Let a, b,c and d be the number of molecules in the balanced equation as shown below.
aAlBr_3 +b K_2SO_4 -> cKBr + dAl_2(SO_4)_3
Inspection reveals that the on the right hand side of the equation number of radical (SO_4) is the highest, i.e., 3. Hence take d as a lowest integer.
Therefore, with d=1, the equation becomes
aAlBr_3 +b K_2SO_4 -> cKBr + Al_2(SO_4)_3
To balance 3 radicals SO_4, on the left hand side b=3.
With this the equation becomes
aAlBr_3 +3 K_2SO_4 -> cKBr + Al_2(SO_4)_3
Now balancing 6 atoms of K on the left hand side, c=6
we get
aAlBr_3 +3 K_2SO_4 -> 6KBr + Al_2(SO_4)_3
To find the value of a, you may either balance Al or Br. We obtain a=2 and get the balanced equation as in the Answer.