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**.