Bézout's identity

Back-substitution.
$$21=462-3\cdot147,$$
$$147=1071-2\cdot462.$$
So
$$21=462-3(1071-2\cdot462)=462-3\cdot1071+6\cdot462=7\cdot462-3\cdot1071.$$

Conclusion. $u=-3$, $v=7$ works: $1071\cdot(-3)+462\cdot7=-3213+3234=21$.

Remark. Bézout's identity exists iff $\gcd(a,b)$ divides the target; the coefficients are not unique (adding multiples of $b/\gcd$ to $u$ and subtracting multiples of $a/\gcd$ from $v$ gives other solutions).