\(\dfrac{1-x}{3}=\dfrac{2y-1}{8}\)
=>8(1-x)=3(2y-1)
=>8-8x=6y-3
=>-8x-6y=-11
=>8x+6y=11
mà 2x+y=6
nên ta có hệ phương trình:
\(\left\{{}\begin{matrix}8x+6y=11\\2x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x+6y=11\\8x+4y=24\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2y=-13\\2x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{13}{2}\\2x=6-y=6+\dfrac{13}{2}=\dfrac{25}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{25}{4}\\y=-\dfrac{13}{2}\end{matrix}\right.\)
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