How do you balance a reaction in basic solution

BrO3- + _Cd <-> _Br2 + _Cd2+
the question asked for sum of coefficient including unknown species
but i just want to know how i can balance it.

Balance each half-reaction separately.

BrO3- ==> Br2 . . .put a 2 in front of BrO3- to balance Br
2BrO3- ==> Br2 . . .put 3H2O on the right side to balance O
2BrO3- ==> Br2 + 3H2O . . .put 6H+ on the left side to balance H
2BrO3- + 6H+ ==> Br2 + 3H2O . . .add 4e- to the left side to balance the charge.
2BrO3- + 6H+ + 4e- ==> Br2 + 3H2O

Cd ==> Cd2+ . . .add 2e- to the right side to balance the charge.
Cd ==> Cd2+ + 2e-
Multiply this entire equation by 2 to give 4e-. Then add it to the first half-reaction (the 4e- will cancel).

..2Cd ==> 2Cd2+ + 4e-
+ 2BrO3- + 6H+ + 4e- ==> Br2 + 3H2O
=============================
= 2Cd + 2BrO3- + 6H+ ==> 2Cd2+ + Br2 + 3H2O . .

Since the reaction is in alkaline solution, add 6OH- to both sides. The idea is to convert the 6H+ to 6H2O (i.e. 6H+ + 6OH- = 6H2O).

2Cd + 2BrO3- + 6H+ + 6OH- ==> 2Cd2+ + Br2 + 3H2O + 6OH-
2Cd + 2BrO3- + 6H2O ==> 2Cd2+ + Br2 + 3H2O + 6OH- . . .cancel 3H2O from both sides.
2Cd + 2BrO3- + 3H2O ==> 2Cd2+ + Br2 + 6OH-