= Solution
Divide the <linear congruence> by the common factor $3$, also dividing its modulus:
$$
237x\equiv21\pmod{300}\quad\Longleftrightarrow\quad79x\equiv7\pmod{100}.
$$
The <Bézout's identity> from part (a) makes $19$ the inverse of $79$ modulo $100$. Multiplication by $19$ therefore gives $x\equiv133\equiv33\pmod{100}$. Thus all <integer> solutions are $\boxed{x=33+100k,\ k\in\mathbb Z}$. Equivalently, the three solution classes modulo $300$ are $33,133,233$; using modulus $300$ must not discard two of these classes.
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