Balance the following redox equation in basic solution

MnO₄^- + C₂O₄^2- → MnO₂ + CO₂

Step #1 - Write the half-reactions

MnO₄^- → MnO₂

C₂O₄^2- → CO₂

Step #2 - Balance the elements that are NOT oxygen and hydrogen

MnO₄^- → MnO₂

C₂O₄^2- → CO₂

Step #3 - Balance oxygen with water

MnO₄^- → MnO₂ + H₂O

C₂O₄^2- → CO₂

Step #4 - Balance hydrogen with hydrogen ions, H⁺

H⁺ + MnO₄^- → MnO₂ + H₂O

C₂O₄^2- → 2CO₂

Step #5 - Balance charges with electrons, e^-

e^- + H⁺ + MnO₄^- → MnO₂ + H₂O

C₂O₄^2- → 2CO₂ + e^-

Step #6 - Balance electrons lost and gained

e^- + H⁺ + MnO₄^- → MnO₂ + H₂O

C₂O₄^2- → CO₂ + e^-

Step #7 - Sum the two half-reactions

H⁺ + MnO₄^- + C₂O₄^2- → MnO₂ + CO₂ + H₂O

Step #8 - Add enough hydroxide to each side of the reaction to cancel the H+ ions (turn them into water)

OH^- + H⁺ + MnO₄^- + C₂O₄^2- → MnO₂ + CO₂ + H₂O + OH^-

Step #9 - Reduce water to one side of the equation

H₂O + MnO₄^- + C₂O₄^2- → MnO₂ + CO₂ + OH^-