\(\left\{{}\begin{matrix}x+2y-2=0\\\dfrac{x}{2}=\dfrac{y}{3}+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2y=2\\\dfrac{1}{2}x-\dfrac{1}{3}y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x+y=1\\\dfrac{1}{2}x-\dfrac{1}{3}y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2y=2\\\dfrac{4}{3}y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2\cdot0=2\\y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
Vậy: ....
\(\left\{{}\begin{matrix}x+2y-2=0\\\dfrac{x}{2}=\dfrac{y}{3}+1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+2y=2\\3x-2y=6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2-2y\\3.\left(2-2y\right)-2y=6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2-2y\\6-6y-2y=6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2-2y\\-8y=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2-2y\\y=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2-2.0\\y=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
Vây ...
