Let $A,B,C\triangleleft R$ be some Ideals. prove that if:
$$\left\{\begin{matrix}x\equiv y \mod C\\z\equiv x \mod B\\y\equiv z \mod A \end{matrix}\right.\Rightarrow \text{There is } t\in R \text{ such that}\left\{\begin{matrix}t\equiv x\equiv y \mod C\\t\equiv z\equiv x \mod B\\t\equiv y\equiv z \mod A \end{matrix}\right.$$ Then:
$$B\cap(A+C)\subseteq C+(A\cap B)$$
In the book the question was taken from there is a hint to pick $z=0$. Yet, I am still struggeling.
Let $x\in B\cap (A+C)$. Then $x\in B$ and $x\in A+C$. Write $x=y+(x-y)$ for some $y\in A$. Note that the chosen $x$ and $y$ together with $z=0$ satisfy the premise of the implication given in the question. So we can find a $t\in R$ with the given properties. Now write $x=(x-t)+t$ as Lozenges suggested in the comment. Observe that by the condition on $t$, we have $x-t\in C$ and $t\in A\cap B$ and hence $x\in C+(A\cap B)$ as desired.