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.)
+128407 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)
