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 $\operatorname{Supp}(R/I)=V(I)$.



vanishing of local cohomology $H^2_{(x,y)}\left(\frac{\Bbb Z[x,y]}{(5x+4y)}\right)=0$
$P=(x,y,p)$ implies $P\cap\mathbb Z=p\mathbb Z$ (here $p$ is a prime or $0$). As localization commutes with local cohomology 
$$H^2\_{(x,y)}\left(\frac{\mathbb{Z}[x,y]}{(5x+4y)}\right)\_P\simeq H^2\_{(x,y)}\left(\frac{\mathbb{Z}[x,y]\_P}{(5x+4y)}\right).$$
But $\mathbb Z[x,y]\_P\simeq\mathbb Z\_{(p)}[x,y]\_{\overline{P}}$, where $\overline{P}$ is the extension of $P$ to $\mathbb Z\_{(p)}[x,y]$. Now let's see what happens with the involved ideals via this isomorphism: $(5x+4y)$ goes to $(5x+4y)\mathbb Z\_{(p)}[x,y]\_{\overline{P}}$ and note that $5$ or $4$ (or both) are invertible in $\mathbb Z\_{(p)}$, so the factor ring $\mathbb Z\_{(p)}[x,y]\_{\overline{P}}/(5x+4y)\mathbb Z\_{(p)}[x,y]\_{\overline{P}}$ is isomorphic to $\mathbb Z\_{(p)}[t]\_Q$, where $Q\cap \mathbb Z\_{(p)}=p\mathbb Z\_{(p)}$. Local cohomology is independent of base ring, so finally we arrive to $H^2\_{(t)}(\mathbb Z\_{(p)}[t]\_Q)=0$ (since the local cohomology in a principal ideal is zero from $2$ onwards).


