How would you go about graphing a set of points (x,y) that satisfies the following equation? Could you explain how this is different from graphing any other equation? 
What confuses me is that there are two y-values and two x-values. 
The equation is: y+|y| = x+|x|
 A: Split it into four equations:
$y,x \geq 0 \implies 2y =2x \implies y=x$
$x \leq 0 \leq y \implies 2y=0 \implies y=0$.
$y \leq 0 \leq x \implies 2x=0 \implies x=0$.
$y,x \leq 0 \implies 0=0$, so here this is true for all values of $x$ and $y$.
Hence, we take the graphs of these four functions and combine them to get the solution set:
$$
S = \{ y \leq 0 ,x \leq 0\} \cup \{ y =0\} \cup \{x=0\} \cup \{x=y\}
$$
The second and third set absorb into the first, so the answer would be:
$$
S = \{ y \leq 0 ,x \leq 0\} \cup \{x=y\}
$$
Hence $|x| + x = |y| + y$ if and only if $(x,y) \in S$.
A: Hint: what is $x+|x|$ when $x<0$?
A: You need to analyse this cases:


*

*y>= 0, x >= 0
Here you have equation y + y = x + x, that is 2y = 2x, y = x. You draw this only in the first quadrant.

*y >= 0, x < 0
Here you have equation y + y = x - x, that is 2y = 0, y = 0. You draw this only in the second quadrant (it is negative X axis)

*y < 0, x >=0
y - y = x + x, 0 = 2x, x = 0. You draw this in fourth quadrant, it is negative Y axis

*y < 0, x < 0
y - y = x - x , 0 = 0. You have to color whole third quadrant, without the axes.
