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June 6th, 2014, 08:31 AM   #11
Joined: Jun 2014
From: rhsfhb

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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...
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June 6th, 2014, 08:54 AM   #12
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Originally Posted by skipjack View Post
If you still can't understand soroban's approach, read the first two paragraphs of this article.
An alternative (although longer) approach is to split the domain of $u$ into intervals in which each of the terms on the left is of constant sign.
(2-u) + (3-u) &= 1 & u &\le 2 \\
(u-2) + (3-u) &= 1 & 2\le u &\le 3 \\
(u-2) + (u-3) &= 1 & 3\le u \\
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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