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March 26th, 2014, 04:12 PM   #1
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System of equation

I want to solve for u and v the following system:



I thought in to resolve by sum of product, so:



Now appears another complicated term: uČ+vČ

Anyway, I don't know how to solve this...
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March 26th, 2014, 05:22 PM   #2
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Re: System of equation

Quote:
Originally Posted by Jhenrique
k = u+v

3ak + b = 0 [1]
3ak^2+ 2bk + c = 0 [2]
-3ak^2 - bk = 0 : [1] * -k
=============
bk + c = 0 ; substitute back in:
b(u + v) + c = 0
bu + bv = -c
bu = -bv - c
u = (-bv - c) / b

Hmmm....dunno....
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March 26th, 2014, 06:16 PM   #3
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Re: System of equation

u and v cannot be uniquely determined by that system. Since both equations are (as Denis noticed) in (u + v) we will find that if is a solution, then so are and .

In other words, the equations must be either linearly dependent or inconsistent.
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March 26th, 2014, 07:45 PM   #4
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Re: System of equation

Quote:
Originally Posted by Jhenrique


How did you get denominator 3^2 a^2? The 3a is not squared:
I think your unnecessary brackets at start and end confused you; yes?

I make that:
-uv = (u^2 + v^2)/2 + b(u + v) / (3a) + c / (6a)
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March 26th, 2014, 10:51 PM   #5
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Re: System of equation

Quote:
Originally Posted by Jhenrique
I want to solve for u and v the following system:
and so and where is any number.
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March 27th, 2014, 03:12 AM   #6
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Re: System of equation

Quote:
Originally Posted by Dacu
..and where is any number.
or v = +-SQRT[c/(3a)] - u where u is any number
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