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January 8th, 2015, 02:17 PM   #1
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Question 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.

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January 9th, 2015, 05:25 AM   #2
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Quote:
Originally Posted by Jopus View Post
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.
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