Solve congruence $8x \equiv 28 \mod30 , 11x \equiv 1 \mod35$

Question

I need to solve the following set of congruences.

\begin{cases} 8x \equiv 28 & \mod30 \\ 11x \equiv 1 & \mod35\end{cases}

Finding the inverse of $11$ in the ring $\mathbb{Z}_{35}$ led to this simplification.

\begin{cases} 8x \equiv 28 & \mod30 \\ x \equiv 16 & \mod35\end{cases}

So we have: $$30m+28=8x=8(35*n+16)=280n+128$$ $$30m-280n=100$$

Now $30\cdot(-9)+280=10$ so could possibly be $m=-90$ and $n=10$.

So $x=350+16=366$.

But $366$ doesn't fit the bill.

What did I do wrong?

Answer 1

Rather: "... so $m=-90$ and $n=-10$".