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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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@ Jasper Loy: No other condition. –  ᴊ ᴀ s ᴏ ɴ Sep 26 '12 at 12:23
    
The language seems ambiguous here. In one interpretation, $b$ is given and you are expected to solve the two equations separately for $y$. In another, you are expected to solve the system of two equations for two unknowns $b$ and $y$. Common naming conventions support the first interpretation, since $a$, $b$, $c$ etc. are often used for given parameters, while $x$, $y$, $z$ etc. are used for variable or unknown quantities. Maybe you can infer what is meant from context. –  Harald Hanche-Olsen Sep 26 '12 at 12:28
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3 Answers

up vote 2 down vote accepted

$b+2=|b+y|$ 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 $b=1,y=2$.

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Thank you! ~ ${}{}{}{}{}$ –  ᴊ ᴀ s ᴏ ɴ Sep 27 '12 at 7:56
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$2\min (b,y)=b+y-|b-y|=2$ 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 $b=1$.

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what happens if $b+y\lt0$? You cannot replace $|b+y|$ with $b+y$ unless you know that $b+y\ge0$. –  robjohn Sep 26 '12 at 18:24
    
correct you are. (+1) –  robjohn Sep 26 '12 at 18:34
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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 $y=2$.

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