Subtraction is as easy as addition (unless you go into negative numbers). The stategy is to walk the subtrahend left-to-right, eliminating from both as you go. If you run into a value that you don't have, then you expand the minuend from the right until you do.
For your "difficult" example of 42 - 13:
XXXXII - XIII
1 Eliminate an X from both.
Now you have XXXII - III.
2 Eliminate I from both
Now you have XXXI - II. Repeating (2) gives you XXX - I.
3 Since there is no I to eliminate, expand the minuend from the right to create one.
Now you have XXVV - I, repeat since you still don't have a I, repeat (3) which gives you XXVIIIII - I
Eliminate the from I both, giving the final result: XXVIIII.
Optional, you can "reduce" this to XXVIV, like you would if you had a fraction, like 3/9, but as stated in the article, this notation was not common place in ancient Rome.
From my example, which was meant for subtraction, not addition or multiplications, we get the following:
XLII - XIII
so we convert to an optimized form for arithmetic (which is, as I've noted a second format you need to know and keep track of, since it's not valid Roman numeral format
XXXXII - XIII
then we note that we have to borrow from the X "place", so let's do that
XXXVIIIIIII - XIII
now we can actually subtract
(XXXX - X) + V + (IIIIIII - III) = XXX + V + IIII = XXXVIIII
Now we have to convert back to valid Roman numerals
XXXVIV = XXXIX
This is using common current arithmetic techniques. I'm not sure whether this is the method they would have employed, or whether it would have been a different method that could have been easier or harder depending on how it combines with the attributes of this number system. Important things to note about the method shown are a pre and post conversion step (or steps, if you don't get every sub-conversion right initially or don't go to a lowest common format), and then the arithmetic itself requires more borrows.