Set theory and intersection For example, if we have some numbers $x,y,z,$ from $x=y-z,$ we can infer $y=x+z.$
Is there such a simple way of looking at sets?
Suppose for some sets $A,B,C$ if we have $A=B \cap C$ how can we think of $B$ in terms of $A$ and $C?$
 A: You see, being 'analogous' in math roughly means generalization of the specific object of study. For e.g. real numbers can be 'analogous' to Hermitian operators and adjoint of an operator can be 'analogous' to conjugate of complex numbers.
In your question, numbers are not generalized by sets. Hence, 'computational' work on sets consists of different rules. There are bunch of set rules introduced in the beginning of any set theory book. You should refer to them and work it out accordingly.
A: I agree with the Chrystomath’s comment, there is not much to say. The best we can do is saying that $B \subseteq A \cup C^{c}$ (where $C^c$ is the complementary set of $C$).
Although I found the following formulation for $B,$ but not just in terms of $A$ and $C.$ I had to add also $B$ because there are several sets that may hold the condition stated above.

Let $A, B,$ and $C$ be sets, such that $A = B \cap C.$ Then $B = A \cup (B \setminus C).$
Proof: Suppose that $A = B \cap C.$ Let $x \in B.$ We have that either $x \in C,$ or $x \notin C.$ In the former case, $x \in B \cap C,$ so $x \in A.$ It follows that $x \in A \cup (B \setminus C).$ In the latter case, $x \in B \setminus C,$ so $x \in A \cup (B \setminus C).$ Therefore $B \subseteq A \cup (B \setminus C).$ Now, let $y \in A \cup (B \setminus C).$ Then $y \in A,$ or $y \in B \setminus C.$ In the former case, $y \in B \cap C,$ so $y \in B$ and $y \in C,$ and in particular, $y \in B.$ In the latter case, $y \in B \setminus C,$ then $y \in B$ and $y \notin C.$ In particular $y \in B.$ Therefore $A \cup (B \setminus C) \subseteq B.$ Since $B \subseteq A \cup (B \setminus C) \subseteq B,$ we deduce that $B = A \cup (B \setminus C).$ $\square$
