\(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2xy+5x-6y-15=2xy-2x+7y-7\\12xy-24x+3y-6=12xy+18x-2y-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7x-13y=8\\-42x+5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}42x-78y=48\\-42x+5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-73y=51\\7x-13y=8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{51}{73}\\x=-\dfrac{79}{511}\end{matrix}\right.\)
Đúng 2
Bình luận (0)
