`{(x+2y=-3),(3x-y=5):}<=>{(x+2y=-3),(6x-2y=10):}<=>{(7x=7),(2y=-3-x):}<=>{(x=1),(y=-2):}`
Vậy.....
\(\left\{{}\begin{matrix}x+2y=-3\\3x-y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3-2y\\3x-y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3-2y\\3\left(-3-2y\right)-y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2y=-3\\-9-6y-y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2y=-3\\-7y=14\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2y=-3\\y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2.\left(-2\right)=-3\\y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-4=-3\\y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy hệ pt có nghiệm duy nhất \(\left(x;y\right)=\left(1;-2\right)\)
