2. Cross out the common symbols in the numerals of different signs.
The matching numerals / symbols are colored the same.
MCMLIII + CCXIX - MCCXLVIII = CML + XIX - XLV;
When crossing out, do not mix the symbols in subtractive and additive groups.
For example, the numerals XI and IV:
XI is a group in additive notation - the largest symbol is to the left, down to the smallest to the right - to calculate the value of the group simply add up the values of the symbols in the group: XI = X + I = 10 + 1 = 11
IV is a group in subtractive notation - a smaller symbol precedes a larger one - to calculate the value of the group subtract the value of the first symbol from the value of the second: IV = V - I = 5 - 1 = 4
When subtracting the two numerals, XI - IV = (uncompact the subtractive) XI - IIII = X - III = VIIIII - III = VII (= 7), CORRECT. But if you cross out the common symbol regardless of the fact that it is part of a subtractive group: XI - IV = X - V = VV - V = V (= 5), WRONG.
3. Replace the groups in subtractive notation.
Replace any groups in subtractive notation in the roman numerals; that is, "uncompact" them using only the additive notation.
A group in subtractive notation = a group of two numerals, one of a lower value preceding another of larger value, the only allowed ones are: IV, IX, XL, XC, CD, CM, M(V), M(X), (X)(L), (X)(C), (C)(D), (C)(M) - to calculate the value of the group subtract the value of the first symbol from the value of the second.
A group in additive notation = a group of two or more numerals, of equal value or sorted in descending order of their value, from high to low, the largest symbol to the left, down to the smallest to the right - to calculate the value of the group add up the values of all the symbols.
CML: CM = M - C = D CCCCC - C = D CCCC = DCCCC; CML = DCCCCL;
XIX: IX = X - I = V IIIII - I = V IIII = VIIII; XIX = XVIIII;
XLV: XL = L - X = XXXXX - X = XXXX; XLV = XXXXV;
4. Put the Roman numerals of the same sign together.
Catenate the positive numerals together.
DCCCCL + XVIIII = DCCCCLXVIIII;
5. Cross out (remove) any symbols occurring in both strings.
DCCCCLXVIIII - XXXXV = DCCCCLIIII - XXX;
6. Using the additive notation, convert larger symbols to smaller ones and cross out the common symbols.
The larger symbol that will be converted to additive notation: L = XXXXX;