\(\left\{{}\begin{matrix}3\left(x+y\right)+5\left(x-y\right)=12\\-5\left(x+y\right)+2\left(x-y\right)=1\end{matrix}\right.\)
Đặt a = x + y, b = x - y
Ta có:
\(\left\{{}\begin{matrix}3a+5b=12\\-5a+2b=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\frac{19}{31}\\b=\frac{63}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+y=\frac{19}{31}\\x-y=\frac{63}{31}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{41}{31}\\y=-\frac{22}{31}\end{matrix}\right.\)