ĐK : \(y\ne2x\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{2x-y}=a\\x+3y=b\end{matrix}\right.\left(a\ne0\right)\)hpt đã cho trở thành \(\left\{{}\begin{matrix}a+b=\dfrac{3}{2}\\4a-5b=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4a-4b=-6\\4a-5b=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-9b=-9\\a+b=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=1\end{matrix}\right.\left(tm\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2x-y}=\dfrac{1}{2}\\x+3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=2\\x+3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-3y=6\\x+3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x=7\\x+3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\left(tm\right)\)
