]

Wow, this is the first time I see a whole quadrant being in the graph, I would never have figured that out myself, thanks a lot for your clear explanation!