We can use Simon's Favourite Factoring Trick to solve this problem.

\(3x + 2y + xy = 18 + 7x + 5y\)

\(xy-4x-3y-18=0\)

\(xy-4x-3y+12=30\)

\(x(y-4)-3(y-4)=30\)

\((x-3)(y-4)=30\)

The pairs of factors of 30 is (1, 30), (2, 15), (3, 10), (5, 6), (6, 5), (10, 3), (15, 2), and (30, 1). We choose x and y such that the values in the parentheses is equal to one of the two factors of 30. This makes all values of (x, y) equal to (4, 34), (5, 19), (6, 14), (8, 10), (9, 9) (13, 7), (18, 6), or (33, 5). The sum of x and y for each pair respectively is 38, 24, 20, 18, 18, 20, 24, and 38. Therefore, all possible sums of x and y are **18, 20, 24, and 38.**