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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?

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2 Answers 2

up vote 2 down vote accepted

Divide into regions like so:

Case 1: Assume $s\ge 1$

Your equation reduces to:

$s-1-3(s+1)+(s+2)=t$

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

Your equation reduces to:

$-(s-1)-3(s+1)+(s+2)=t$

and so on.

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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? –  Jobin Idiculla 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. –  Jobin Idiculla Jun 5 '13 at 18:14
    
That sounds about right. –  response Jun 5 '13 at 18:24
    
Thank you for the help, response! :) –  Jobin Idiculla 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)….$$CONCLUSION…~~$$SO.now 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...:)

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