có
\(\left\{{}\begin{matrix}3x-y=2\\5x+y=6\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}y=3x-2\\y=6-5x\end{matrix}\right.\\ =>3x-2=6-5x\\ < =>8x=8\\ < =>x=1\\ =>y=3\cdot1-2=1\\ =>\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}3x-y=2\\5x+y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-y=2\\8x=8\end{matrix}\right.\) (Ta có: \(\left(3x+5x\right)+\left(-y+y\right)=2+6\))
\(\Leftrightarrow\left\{{}\begin{matrix}3x-y=2\\x=\dfrac{8}{8}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\cdot1-y=2\\x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3-y=2\\x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3-2\\x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)
Vậy căp (x;y) thỏa mãn là (1;1)
