First tally the atoms. Since #SO_4^"2-"# is an ion, I'm going to consider it as one "atom" in order to not confuse myself.

#CaSO_4# + #AlCl_3# #rarr# #CaCl_2# + #Al_2(SO_4)_3# (unbalanced)

Based on the subscripts,

*left side*:

#Ca# = 1

(#SO_4#) = 1

#Al# = 1

#Cl# = 3

*right side*:

#Ca# = 1

(#SO_4#) = 3

#Al# = 2

#Cl# = 2

#color (red) 3CaSO_4# + #AlCl_3# #rarr# #CaCl_2# + #Al_2(SO_4)_3#

Let's start balancing the most complicated 'atom', #SO_4^"2-"#

*left side*:

#Ca# = (1 x #color (red) 3#) = 3

(#SO_4#) = (1 x #color (red) 3#) = **3**

#Al# = 1

#Cl# = 3

*right side*:

#Ca# = 1

(#SO_4#) = **3**

#Al# = 2

#Cl# = 2

Since #CaSO_4# is a substance, we need to also multiply the coefficient with its #Ca# atom. Now that there are 3 #Ca# atoms on the left, there must be 3 #Ca# atoms on the right.

#3CaSO_4# + #AlCl_3# #rarr# #color (blue) 3CaCl_2# + #Al_2(SO_4)_3#

*left side*:

#Ca# = (1 x 3) = **3**

(#SO_4#) = (1 x 3) = **3**

#Al# = 1

#Cl# = 3

*right side*:

#Ca# = (1 x #color (blue) 3#) = **3**

(#SO_4#) = **3**

#Al# = 2

#Cl# = (2 x #color (blue) 3#) = 6

Again notice that since #CaCl_2# is a substance, the coefficient 3 should also be applied to its #Cl# atoms. Since there are 6 atoms of #Cl# on the right, we need to also have the same number of #Cl# atoms on the left.

#3CaSO_4# + #color (green) 2AlCl_3# #rarr# #3CaCl_2# + #Al_2(SO_4)_3#

*left side*:

#Ca# = (1 x 3) = **3**

(#SO_4#) = (1 x 3) = **3**

#Al# = (1 x #color (green) 2#) = **2**

#Cl# = (3 x #color (green) 2#) = **6**

*right side*:

#Ca# = (1 x 3) = **3**

(#SO_4#) = **3**

#Al# = **2**

#Cl# = (2 x 3) = **6**

Again, since #AlCl_3# is a substance, the coefficient should also apply to the bonded #Al# atom.

Now the equation is balanced.