How do you balance this redox reaction (in a basic solution) with the half-reactions method

Here is the equation: Cl2 + OH- → Cl- + ClO3-
(1 diatomic chlorine molecule + 1 hydroxide anion → 1 chloride anion + 1 chlorate anion)

Here is the answer: 3Cl2 + 6OH- → 5Cl- + ClO3- + 3H2O
(3 diatomic chlorine molecules + 6 hydroxide anions → 5 chloride anions + 1 chlorate anion + 3 water molecules)

I figured out how to do it with about 20 minutes of trial and error, but I'm curious if there's a way to balance the equation with the half-reaction method. I'm not sure how to split it up.

This type of reaction is called a disproportionation and I have a number of redox examples here:

http://www.chemteam.info/Redox/What-if-o…

Example #5 is your reaction and a bit down the file I provide a solution using Br rather than Cl.

You spent 20 minutes balancing by trial-and-error? Yikes. Man, you owe it to yourself to have a big ole honkin' bowl of ice cream. I bet you were pretty frustrated and then, suddenly, very, very happy.

This is a chemistry question and not a math problem in itself.
It is a redox equation. It is possible to have one substance act as an oxizing agent and a reducing agent.
Cl^0 +1e --> Cl 1-
Cl^0 -5e --> Cl+5 The Cl2 is the problem. and when we balance it we may require fractional molecule,so we have to double up.
3Cl2 --> 5Cl1- + CLO3- Mathematically it worked but it wasn't chemistry. This often happens.
If you had to show your work you may get 12 mark of 6.