How would you go about graphing a set of points (x,y) that satisfies the following equation?

Question

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|

Answer 1

Hint: what is $x+|x|$ when $x<0$?

Answer 2

You need to analyse this cases:

1. 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.

2. 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)

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

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

Answer 3

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 $(x,y) \in S$.