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June 6th, 2014, 07:31 AM  #11 
Newbie Joined: Jun 2014 From: rhsfhb Posts: 24 Thanks: 0 
Well, i understood what he said, but i thought that there are more solutions, beacause when i finished whole equation his solution was just one of many, anyway i think i got it now. Probably... 
June 6th, 2014, 07:54 AM  #12  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,634 Thanks: 2620 Math Focus: Mainly analysis and algebra  Quote:
\begin{align*} (2u) + (3u) &= 1 & u &\le 2 \\ (u2) + (3u) &= 1 & 2\le u &\le 3 \\ (u2) + (u3) &= 1 & 3\le u \\ \end{align*} The middle equation is valid for all values of $u$ (for which the equation applies) since $1 \equiv 1$. Sometimes in systems like this (but not in this case), one or more of the equations produces a value for $u$ which is outside the interval for which the equation is valid. In this case the value of $u$ produced is not a valid solution. You also have to be careful about whether you use $\le$ or $\lt$.  

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