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.
