Show that the following linear system has no solution:
$x ≡ 1 \pmod 2$ and $x ≡ 2 \pmod 4$.
The equivalent linear equation are, for some integers $a,b$,
$$ x = 1 + 2a, \quad x = 2 + 4b $$
Equating
$$ 2a - 4b = 1 $$
By proposition 1.17, this has integer solutions if $\gcd(2,4)=2$ divides 1, which it doesn't, and so there is no solution.