Answer
See graph and explanations.
Work Step by Step
To graph the equation, we need to consider several different cases:
Case 1. $x\geq0$, we have $y+|y|=x+x$ which leads to two more cases:
(i) $y\geq0$, we have $y+y=2x$ or $y=x$ which represent a line in the figure.
(ii) $y\leq0$, we have $y-y=2x$ or $x=0$. However, as $x=0$ leads to $y=0$, this gives a point $(0,0)$ or the origin.
Case 2. $x\leq0$, we have $y+|y|=x-x=0$ which leads to two more cases:
(i) $y\geq0$, we have $y+y=0$ or $y=0$ which represents the negative x-axis.
(ii) $y\leq0$, we have $y-y=0$ or $0=0$. This means that any point in Quadrant III will satisfy the equation.
