Points on a plane satisfying an equation

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SUMMARY

The equation y + |y| = x + |x| can be analyzed under four distinct conditions based on the signs of x and y. For both x and y non-negative, the solution is y = x. If y is non-negative and x is negative, the solution simplifies to y = 0. Conversely, if y is negative and x is non-negative, the solution yields x = 0. Lastly, when both x and y are negative, all values satisfy the equation, resulting in the trivial solution 0 = 0. The discussion emphasizes the correctness of these conclusions and suggests simplifying the reasoning by focusing on two cases regarding x + |x|.

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Homework Statement
Indicate the points on the place ##xOy## which satisfy the equation : ##\mathbf{y+|y|-x-|x| = 0}##.
Relevant Equations
The modulus of a variable ##|x| = x\; \text{if}\; x\geq 0## and ##-x\; \text{if}\; x\leq 0##
We can write the equation given as ##y+|y| = x+|x|##

This has a few conditions.

(1) If ##\underline{y\geq 0\; \text{and}\; x\geq 0}##, we obtain ##2y = 2x \Rightarrow \boxed{y = x}##.
(2) If ##\underline{y\geq 0\; \text{and}\; x < 0}##, we obtain ##2y = 0 \Rightarrow \boxed{y = 0}##.
(3) If ##\underline{y < 0\; \text{and}\; x\geq 0}##, we obtain ##2x = 0\Rightarrow \boxed{x = 0}##
(4) If ##\underline{y < 0\; \text{and}\; x < 0}##, we obtain the trivial solution ##0 = 0##. But since 0 =0 always, this means all values of ##y<0\;\text{and}\; x<0## are solutions to the equation.

1607437660789.png
I plot the graph alongside for the solution and regions.

1607437791599.png

On the left is the solution from the book.

Though my answer looks correct, is the reasoning alright?
 
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That looks good to me.
 
You can slightly simplify the logic by considering just two cases, x+|x| zero or nonzero.
 

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