Let a, b, m, n for:
$x\equiv a\pmod m$
$x\equiv b\pmod n$
satisfy the Chinese Remainder Theorem.
Let a, m be constant.
Since the Chinese Remainder Theorem can be applied, there exists a solution x. Consider that $x\equiv b\pmod n$ represents the set $[..., b - 2n, b - n, b, b + n, b + 2n, b + kn, ...]$. If $b+kn\equiv a\pmod m$ for any for any given b, n, is it rigorous enough to suggest $b+kn$ is a always a solution?