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.
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$b+2=|b+y|$ 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 $b=1,y=2$.
$2\min (b,y)=b+y-|b-y|=2$ so that $\min (b,y)=1$. This implies that $b$ and $y$ are both positive so that $b+y=b+2$. Hence $y=2$ and $b=1$.
We always have $b+y-|b-y|=2\min(b,y)$ and from the first of the equations given, this is $2$. Therefore, we know that $$ \min(b,y)=1 $$ Since we know that $\min(b,y)=1$, we know that $b+y>0$ and so $|b+y|=b+y$. Therefore, the second equation is $b+y=b+2$, which gives us $$ y=2 $$ So the solution is $b=1$ and $y=2$.