If the two relations hold $5x \ge 7y$ comes to $5a + 3b + c \ge 0$ which is true. Conversely, write $y = 5d + r$, $r \in \{0, 1, 2, 3, 4\}$. For $r = 0$, $x \ge 7d$ and just take $a = x - 7d$, $b = c = 0$. For $r = 1$, $5x \ge 35d + 7 > 5(7d + 1)$ whence $x \ge 7d + 2$ (since it is an integer) and take $a = x - 7d - 2$, $b = 1$ and $c = 0$. For $r = 2, 3, 4$ you will get analogously $(b, c) = (0, 1); (1, 1)$ and $(0, 2)$ respectively and $a = x - 7d - 2b - 3c$.