$x \equiv 2 \pmod 3$  

$2x \equiv 4 \pmod 7$  

$x \equiv 9 \pmod {11}$  

What is troubling me is the $2x$. 

I know of an algorithm (not sure what it's called) that would work if all three equations were $x \equiv$ by using euclidean algorithm back substitution three times but since this one has a $2x$ it won't work.

Would it be possible to convert it to $x\equiv$ somehow? Could you divide through by $2$ to get

$x \equiv 2 \pmod{\frac{7}{2}}$

Though even if you could I suspect this wouldn't really help..

What would be the best way to do this?