My Math Forum Equation problem

 Algebra Pre-Algebra and Basic Algebra Math Forum

 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
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Quote:
 Originally Posted by skipjack 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.
\begin{align*}
(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 \\
\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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