A + B + C = 1
A + B = 0.7
B + C = 0.6.
Which is a set of three simultaneous linear equations in three unknowns,
which can be solved by the usual methods.
Eq(2) + Eq(3) - Eq(1) gives:
(A + B) + (B + C) - (A + B + C) = 0.7 + 0.6 - 1
B = 0.3.
The substituting this in to Eq2 gives A=0.4, and into Eq3 gives C=0.3.