on the proof of a simple inequality in dimension theory

Let $(R,m)$ be a local ring and $M \neq 0$ a finite $R$-module. Let $x \in m$ and set $\bar{M}=M/xM$. Then $\dim M/xM \ge \dim M -1$. One way to see this is as follows: let $\dim M/xM = s$ and let $y_1,\dots,y_s$ be a system of parameters for $M/xM$ in $m$, i.e. $l(\bar{M}/(y_1 \bar{M}+\cdots+y_s \bar{M})) < \infty$. Then this implies that $l(M/(xM + y_1M+\cdots+y_s M)) < \infty$ and so $\dim M \le \dim M/xM + 1$ (see e.g. Theorem 13.4 in Matsumura, CRT).

But the proof of proposition A.4 in Bruns and Herzog (CMR), implies that we can show the inequality $\dim M/xM \ge \dim M -1$ by replacing $M$ with $R/I, \, I = ann(M)$. But if we do that we obtain the inequality $\dim R/(Rx+I) \ge \dim R/I -1$. My objection is that since $Rx+I \subset ann(M/xM)$ in general we will have $\dim R/(Rx+I) \ge \dim M/xM$ and so proving $\dim R/(Rx+I) \ge \dim R/I -1$ does not imply $\dim M/xM \ge \dim M -1$. Am i missing something?

$\operatorname{Supp}(M/xM)=\operatorname{Supp}(M)\cap V(xR)=V(I)\cap V(xR)=V(I+xR)$, so $\dim M/xM=\dim R/(I+xR)$.