Find the smallest positive integer satisfying 3 simultaneous congruences $x \equiv 2 \pmod 3$
$2x \equiv 4 \pmod 7$
$x \equiv 9 \pmod {11}$
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?

To supplement the other answers, if your equation was
$$ 2x \equiv 4 \pmod 8 $$
then the idea you guessed is actually right: this equation is equivalent to
$$ x \equiv 2 \pmod 4 $$
More generally, the equations
$$ a \equiv b \pmod c $$
and
$$ ad \equiv bd \pmod {cd}$$
are equivalent. Thus, if both sides of the equation and the modulus share a common factor, you can cancel it out without losing any solutions or introducing spurious ones. However, this only works with a common factor.
