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March 15th, 2015, 09:09 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, 02:32 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,046 Thanks: 1618 
$42axy = 3 = 7  4 = 7(10(y+1)(x+4)a)  4(7(y+2)(x+10)a) = 42axy + 14a(x  20) \\ x = 20$ Can you finish from there? 
March 16th, 2015, 10:02 AM  #3 
Member Joined: Aug 2014 From: Lithuania Posts: 60 Thanks: 3  
March 16th, 2015, 11:28 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,046 Thanks: 1618 
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: 3,188 Thanks: 871 
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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