Làm đại bạn nhé, tại chưa có học hệ pt:Đ
\(2x+3y=3\Rightarrow3y=3-2x\)
\(\Rightarrow5x-6y=12\Leftrightarrow5x-2.3y=12\Leftrightarrow5x-2\left(3-2x\right)=12\Leftrightarrow5x-6+4x=12\Leftrightarrow9x-6=12\Leftrightarrow x=2\)Thay x=2, ta được y=\(-\dfrac{1}{3}\)
Chắc là sai đó anh/chị)):
\(\left\{{}\begin{matrix}2x+3y=3\\5x-6y=12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4x+6y=6\left(1\right)\\5x-6y=12\left(2\right)\end{matrix}\right.\)
Lấy \(\left(1\right)+\left(2\right)\Leftrightarrow9x=18\Rightarrow x=2\Rightarrow y=\dfrac{3-4}{3}=-\dfrac{1}{3}\)
\(\left\{{}\begin{matrix}2x+3y=3\\5x-6y=12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=6\\5x-6y=12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9x=18\\2x+3y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2.2+3y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\3y=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{1}{3}\end{matrix}\right.\)
