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 March 15th, 2015, 10:09 AM #1 Member   Joined: Aug 2014 From: Lithuania Posts: 60 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, 03:32 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,104 Thanks: 1907 $42axy = 3 = 7 - 4 = 7(10(y+1)(x+4)a) - 4(7(y+2)(x+10)a) = 42axy + 14a(x - 20) \\ x = 20$ Can you finish from there? Thanks from topsquark
March 16th, 2015, 11:02 AM   #3
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 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, 12:28 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,104 Thanks: 1907 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, 01:32 PM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 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.

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