x=2 (mod 5), 2x=3 (mod 7), 3x=4 (mod 11)
The congruence, $\displaystyle 2x\equiv 3(\bmod 7)$ is equivalent to $\displaystyle 4(2x)\equiv 4\cdot 3 (\bmod 7)$ which becomes $\displaystyle x\equiv 5(\bmod 7)$. The congruence, $\displaystyle 3x\equiv 4(\bmod 11)$ is equivalent to $\displaystyle 4(3x)\equiv 4\cdot 4(\bmod 11)$ which becomes $\displaystyle x\equiv 5(\bmod 11)$.
Thus, you have to solve,
$\displaystyle x\equiv 2(\bmod 5)$
$\displaystyle x\equiv 5(\bmod 7)$
$\displaystyle x\equiv 5(\bmod 11)$
Combine the last two congruences and so,
$\displaystyle x\equiv 2(\bmod 5)$
$\displaystyle x\equiv 5(\bmod 77)$
This is equivalent to,
$\displaystyle x\equiv 2 + 16\cdot 5(\bmod 5)$
$\displaystyle x\equiv 5 + 77(\bmod 77)$
We get,
$\displaystyle x\equiv 82(\bmod 5)$
$\displaystyle x\equiv 82(\bmod 77)$
Combine,
$\displaystyle x\equiv 82(\bmod 385)$