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

I saw this interesting problem in a math puzzle forum:-

Find all integral values of $t$ such that the equation $|s-1| - 3|s+1| + |s+2| = t $ has no solutions.

How does one approach these kind of problems?

share|cite|improve this question
up vote 2 down vote accepted

Divide into regions like so:

Case 1: Assume $s\ge 1$

Your equation reduces to:


Case 2: Assume $-1 \le s \le 1$

Your equation reduces to:


and so on.

share|cite|improve this answer
Thank you for the quick response. So there are 4 cases, from which I get 4 equations as:- $ -s-2 = t$, $ -3s=t $, $ 3s+6=t$ and $ s+2=t $. But this does not seem to make any sense to me. To get a unique value for t, only two of those equations are needed, right? And how do you find those values of $t$ which do not give a solution? – Anton Schigur Jun 5 '13 at 18:03
These are four mutually exclusive cases. Consider Case 1: We know that $-s-2=t$ and that $s \ge 1$ For which values of $t$ will the above system not have any solutions for $s$? – response Jun 5 '13 at 18:05
Oh, I get it now! Sheesh, that was rather silly of me, sorry! So, for $ -s-2=t$ and $s \ge 1$, for the above system to not have any solution, $t>-3$, right? And I do this for all the cases and get the corresponding intervals for $t$, and the union of those intervals would form all the values of $t$, for which there are no solutions to the original equation. – Anton Schigur Jun 5 '13 at 18:14
That sounds about right. – response Jun 5 '13 at 18:24
Thank you for the help, response! :) – Anton Schigur Jun 5 '13 at 18:26

First you have to know the properties of absolute value to confront these kind of problems.When x is negative, $ |x| =-x $ and when x is positive $ |x|=(+x) $. Then you check if you are given $|x-a|$ then for $x>a$ ,$|x-a|$=positive or for $x<a$ ,$|x-a|$ is negative.

Now look to your problem:$|s-1|-3|s+1|+|s+2|=t$,Consider the critical points of the curve $y=|x-1| - 3|x+1| + |x+2|$ are the points where $x=(-2),x=(-1),x=1$.So there are some distinctive parts to be considered.$$ case 1. $$When $s<-2$;$|s-1|=(-ve),|s+1|=(-ve),|s+2|=(-ve)$.Then the $|s-1|-3|s+1|+|s+2|$ becomes $=-(s-1)-3*{-(s+1)}+{-(s+2)} $=$(1-s+3s+3-s-2)$=$(s+2)$...So “t” can take value within the range (-infinity,0) for all$ s<-2$. $$ case 2.$$When $-2<s<-1$;$|s-1|=(-ve),|s+1|=(-ve),|s+2|=(+ve)$.Then the”t” becomes=$(|s-1|-3|s+1|+|s+2|)$=$(1-s+3s+3+s+2)$=$(3s+6)$ .Therefore “t” can take value within the range (0,3) for all $-2<s<-1$..$$case 3.$$when $-1<s<1$;$|s-1|=(-ve),|s+1|=(+ve),|s+2|=(+ve)$.Then the “t” becomes=$(1-s-3(s+1)+s+2)=1-s-3s-3+s+2=-3s $.So “t” can take value within the range (-3,3) for all$ -1<s<1$...$$case 4.$$ When $x>1$;$|s-1|=(+ve),|s+1|=(+ve),|s+2|=(+ve)$.Then the “t” becomes=$(s-1-3s-3+s+2)=(-s-2)$.So, “t” can take value within range (-3,-infinity)….$$CONCLUSION…~~$$ you can see that as “s” varies over real values $“t” can get its maximum value as [+3](including)$ and $”t” can have minimum value as (-infinity)$...Then to find all such integral values of “t” such that the equation given has no solution “t” must lie within the range (+3,+infinity]...I think I could help you with my knowledge...Thanks in advance if you point out any of my fault to be rectified...:)

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.