= Solution
Eliminate $x$ by twice the first <modular congruence> minus three times the second. This gives
$$
3y\equiv2(4)-3(14)=-34\equiv13\pmod{47}.
$$
The <modular inverse> of $3$ is $16$, because $3\cdot16=48\equiv1$. Hence $y\equiv16\cdot13\equiv20$. Substitution into the second <modular congruence> gives $6x\equiv14-7(20)\equiv15$, and $6^{-1}\equiv8$, so $x\equiv8\cdot15\equiv26$. Thus
$$
\boxed{x=26+47r,\qquad y=20+47s,\qquad r,s\in\mathbb Z.}
$$
In particular, $(x,y)=(26,20)$ works: $9(26)+12(20)=474=10(47)+4$ and $6(26)+7(20)=296=6(47)+14$. The elimination is reversible because $3$ and $6$ have <modular inverses>, so the displayed classes contain all solutions.
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