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.
Solve a simultaneous equation.
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0The 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. – 2012-09-26
3 Answers
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 $y=2$.
$2\min (b,y)=b+y-|b-y|=2$ 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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0correct you are. (+1) – 2012-09-26
$b+2=|b+y|$ 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
So, $|1+y|=3\implies y+1=3\implies y=2$
The only solution $b=1,y=2$.