n(A) = x+y
n(B) = y+z
n(A-B) = (x+y)-y = x, 
n(B-A) = (y+z)-y = z,
n[(A-B) U (B-A)] = x+z since (A-B) and (B-A) are disjoint

n(A) = x+y = 80
n(B) = y+z = 30
n(A U B) = x+y+z = 100 

So we have the system of three equations in three unknowns:

x + y     =  80
    y + z =  30
x + y + z = 100

Subtracting the first equation from the third gives z = 20
Substituting that in the second equation gives y = 10
Substituting that in the first equation gives x = 70

Now our Venn diagram becomes




n[(A-B) U (B-A)] = x+z = 70+20 = 90

Edwin