Explain how can we graph the equation $y+|y|=x+|x|$.( Relation involving absolute values) According to my answer it's graph would be $x=y$ when $x,y\ge0$ and the whole third quadrant including $x=0$ and $y=0$, when $y$ is not a function of $x$. When it is a function of $x$ then the graph is $x=y$ when $x,y\ge0$ and the negative $x$-axis. According to a graphing utility (like Desmos) my answer is wrong. Please can someone correct me? I would be thankful. 
 A: Without thinking too much :), just look exhaustively
at all 4 possible cases and see what they give you.  
1) $x \ge 0, y \ge 0$
Then obviously you get $2y = 2x$ and then $y=x$
This when plotted is a ray: the bisector of 1st quadrant.   
2) $x \ge 0, y \le 0$
Then you get $x = 0$ 
So any pair $(0, y \le 0)$ is a solution here.   
3) $x \le 0, y \le 0$
Here you get $0=0$ which is always true.
So any pair $(x \le 0, y \le 0)$ is a solution.   
4) $x \le 0, y \ge 0$
Then you get $y = 0$
So any pair $(x \le 0, 0)$ is a solution here.    
So eventually you get this plot below
/ or maybe a slightly better one because I am bad painter :) /.        
Final answer:
All points from 3rd quadrant, the contour of 3rd quadrant,
and the bisector of 1st quadrant. This is the graph of your
equation.   
If some software gives you a different thing - don't trust it :) 

A: When $x,y \geq 0$, the equation becomes:
$$y+y=x+x \leftrightarrow x=y$$
When $x<0 \land y>0$ or $x>0 \land y<0$ we have:
$$y+y=x-x\leftrightarrow y=0$$
When both $x,y <0$, we can write:
$$y-y=x-x$$
which is always correct, and it has a graph in $3$rd quadrant. Here's a graph:

Obsiously, here the $3$rd quadrantis not highlined, but it's correct.
A: Rewriting we get
$$y+|y| = x+|x| \Leftrightarrow y-x = |x| -|y|$$
Squaring gives 
$$\Rightarrow xy = |xy| \Rightarrow xy \geq 0$$
So, two cases:
$$x,y> 0 \Rightarrow y=x$$
$$x,y \leq 0\Rightarrow 0=0 \Rightarrow \text{the whole 3rd quadrant}$$
A: Our goal would be to get an equation of the form $y = ...$ , hoping the equation defines a function ( though it may not be the case). 
So we have to get rid of the absolute value expression on the LHS. 
Suppose $y\geq0$, in that case, $|y|=y$, and we have : 
$y+y = x+|x| $
$\iff 2y = x+|x| $
$\iff y = \frac {x+|x| } {2}$
$\iff (x \geq 0 \rightarrow y = \frac {x+x } {2} = x) \land (x\lt 0 \rightarrow y = \frac {x+ (-x) } {2} = 0)$
So the left part of the X-axis ( the line $y=0$ with $x\lt0$) and the line $y=x$(  with $x\geq0$) are two subsets of the solution set. 
Now, if $y\lt 0$ , we have $|y| = -y$ , and the equation becomes 
$y+|y| = x+|x| $
$\iff y+ (-y) = x+|x|$
$ \iff 0 = x+|x|$
$\iff  -x = |x|$
$\iff  |x| = -x$ 
$ \iff x\leq 0$ 
So, every point $(x,y)$ with a negative $y$ and an $x$ negative or null  satisfies the equation ( which, therefore, is not the equation of a function). 
So the set $\{(x,y) | (y\lt 0 \land x = 0)  \lor (y \lt 0  \land x\lt 0)\}$ is a subset of the solution set. 
It means that the below part of the Y axis is a part ( subset) of the solution set, as well as the whole south west quadrant of the X-Y plane . 
Consequently, the solution set is the union of 4 sets : 
(1) the strictly negative part of the X axis 
(2) the line $y=x$ in the north east quandrant ( with point $(0,0)$ included). 
(3) the strictly negative part of the Y axis 
(4) the whole south west quadrant ( the axis parts being excluded) 
