Does there exist an integer $x$ such that $x \equiv 1(\bmod 6), x \equiv 9(\bmod 14)$ and $x \equiv 7(\bmod 15)$?
We would like to apply the Chinese Remainder Theorem directly, but we cannot, because 6 and 14 are not coprime (and 6 and 15 are not coprime either).

We then need to be clever in using the Chinese Remainder Theorem: by factoring $6=2 \times 3, 14=2 \times 7$ and $15=3 \times 5$, we notice that

$-x \equiv 1(\bmod 6)$ if and only if $x \equiv 1(\bmod 2)$ and $x \equiv 1(\bmod 3)$;

$-x \equiv 9(\bmod 14)$ if and only if $x \equiv 9 \equiv 1(\bmod 2)$ and $x \equiv 9 \equiv 2(\bmod 7)$;
$-x \equiv 7(\bmod 15)$ if and only if $x \equiv 7 \equiv 1(\bmod 3)$ and $x \equiv 7 \equiv 2(\bmod 5)$.

We have thus shown that $x \equiv 1(\bmod 6), x \equiv 9(\bmod 14)$ and $x \equiv 7(\bmod 15)$ if and only if $x \equiv 1(\bmod 2), x \equiv 1(\bmod 3), x \equiv 2(\bmod 5)$ and $x \equiv 2(\bmod 7)$. The Chinese Remainder Theorem then indicates that there indeed exists such an integer $x$, and moreover, after some work, that these integers are the integers congruent to $37(\bmod 210)$.