solve in Z :
The conclusion is valid, the argument less so. If once cites the Chinese Remainder theorem, that argument would be sufficient
Alternatively:
$\displaystyle x \equiv -3 \pmod{11} \Rightarrow x = 11k-3$
Therefore,
$\displaystyle 11k-3 \equiv -1 \pmod{9} \Rightarrow 11k \equiv 2 \pmod{9} \Rightarrow k \equiv 2(11)^{-1} \pmod{9}$
$\displaystyle \Rightarrow k \equiv 1 \pmod{9} \Rightarrow k = 9l + 1 \Rightarrow x = 99l+8$