Ta có : \(\left\{{}\begin{matrix}\frac{2x-3y}{4}-\frac{x+y-1}{5}=2x-y-1\\\frac{4x+y-2}{4}=\frac{2x-y-3}{6}-\frac{x-y-1}{3}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{5\left(2x-3y\right)}{20}-\frac{4\left(x+y-1\right)}{20}=\frac{20\left(2x-y-1\right)}{20}\\\frac{3\left(4x+y-2\right)}{12}=\frac{2\left(2x-y-3\right)}{12}-\frac{4\left(x-y-1\right)}{12}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}5\left(2x-3y\right)-4\left(x+y-1\right)=20\left(2x-y-1\right)\\3\left(4x+y-2\right)=2\left(2x-y-3\right)-4\left(x-y-1\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}10x-15y-4x-4y+4=40x-20y-20\\12x+3y-6=4x-2y-6-4x+4y+4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}10x-15y-4x-4y+4-40x+20y+20=0\\12x+3y-6-4x+2y+6+4x-4y-4=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}-34x+y=-24\\12x+y=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=-24+34x\\12x-24+34x=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=-24+34x\\46x=28\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=-\frac{76}{23}\\x=\frac{14}{23}\end{matrix}\right.\)
Vậy hệ phương trình trên có nghiệm là ( x;y ) = \(\left(\frac{14}{23};-\frac{76}{23}\right)\)
