Teacher2Teacher |
Q&A #3962 |
From: Pat Ballew
To: Teacher2Teacher Public Discussion
Date: 2000122121:41:35
Subject: Re: Re: A special linear equation
I will proceed with your assumption that x>y>0 although I see no reason to exclude the negatives in this problem (probably an oversight on my part).... but Given that x+y = k(x-y) we proceed to x+y = kx-ky which leads to ky+y = kx-x and (k+1)y= (k-1)x k-1 to produce y= ------ x (k+1) at this point we see that for every value of k, we can let x = k+1, and y= k-1 and we get a solution. Since this will work for all k>1 you have your infinite solution set. Hope I wrote that correctly... Pat Ballew, Misawa, Jp Math Words, and other words of interest http://www.geocities.com/etyindex.html
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