\(\left\{{}\begin{matrix}4x+4y=104\\2x-2y=28\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+4y=104\\4x-4y=56\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x=160\\2x-2y=28\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=\dfrac{20.2-28}{2}=\dfrac{12}{2}=6\end{matrix}\right.\)
Vậy...
\(\left\{{}\begin{matrix}4x+4y=104\\2x-2y=28\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}4\left(x+y\right)=104\\2\left(x-y\right)=28\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+y=26\\x-y=14\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\dfrac{26+14}{2}\\y=\dfrac{26-14}{2}\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=20\\y=6\end{matrix}\right.\)
Vậy (x;y)=(20;6)
