I've got three simultaneous congruence to solve, which I now know how to do, but in this particular question, one of them is in the form of $2x \equiv$ instead of of $x \equiv$ that I'm used to: $\begin{align*} x &\equiv 2 \pmod 3\\ 2x &\equiv 4 \pmod 7\\ x &\equiv 9 \pmod{11} \end{align*}$
How do I divide through the second congruence to get just $x \equiv $ something (mod something) ?
Is it just $x \equiv 2\pmod7$ ?