Giải hệ phương trình \(\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=-4\\4\left(x+y\right)+5\left(x-y\right)=-2\end{matrix}\right.\) bằng phương pháp thế, ta được nghiệm là
\(\left(x,y\right)=\left(\dfrac{1}{2},\dfrac{13}{2}\right)\).\(\left(x,y\right)=\left(\dfrac{3}{2},2\right)\).\(\left(x,y\right)=\left(1,\dfrac{13}{2}\right)\).\(\left(x,y\right)=\left(-\dfrac{3}{2},-2\right)\).Hướng dẫn giải:\(\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=-4\\4\left(x+y\right)+5\left(x-y\right)=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x-y=-4\\9x-y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=5x+4\\9x-5x-4=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=5x+4\\4x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{13}{2}\end{matrix}\right.\)
