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Given positive integers $x$ and $y$ such that $\frac{1}{x} + \frac{1}{2y} = \frac{1}{7}$, what is the least possible value of $xy$?

 Jun 21, 2019
 #1
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\(\frac{1}{x} + \frac{1}{2y} = \frac{1}{7}\)

 

[ x + 2y ]              1

_______   =       ___

    2xy                  7

 

7 [ x + 2y ]  =  2xy

 

7x + 14y  = 2xy

 

14y - 2xy  = - 7x

 

2xy - 14y  =  7x

 

y [2x - 14 ]  =  7x

 

y   =                7x

                    _______

                     2x  -14

 

If

 

x  = 8   y  =  28         

x = 14  y   = 7

x = 56  y  = 4

 

We can stop here since if y  = 3   then   

3(2x - 14 = 7x

6x - 42 = 7x

-42  = x

And x will also be negative if y = 2 or y = 1

 

So.....the minimum value of xy  =  14 *  7  =   98

 

 

cool cool cool

 Jun 21, 2019
edited by CPhill  Jun 21, 2019

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