(i) Given that 1/x + 1/y = 1. Multiplying both sides by xy, we get:
y + x = xy
Rearranging, we have:
xy - x - y = 0
(x - 1)(y - 1) = 1
Since x and y are natural numbers, x - 1 = y - 1 = 1, which gives us x = 2 and y = 2.
Therefore, the only solution to the equation 1/x + 1/y = 1 is (x, y) = (2, 2).
(ii) We are given that xy + 5x = 4y + 38.
Rearranging, we get:
xy - 4y = -5x + 38
y(x - 4) = -5x + 38
y = (-5x + 38)/(x - 4)
Substituting this value of y in the equation 1/x + 1/y = 1, we get:
1/x + (x - 4)/(-5x + 38) = 1
Multiplying both sides by (-5x + 38)x, we get:
(-5x + 38) + x(x - 4) = x(-5x + 38)
Simplifying, we get:
x^2 - 6x + 38 = 0
Using the quadratic formula, we get:
x = (6 ± √(6^2 - 4*1*38))/2
x = 3 ± i√5
Since x is a natural number, the only possible value is x = 3.
Substituting x = 3 in the equation xy + 5x = 4y + 38, we get:
3y + 15 = 4y + 38
y = 23
Therefore, the only solution to the system of equations xy + 5x = 4y + 38 and 1/x + 1/y = 1 is (x, y) = (3, 23).