How is the graph of x + |x| = y + |y| drawn?

[ialink]

x + |x| = y + |y| ??

Homework Statement

Draw the graph of x + |x| = y + |y|

The Attempt at a Solution

x + |x| = y + |y|
2x = y + |y| for x \geq 0
0 = y + |y| for x < 0

2x = y + |y|
2x = 2y which is x = y for y \geq 0
2x = 0 for y < 0

0 = y + |y|
0 = 2y for y \geq 0
0 = 0 for y < 0

The answer is the graph y = x for x > 0 which i can find in my work but it is also the entire quadrant formed by x < 0 and y < 0. That quadrant i can't find in my work. Who knows how this quadrant is found?

greetz
Ivar

 

[Quinzio]

Basically , \forall(x,y): x&lt;0, y&lt;0 satisfy the equation giving 0=0

 

[ArcanaNoir]

You addressed:
x>0 and y>0
x>0 and y<0

You did not specifically address the two separate cases when x<0.
if x<0 and y>0? I know, I'm just saying make sure you've thought about it...
and x<0 and y<0?
And then of course, what about y is 0 or x is 0?

 

[ialink]

You did not specifically address the two separate cases when x<0.
if x<0 and y>0? I know, I'm just saying make sure you've thought about it...
and x<0 and y<0?
And then of course, what about y is 0 or x is 0?

I have added the missing cases that indeed were missing. Thank you.

Basically , ∀(x,y):x<0,y<0 satisfy the equation giving 0=0

In reply quinzo's comment I indeed understand that for each negative value for x and y results in 0 = 0 so the entire quadrant is a valid combination of x and y.

Still I'm unsure having proved that the entire quadrant is consists of possible solutions. Though I'm Not questioning they are. Have i Proved it with the added cases?

 

[ialink]

Okay thanks guys i figured it out. The values for x < 0 and y < 0 are only valid if the combination meets the requirement 0 = 0 which offcourse is for all values in this domain.

Thank you!