Đặt \(\left\{{}\begin{matrix}x+1=a\left(a\ne0\right)\\y-3=b\left(b\ne0\right)\end{matrix}\right.\)

Hệ pt trở thành \(\left\{{}\begin{matrix}\dfrac{1}{a}-\dfrac{2}{b}=-\dfrac{1}{2}\\\dfrac{3}{a}+\dfrac{1}{b}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{a}-\dfrac{2}{b}=-\dfrac{1}{2}\\\dfrac{6}{a}+\dfrac{2}{b}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{a}=\dfrac{7}{2}\\\dfrac{1}{a}-\dfrac{2}{b}=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=2\left(tm\right)\\\dfrac{1}{2}-\dfrac{2}{b}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=2\left(tm\right)\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2\\y-3=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)

Vậy x=1;y=5

bạn ơi điểm E là điểm nào vậy?