Let $R$ be commutative ring with $1$. One says that an ideal $I$ is unmixed if $I$ has no embedded prime divisors (in other words, if the associated prime ideals of $R/I$ are the minimal prime ideals of $I$).
Let $I$ be an unmixed radical ideal of $R$. I want a hint to prove $(I:x)$ is unmixed ($x\in R\setminus I)$.
Thank you
$\newcommand{\Ass}{{\mathrm{Ass}}}$
I assume $R$ noetherian.
We write $I=Q_1 \cap \cdots \cap Q_r$ as the representation of $I$ as an intersection of primary ideals. Then $(I:x) = (Q_1:x) \cap \cdots \cap (Q_r:x)$ as this is a general property of intersection of ideals.
Now $(Q_i:x)$ is equal to $R$ if $x \in Q_i$ and $(Q_i:x) \subseteq \sqrt{Q_i} = P_i$ for $x \notin Q_i$. (Use the definition of primary ideal: $y z \in Q_i$ and $z \notin Q_i$ then $y^m \in Q_i$).
Additionally if $(Q_i:x) \neq (1)$ then $(Q_i:x)$ is $P_i$-primary: Say $z w \in (Q_i:x)$ and $z \notin (Q_i:x)$, so $z w x \in Q_i$ and $z x \notin Q_i$ so $w^n \in Q_i \subseteq (Q_i:x)$.
So we have a primary decomposition
$$(I:x) = (Q_{i_1}:x) \cap \cdots \cap (Q_{i_s}:x)$$
with $(Q_j:x) \neq (1)$ exactly when $j = i_\nu$ for a certain $\nu$.
So $\Ass (R/(I:x)) = \{P_{i_1},\ldots,P_{i_s}\}$
As $P_i$ is minimal over $I$ by assumption so $P_{i_\nu} \supseteq (I:x) $ is minimal over $(I:x) \supseteq I$ a fortiori.
This proves your proposition.