Since this was originally posted two months ago:
We have aBaCl2(aq)+bK2SO4(aq)→nBaSO4↓+mKCl(aq)
Looking at that I see that there are "a" Ba atoms on the left and "n" Ba atoms on the right so we must have a= n. I see that there are "2a" Cl atoms on the left and "m" Cl atoms on the right so we must have 2a= m. Is that there are "2b" K atoms on the left and "m" K atoms on the right so we must have 2b= m. Finally I see that there are "b" SO4 ions on the left and "n" SO4 ions on the right so we must have b= n.
That is, we have the four equations a= n, 2a= m, 2b= m and b= n. That is four equations in the four unknowns, a, b, m, and n. One difficult is that when the number of equations is equal to the number of unknowns there is typically a unique solution. But it is obvious that a= b= m= n= 0 is solution while we want non-zero values! If there is not a unique solution we can try to solve for all but one unknown in terms of that one. Let's try to solve for b, m, and n in terms of a. We have immediately that n= a and m= 2a. 2b= m= 2a so b= a and n= b= a.
So if we arbitrarily take a= 1, b= 1, m= 2, and n= 1. We have
BaCl2(aq)+K2SO4(aq)→ BaSO4↓+2KCl(aq)