Divide through by 2: \(3x + 5y = 16\text{.}\) Convert to a congruence, modulo 3: \(5y \equiv 16 \pmod 3\text{,}\) which reduces to \(2y \equiv 1 \pmod 3\text{.}\) So \(y \equiv 2 \pmod 3\) or \(y = 2 + 3k\text{.}\) Plug this back into \(3x + 5y = 16\) and solve for \(x\text{,}\) to get \(x = 2-5k\text{.}\) So the general solution is \(x = 2-5k\) and \(y = 2+3k\) for \(k \in \Z\text{.}\)