Linear Diophantine EquationsDate: 11/27/97 at 15:12:50 From: Ulrich Richers Subject: Linear Diophantus Equations Recently my son's homework has included several problems with a number of linear equations smaller than the number of unknown parameter. All solutions had to be integer and greater than zero. I have a math book from East Germany with an example but without directions for how to get there. 3x - 2y -5 = 0 there are several solution pairs: x = 2t + 1 y = 3t - 1 How was the t-variable introduced, and what is the general method? Thank you and apologies for my rusty math, Ulrich Richers Date: 11/27/97 at 19:24:08 From: Doctor Anthony Subject: Re: Linear Diophantus Equations Divide through 2, the smaller coefficient, and so 3x/2 - y = 5/2 x + x/2 - y = 2 + 1/2 x-y + (x-1)/2 = 2 .........(1) and so (x-1)/2 must be an integer, say t, and we get (x-1)/2 = t x-1 = 2t x = 2t + 1 Substitute this for x in (1) 2t+1 - y + t = 2 3t - 1 = y We therefore have x = 2t+1, y = 3t-1 where t is any integer. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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