Edited:

I guess

$$H^2_{(x,y)}\left(\frac{\Bbb Z[x,y]}{(5x+4y)}\right)=0$$

We know that if $\operatorname{Supp} H^i_I(M)\subseteq V(I)\cap \operatorname{Supp}(M)$, then $$\operatorname{Supp} H^2_{(x,y)}\frac{\Bbb Z[x,y]}{(5x+4y)})\subseteq V((x,y))\cap V((5x+4y))=V((x,y))=\lbrace(x,y) \rbrace\cup \lbrace(x,y,p) \rbrace$$ where $p$ is prime number.

If we could show that for every $P\in V((x,y))$, $$\left(H^2_{(x,y)}\left(\frac{\Bbb Z[x,y]}{(5x+4y)}\right)\right)_P=0$$

then it is done.

**background:**
$H^i_I(M)$ means $i$-th local cohomology module of $M$ with respect to ideal $I$.
$V(I)=\lbrace P\in Spec(R); I\subseteq P\rbrace $ and $\operatorname{Supp}(M)=\lbrace P\in \operatorname{Spec}(R) ; M_p\neq 0\rbrace$and $M_P$ means $M$ localized at prime ideal $P$. Furthermore $\operatorname{Supp}(R/I)=V(I)$.