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 January 8th, 2015, 02:17 PM #1 Member   Joined: Feb 2013 From: London, England, UK Posts: 38 Thanks: 1 Gaussian Elimination Type Method Needed Hi, I'm struggling a bit with the following problem: 3 + 14*x = 1 + 25*y = 9 + 288*z I have a series of these equations which I need to solve, with different first terms in each case and one of these first terms changes for each equation: e.g. the second such equation is: 3 + 14*x = 1 + 25*y = 47 + 288*z Obviously, I'd prefer not to brute force this problem, or I wouldn't be posting here. But I just thought there might be an analytical solution, because as I progress further with this, the x, y and z coefficients will get larger. Also, if you could advise as to a general solution for any number of terms, I'd welcome it. Muchos gracias amigos.  January 9th, 2015, 05:25 AM   #2
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Quote:
 Originally Posted by Jopus Hi, I'm struggling a bit with the following problem: 3 + 14*x = 1 + 25*y = 9 + 288*z
This is the same as the two equations 3+ 14x= 1+ 25y and 1+ 25y= 9+ 288z. (The "third" equation, 3+ 14x= 9+ 288z is not independent or the first two.) With two equation in three unknowns you cannot solve for all three but can solve for two of the unknowns in terms of the third. For example, if I choose to solve for x and y in terms of z, I see that subtracting 1 from each side of 1+ 25y= 9+ 288z I have 25y= 8+ 288z. Dividing by 28, y= 8/25+ (288/25)z. I could now put that into 3+ 14x= 1+ 25z but, since the "third" equation, 3+ 14x= 9+ 288z, can be derived from the first two, it is easier to solve that for x: 14x= 6+ 288z so x= 6/14+ (288/14)z= 3/7+ (144/7)z.

Quote:
 I have a series of these equations which I need to solve, with different first terms in each case and one of these first terms changes for each equation: e.g. the second such equation is: 3 + 14*x = 1 + 25*y = 47 + 288*z Obviously, I'd prefer not to brute force this problem, or I wouldn't be posting here. But I just thought there might be an analytical solution, because as I progress further with this, the x, y and z coefficients will get larger. Also, if you could advise as to a general solution for any number of terms, I'd welcome it. Muchos gracias amigos. If a+ bx= c+ dy= e+ fz then we have a+ bx= e+ fz and c+ dy= e+ fz. So bx= (e- a)+ fz and then x= (e- a+ fz)/b. Similarly, dy= (e- c)+ fz so y= (e- c+ fz)/d. Tags elimination, gaussian, method, needed, type 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 Maurux Linear Algebra 3 January 20th, 2015 07:03 AM ewuc Algebra 2 September 17th, 2011 12:59 PM wulfgarpro Algebra 5 June 12th, 2010 03:00 PM allanc16 Linear Algebra 2 August 27th, 2009 08:14 PM julian21 Calculus 1 May 10th, 2009 10:35 PM

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