Vì $\frac{x}{4}=$$\frac{y}{7}$=$k $ $( k\in \mathbb{Z})$
$\rightarrow \left\{\begin{matrix}
x=k.4 & & \\
y=k.7& &
\end{matrix}\right.$ (1)
Từ (1),ta có:
$x.y=112=k.4.k.7$
$\rightarrow k^{2}.28=112$
$\rightarrow k^{2}=112:28=4$
$\rightarrow k^{2}=2;-2$
+,Với k=2:
$\rightarrow \left\{\begin{matrix}
x=4.2=8 & & \\
y=7.2=14 & &
\end{matrix}\right.$
+,Với k=-2:
$\rightarrow \left\{\begin{matrix}
x=4.-2=-8 & & \\
y=7.-2=-14 & &
\end{matrix}\right.$