How would you balance: AlBr3 + K2SO4 --> KBr + Al2(SO4)3?
The first thing I notice is that the product side has 3 sulfate groups, while the reactant side has only 1. We can multiply the potassium sulfate group by 3 so there are 3 on either side.
However, this leaves us with 6 potassium atoms on the reactant side and only 1 on the product side, so we multiply the potassium bromide by 6.
Now, we have the imbalance of 6 bromine atoms on the right and only 3 on the right, so we multiply the aluminum bromide on the right by 2. Finally, this leaves 2 aluminum atoms on the reactant side and on the product side, so we are done balancing.