Intersection of ideals equal to their product on noetherian ring

Question

Let $R$ be a noetherian ring, $I\le R$ and $x\in R$ such that $\forall\frak{p}\in$Ass$(R/I)$, $x\not\in\frak{p}$. Show that $(x)\cap I=(x)I$.

Obviously, we have the inclusion $(x)I\subseteq (x)\cap I$. For the other inclusion, we are supposed to use the following result:

Let $R$ be a noetherian ring, $I,J\le R$ such that $\forall \frak{p}\in$Ass$(R/I)$, $JR_\frak{p}\subseteq$$IR_\frak{p}$. Then, $J\subseteq I$.

I've tried choosing $(x)\cap I$ and $(x)I$ to play the role of $J$ and $I$ in the previous result, but it seems to lead me nowhere.