Solve a simultaneous equation. How do we solve $|b-y|=b+y-2\;and\;|b+y|=b+2$? I have tried to square them and factorize them but got confused by and and or conditions.

$b+2=|b+y|$ which is real, so is $b$
$y+b-2=|b-y|$ which is real, so is $y+b-2$ and $y$
(1)If $b \ge y, b-y=b+y-2\implies y=1 \implies |b+1|=b+2$ and $b \ge y=1$
So, $b+1 >0\implies |b+1|=b+1=b+2$ which has no finite solution.
(2) If $b<y, y-b=b+y-2\implies b=1, y>b=1$ 
So, $|1+y|=3\implies y+1=3\implies y=2$
The only solution $b=1,y=2$.
