Knowing that elemental molecules (made of only one element, like #O_2, Cl_2, Br_2#) etc have #0# enthalpy, and the equation for enthalpy
#DeltaH_("rxn")=Sigman*"products"-Sigman*"reactants"#,
or overall enthalpy (#DeltaH_("rxn")#) is the enthalpy of products minus enthalpy of reactants, we can rearrange equation #3.#.
#TiCl_4->TiCl_2+Cl_2, DeltaH=+273kJ#
#273=(TiCl_2+Cl_2)-TiCl_4# (products minus reactants)
#273=TiCl_2-TiCl_4# (enthalpy of #Cl_2# is #0#)
#TiCl_4=TiCl_2-273# (rearranged algebraically)
Now you can substitute this into equation #2.#, again using the products minus reactants equation
#TiBr_2+SnCl_4->SnBr_2+TiCl_4, DeltaH=-74kJ#
#-74=(SnBr_2+TiCl_4)-(TiBr_2+SnCl_4)#
#-74=SnBr_2+TiCl_2-273-TiBr_2-SnCl_4# (substituting in equation #3.# and expanding out everything)
#199=SnBr_2+TiCl_2-TiBr_2-SnCl_4# (rearranging, adding #273# to both sides)
You also have equation #1.#, which you can use products minus reactants to find
#SnCl_2+TiBr_2->SnBr_2+TiCl_2, DeltaH=+4.2kJ#
#4.2=(SnBr_2+TiCl_2)-(SnCl_2+TiBr_2)# (products minus reactants)
#4.2=SnBr_2+TiCl_2-SnCl_2-TiBr_2# (expanding out brackets)
Now you can equate equations #1.# and #2.# in their new form, as so
#SnBr_2+TiCl_2-SnCl_2-TiBr_2+194.8=SnBr_2+TiCl_2-TiBr_2-SnCl_4#
because if you add #194.8# to equation #1# it equals #199# in total, the same as equation #2.# With these two equal to each other, you can cancel out the similarities on both sides and find what's left.
Both sides have a #SnBr_2#, so you can cancel that. They also have a #TiCl_2# and a #-TiBr_2#, so get rid of those from each side, leaving
#-SnCl_2+194.8=-SnCl_4#
#SnCl_4-SnCl_2=-194.8kJ#
This is the answer to the question in the form of products minus reactants, because you can see the product is #SnCl_4# and the reactants are #SnCl_2# and #Cl_2#, though #Cl_2# has an enthalpy of #0# so it can be ignored.
