= Solution
{c}
Apply the <Euclidean algorithm> to $19$ and $11$:
$$
19=11+8,\quad 11=8+3,\quad 8=2\cdot3+2,\quad 3=2+1.
$$
Back-substitution gives the <Bézout's identity>
$$
1=3-2=3\cdot3-8=3\cdot11-4\cdot8=7\cdot11-4\cdot19.
$$
Thus $(x,y)=(-4,7)$ is one solution of the <Diophantine equation>. Subtracting this solution from any other gives
$$
19(x+4)+11(y-7)=0.
$$
Since $19$ and $11$ are <coprime>, $11$ divides $x+4$. To justify that step directly, multiply $1=-4\cdot19+7\cdot11$ by $x+4$: both terms on the right are divisible by $11$, because $19(x+4)=-11(y-7)$. Write $x+4=11t$; substitution then gives $y-7=-19t$. Therefore the complete solution set is
$$
\boxed{x=-4+11t,\qquad y=7-19t,\qquad t\in\mathbb Z.}
$$
Every such pair satisfies the original <Diophantine equation>, since the $t$ terms cancel.
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