Algebra Pre-Algebra and Basic Algebra Math Forum

 June 6th, 2014, 08:31 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, 08:54 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 $\lt$. Tags equation, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post chuackl Linear Algebra 5 May 30th, 2014 01:02 AM p1xel Algebra 5 June 5th, 2012 01:31 PM FreaKariDunk Applied Math 8 April 25th, 2012 08:25 AM Sanju Elementary Math 4 December 23rd, 2010 10:39 AM KranImpire Algebra 9 April 7th, 2010 09:38 AM

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