# X + |x| = y + |y| ?

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 cant find in my work. Who knows how this quadrant is found?

greetz
Ivar

Last edited:

## Answers and Replies

Quinzio

Basically , $\forall(x,y): x<0, y<0$ satisfy the equation giving $0=0$

ArcanaNoir

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?

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?