( x-3)(2y+1) = 7
=> \(\hept{\begin{cases}x-3=1\\2y+1=7\end{cases}}\)=>\(\hept{\begin{cases}x=4\\y=3\end{cases}}\)
=> \(\hept{\begin{cases}x-3=7\\2y+1=1\end{cases}}\)=> \(\hept{\begin{cases}x=10\\y=0\end{cases}}\)
b, xy + 3x - 7y = 21
=> x(y+3) - 7y - 21 = 0
=> x(y+3) - 7(y+3) = 0
=> (x-7)(y+3) = 0
=> \(\orbr{\begin{cases}x-7=0\\y+3=0\end{cases}}\)=> \(\orbr{\begin{cases}x=7\\y=-3\end{cases}}\)
Đúng 0
Bình luận (0)
