Cách 1:
\(\left\{{}\begin{matrix}x+y=\dfrac{4x-3}{5}\\x+3y=\dfrac{y-1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{4x-3}{5}-x=\dfrac{4x-3-5x}{5}=\dfrac{-x-3}{5}\\x+3\cdot\dfrac{-x-3}{5}=\dfrac{1}{2}\left(\dfrac{-x-3}{5}-1\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{-x-3}{5}\\x+\dfrac{-3x-9}{5}=\dfrac{1}{2}\cdot\dfrac{-x-8}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{-x-3}{5}\\\dfrac{5x-3x-9}{5}=\dfrac{-x-8}{10}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-x-3}{5}\\\dfrac{4x-18}{10}=\dfrac{-x-8}{10}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x-18=-x-8\\y=\dfrac{-x-3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=-8+18=10\\y=\dfrac{-x-3}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=\dfrac{-2-3}{5}=-1\end{matrix}\right.\)
Cách 2:
\(\left\{{}\begin{matrix}x+y=\dfrac{4x-3}{5}\\x+3y=\dfrac{y-1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5\left(x+y\right)=4x-3\\2\left(x+3y\right)=y-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5x+5y=4x-3\\2x+6y=y-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+5y=-3\\2x+5y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-x=-1-\left(-3\right)\\x+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\5y=-3-x=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+y=\dfrac{4x-3}{5}\\x+3y=\dfrac{y-1}{2}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=\dfrac{4}{5}x-\dfrac{3}{5}\\x+3y=\dfrac{y}{2}-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(x-\dfrac{4}{5}x\right)+y=-\dfrac{3}{5}\\x+\left(3y-\dfrac{y}{2}\right)=-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{5}x+y=-\dfrac{3}{5}\\x+\dfrac{5}{2}y=-\dfrac{1}{2}\end{matrix}\right.\)
Cách 1 thế:
\(\left\{{}\begin{matrix}\dfrac{1}{5}x+y=-\dfrac{3}{5}\\x+\dfrac{5}{2}y=-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{3}{5}-\dfrac{1}{5}x\\x+\dfrac{5}{2}\left(-\dfrac{3}{5}-\dfrac{1}{5}x\right)=-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{3}{5}-\dfrac{1}{5}x\\x-\dfrac{3}{2}-\dfrac{1}{2}x=-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{3}{5}-\dfrac{1}{5}x\\\dfrac{1}{2}x=-\dfrac{1}{2}+\dfrac{3}{2}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{3}{5}-\dfrac{1}{5}x\\\dfrac{1}{2}x=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{3}{5}-\dfrac{1}{5}\cdot2=-1\\x=2\end{matrix}\right.\)
Cách 2 cộng:
\(\left\{{}\begin{matrix}\dfrac{1}{5}x+y=-\dfrac{3}{5}\\x+\dfrac{5}{2}y=-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{5}x+y=-\dfrac{3}{5}\\\dfrac{1}{5}x+\dfrac{1}{2}y=-\dfrac{1}{10}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y-\dfrac{1}{2}y=-\dfrac{3}{5}+\dfrac{1}{10}\\\dfrac{1}{5}x+y=-\dfrac{3}{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}y=-\dfrac{1}{2}\\\dfrac{1}{5}x+y=-\dfrac{3}{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-1\\\dfrac{1}{5}x=-\dfrac{3}{5}+1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-1\\\dfrac{1}{5}x=\dfrac{2}{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=2\end{matrix}\right.\)
