\(3x+4x-xy=15\)
\(x\left(3+4-y\right)=15\)
\(x\left(7-y\right)=15=1\cdot15=15\cdot1=3\cdot5=5\cdot3=\left(-1\right)\cdot\left(-15\right)=\left(-15\right)\cdot\left(-1\right)=\left(-3\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-3\right)\)
+, \(\left\{{}\begin{matrix}x=1\\7-y=15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=7-15=-8\end{matrix}\right.\)
+, \(\left\{{}\begin{matrix}x=15\\7-y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=15\\y=7-1=6\end{matrix}\right.\)
+, \(\left\{{}\begin{matrix}x=3\\7-y=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=7-5=2\end{matrix}\right.\)
+, \(\left\{{}\begin{matrix}x=5\\7-y=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\y=7-3=4\end{matrix}\right.\)
+, \(\left\{{}\begin{matrix}x=-1\\7-y=-15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\y=7-\left(-15\right)=22\end{matrix}\right.\)
+, \(\left\{{}\begin{matrix}x=-15\\7-y=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-15\\y=7-\left(-1\right)=8\end{matrix}\right.\)
+, \(\left\{{}\begin{matrix}x=-3\\7-y=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\y=7-\left(-5\right)=12\end{matrix}\right.\)
+, \(\left\{{}\begin{matrix}x=-5\\7-y=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\y=7-\left(-3\right)=10\end{matrix}\right.\)
Vậy \(\text{(x;y)=(1;-8);(15;6);(3;2);(5;4);(-1;22);(-15;8);(-3;12);(-5;10)}\)
