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
