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 March 15th, 2015, 09:09 AM #1 Member   Joined: Aug 2014 From: Lithuania Posts: 62 Thanks: 3 System of equations I need to solve these equations for x and y: $\displaystyle 14yxa=1$ $\displaystyle 10(y+1)(x+4)a=1$ $\displaystyle 7(y+2)(x+10)a=1$ From Wolfram Alpha I get x=20 and y=6 , which is correct. Can somebody show me how to solve it? March 15th, 2015, 02:32 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 $42axy = 3 = 7 - 4 = 7(10(y+1)(x+4)a) - 4(7(y+2)(x+10)a) = 42axy + 14a(x - 20) \\ x = 20$ Can you finish from there? Thanks from topsquark March 16th, 2015, 10:02 AM   #3
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Quote:
 Originally Posted by skipjack $42axy = 3 = 7 - 4 = 7(10(y+1)(x+4)a) - 4(7(y+2)(x+10)a) = 42axy + 14a(x - 20) \\ x = 20$ Can you finish from there?
How did you know that? I don't think this kind of reasoning will be accepted. Is there another way to get the answer? March 16th, 2015, 11:28 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 It's valid reasoning (if you mention that a can't be zero before dividing by a), so it will be accepted. I found by inspection that the multipliers 7 and 4 are particularly useful. March 16th, 2015, 12:32 PM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 An obvious point is that if a= 0 then any values of x and y work. If a is not 0, then we have each of 14xy, 10(y+1)(x+ 4), and 7(y+2)(x+ 10) all equal to 1/a so equal to each other. Tags equations, system Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post gabemreis Algebra 2 February 8th, 2015 03:26 PM Regnes Algebra 2 July 19th, 2014 10:33 PM ZeusTheMunja Linear Algebra 20 January 25th, 2013 05:52 AM proglote Algebra 8 April 26th, 2011 06:59 PM outsos Abstract Algebra 22 June 1st, 2010 11:45 AM

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