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March 15th, 2015, 09:09 AM   #1
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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?
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March 15th, 2015, 02:32 PM   #2
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$42axy = 3 = 7 - 4 = 7(10(y+1)(x+4)a) - 4(7(y+2)(x+10)a) = 42axy + 14a(x - 20) \\
x = 20$

Can you finish from there?
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March 16th, 2015, 10:02 AM   #3
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Quote:
Originally Posted by skipjack View Post
$42axy = 3 = 7 - 4 = 7(10(y+1)(x+4)a) - 4(7(y+2)(x+10)a) = 42axy + 14a(x - 20) \\
x = 20$

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?
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March 16th, 2015, 11:28 AM   #4
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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.
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March 16th, 2015, 12:32 PM   #5
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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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