How do you balance REDOX reactions using the Oxidation Number method not the half reaction method step by step in the question below?

Multiply CO₂ by 2 because there are 2 C atoms on the left. The unbalanced equation is :
MnO₄⁻ + HO₂CCO₂H → Mn²⁺ + 2CO₂

Consider MnO₄⁻ → Mn²⁺ …… [1]
The oxidation number of Mn decreases from +7 to +2.
5e⁻ are gained in the conversion.

Consider HO₂CCO₂H → 2CO₂ …… [2]
The oxidation number of C increases from +3 to +4.
As there are 2 C atoms, 2e⁻ are lost in the conversion.

To eliminate the e⁻, [1]×2 + [2]×5 :
2MnO₄⁻ + 5 HO₂CCO₂H → 2Mn²⁺ + 10CO₂

Add 8H₂O to the right in order to make 28 O atoms on the both sides.
2MnO₄⁻ + 5 HO₂CCO₂H → 2Mn²⁺ + 10CO₂ + 8H₂O

Finally, add 6H⁺ to the left in order to made 16 H atoms on the both sides.
2MnO₄⁻ + 5 HO₂CCO₂H + 6H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O
(Check: There are 4 positive charges on the both sides.)

The answer:
2MnO₄⁻ + 5 HO₂CCO₂H + 6H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O