Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
@ 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
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 $b=1,y=2$.

share|cite|improve this answer
Thank you! ~ ${}{}{}{}{}$ – ᴊ ᴀ s ᴏ ɴ Sep 27 '12 at 7:56

$2\min (b,y)=b+y-|b-y|=2$ 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 $b=1$.

share|cite|improve this answer
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

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 $y=2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.