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March 26th, 2014, 04:12 PM  #1 
Senior Member Joined: Nov 2013 Posts: 137 Thanks: 1  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... 
March 26th, 2014, 05:22 PM  #2  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,587 Thanks: 1038  Re: System of equation Quote:
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....  
March 26th, 2014, 06:16 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra  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. 
March 26th, 2014, 07:45 PM  #4  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,587 Thanks: 1038  Re: System of equation Quote:
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)  
March 26th, 2014, 10:51 PM  #5  
Senior Member Joined: Apr 2013 Posts: 425 Thanks: 24  Re: System of equation Quote:
 
March 27th, 2014, 03:12 AM  #6  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,587 Thanks: 1038  Re: System of equation Quote:
 

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