Giải hệ phương trình \(\left\{{}\begin{matrix}x+\sqrt{3}y=0\\\sqrt{3}x+2y=1+\sqrt{3}\end{matrix}\right.\) bằng phương pháp thế, ta được nghiệm là
\(\left(x,y\right)=\left(3+\sqrt{3},-1-\sqrt{3}\right)\). \(\left(x,y\right)=\left(3-\sqrt{3},1+\sqrt{3}\right)\). \(\left(x,y\right)=\left(-3+\sqrt{3},1-\sqrt{3}\right)\). \(\left(x,y\right)=\left(3+2\sqrt{3},1+\sqrt{3}\right)\). Hướng dẫn giải:\(\left\{{}\begin{matrix}x+\sqrt{3}y=0\\\sqrt{3}x+2y=1+\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\sqrt{3}y\\-y=1+\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\sqrt{3}\left(-1-\sqrt{3}\right)\\y=-1-\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3+\sqrt{3}\\y=-1-\sqrt{3}\end{matrix}\right.\)