\(\left\{{}\begin{matrix}x+3y=10\\x-y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+3y=10\\x=-2+y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2+y+3y=10\\x-y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+3y=10+2\\x-y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4y=12\\x-y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3\\x-3=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=1\end{matrix}\right.\)
Vậy hệ pt có nghiệm duy nhất \(\left(x;y\right)=\left(1;3\right)\)
\(\left\{{}\begin{matrix}x+3y=10\\x-y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+3y=10\\4y=12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+3.3=10\\y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+9=10\\y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
