How to balance the redox reaction between bismuth(III) sulfide and nitrate anion? I have to solve the following redox reaction which occurs in an acidic environment, using the half-equation method.

$$\ce {Bi2S3_{(s)} + NO3-_{(aq)} -&gt; BiO3-_{(aq)} + SO4^2-_{(aq)} + NO_{(g)}}$$

There seems to be a lot elements that change their oxidation numbers, except for oxygen:


$\ce{Bi}$ from $\ce{+III}$ to $\ce{+V}$
$\ce{S}$ from $\ce{-II}$ to $\ce{+VI}$
$\ce{N}$ from $\ce{+V}$ to $\ce{+II}$


What should I begin with? How would one go about grouping three redox pairs?
While in general a reaction involving three independent redox pairs may not have unique coefficients, this particular one contains one coupled oxidation half-reaction and one regular reduction half-reaction. 

The key is to note that both elements in $\ce{Bi2S3}$ are oxidized, and hence we have a stoichiometric relationship between the two oxidation half-reactions. This suggests we write the oxidation half-reaction as follows: $$\ce{18H2O + Bi2S3 -&gt; 2BiO3- + 3SO4^{2-} + 36H+ + 28e-,}$$ and we can proceed with the usual algorithm for balancing half-reactions.