A Chinese remainder system

Reformulation. We need $17h-33k=1$ for some integers $h,k$.

Bézout. $\divisibility(17,33)=1$, and $2\cdot17-33=1$. So $h=2$, $k=1$ works.

Particular solution.
$$x_0=2+17\cdot2=36.$$

General solution. By the Chinese remainder theorem,
$$x\equiv36\pmod{17\cdot33=561}.$$
So $x=36+561m$, $m\in\mathbb{Z}$.