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Find all integers x for which there exists an integer y such that 1/x + 1/y = 1/7

(In other words, find all ordered pairs of integers (x,y)  that satisfy this equation, then enter just the x's from these pairs.)

 Aug 26, 2019
 #1
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1/x  + 1/y  = 1/7

 

Since both x and y  will be greater than 7,   let z  =7 and we can write this

 

1/ ( z + a)  +   1 / (z + b)    =   1 / z        get a common denominator on the right

 

[ z + b + z + a ]               1

_____________   =     ____        simplify     

(z + a) (z + b)                 z

 

[2z + a + b]                         1

_______________   =      _____           cross-multiply

z^2 + az + bz+ ab               z

 

 

2z^2 + az + bz =      z^2 + az + bz  + ab   

 

z^2    =    ab

 

7^2  = ab

 

49  = ab

 

We have these pairs  of integers  ab  that  multiply to  49

 

(a,b)   =  (-7, -7)     So    z + a  = x =  0      and z + b = y = 0      we cannot divide by 0 so reject this pair

 

(a, b)  =  (7, 7)     so  z + a  = x =  14   qnd z + b  = y =14       so x = 14   and b  = 14

 

(a, b)  = (1, 49)  so   z + a = x =  8   and z + b =  = y = 56     so x = 8  and b  = 56

 

So

 

(x, y)  =  ( 14, 14)   and ( 8, 56)   or (56, 8)

 

 

cool cool cool

 Aug 26, 2019
edited by CPhill  Aug 26, 2019

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