Hi, Chantal Lewis--
First, you will need to recognize this is a redox problem. Once you know that, determine which species is being reduced and which one is being oxidized. This is quickly accomplished by writing the oxidation numbers above each atom.
Once you do that, you will see that iodine is being both reduced and oxidized. The reduction occurs because it is going from +5 in iodate to -1/3 in triiodide. The oxidation occurs because it is going from -1 in iodide to -1/3 in triiodide.
For iodine in iodate to go from +5 to -1/3, it has a change of -5 1/3 (or -16/3).
For iodine in iodide to go from -1 to -1/3, it has a change of +2/3.
You can now see that we need 8 iodides for every 1 iodate (2/3 x 8 = 16/3). The rest balances easily after that.
It might help to see it done pictorially:
Good luck!