\(\left\{{}\begin{matrix}2x-y=3\\3x+2y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-3y=9\\6x+4y=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-7y=-7\\6x+4y=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\6x+4.1=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)
Vậy phương trình có n0 duy nhất \(\left(x;y\right)=\left(2;1\right)\)