How to find new number congruent to another one with respect to some given congruences? If $a \equiv 4 (mod 13)$ and $b \equiv 9 (mod 13)$, then how can we find $0 \leq c \leq 12$  such that $c \equiv 9a (mod 13)$ and $ c \equiv 2a+3b (mod 13)$?

We need to find the residue of $c$ modulo $13$ so because $c\equiv9a(\bmod13)$, $c\equiv9\cdot4\equiv36\equiv10(\bmod13)$
$c\equiv2a+3b\equiv2\cdot4+3\cdot9\equiv8+27\equiv35\equiv9(\bmod13)$
So we got that $c\equiv9\equiv10(\bmod13)$ which is obviously false, so we have no solutions.
