$\therefore \exists\ m,\ n \in \mathbb Z\ such\ that\ 7m = 2x - 3y\ and\ 7n = 4x + 5y.$
$7m = 2x - 3y \implies x = 0.5(7m + 3y).$
$\therefore 7n = 4x + 5y \implies 7n = 14m + 11y \implies y = \dfrac{7(n - 2m)}{11} = \dfrac{7}{11} * (n - 2m).$
$But\ y,\ (n - 2m) \in \mathbb Z \implies \exists\ k \in \mathbb Z\ such\ that\ 11k = (n - 2m) \implies$
$y = \dfrac{7}{11} * 11k = 7k$