Let $(X,\mathcal{O}_X)$ be a ringed space. Show that the tensor product $\mathcal{F} \otimes_{\mathcal{O}_X} \mathcal{G}$ of two quasi-coherent $\mathcal{O}_X$-modules $\mathcal{F}, \mathcal{G}$ is again quasi-coherent.
My attempt: since $\mathcal{F},\mathcal{G}$ are quasi-coherent, then for every $x \in X$ we can find two open neighborhoods $U_1, U_2$ such that
$$ \mathcal{O}_{U_1}^{(J_1)} \rightarrow \mathcal{O}_{U_1}^{(I_1)} \rightarrow \mathcal{F}|_{U_1} \rightarrow 0 \\ \mathcal{O}_{U_2}^{(J_2)} \rightarrow \mathcal{O}_{U_2}^{(I_2)} \rightarrow \mathcal{G}|_{U_2} \rightarrow 0 $$
are exact.
Define $U:= U_1 \cap U_2$; $U$ is an open neighborhood of $x$ and we can restrict our two exact sequences to $U$ since exactness is equivalent to exactness at level of stalks. And here I'm stuck because I don't know ho to form a new exact sequence for the $\mathcal{O}_X$-module $\mathcal{F} \otimes_{\mathcal{O}_X} \mathcal{G}$, maybe using $U$ as open neighborhood of $x$.
Any ideas?
EDIT:
This sequence should be still exact:
$$(\mathcal{O}_{U}^{(J_1)} \otimes \mathcal{O}_{U}^{(I_2)}) \oplus (\mathcal{O}_{U}^{(J_2)} \otimes \mathcal{O}_{U}^{(I_1)}) \rightarrow \mathcal{O}_{U}^{(I_1)} \otimes \mathcal{O}_{U}^{(I_2)} \rightarrow \mathcal{F}|_{U} \otimes \mathcal{G}|_{U} \rightarrow 0 $$
and it can be rewritten as $$\mathcal{O}_U^{(J)} \rightarrow \mathcal{O}_U^{(I)} \rightarrow (\mathcal{F} \otimes \mathcal{G})|_U \rightarrow 0$$
where $J=(J_1 \times I_2) \sqcup (J_2 \times I_1)$ and $I=I_1 \times I_2$.
If this is true, then $\mathcal{F} \otimes \mathcal{G}$ would be quasi-coherent.