Explain how can we graph the equation $y+|y|=x+|x|$.( Relation involving absolute values)

Question

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.

Answer 1

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 :)

Answer 2

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.

Answer 3

Rewriting we get

$$y+|y| = x+|x| \Leftrightarrow y-x = |x| -|y|$$

Squaring gives $$\Rightarrow xy = |xy| \Rightarrow xy \geq 0$$

So, two cases:

$$x,y> 0 \Rightarrow y=x$$

$$x,y \leq 0\Rightarrow 0=0 \Rightarrow \text{the whole 3rd quadrant}$$

Answer 4

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)