Question

Giải các hệ pt, bất pt sau:

a, \(\left\{{}\begin{matrix}2x-2y+z=3\\2x+y-2z=-3\\3x-4y-z=4\end{matrix}\right.\)

b, \(\left\{{}\begin{matrix}2x-3y\ge2\\3x+2y< 4\\x-2y\ge5\end{matrix}\right.\)

Answer 1

a: \(\left\{{}\begin{matrix}2x-2y+z=3\\2x+y-2z=-3\\3x-4y-z=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-4y+2z=6\\8x+4y-8z=-3\\3x-4y-z=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}12x-6z=3\\11x-9z=1\\3x-4y-z=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\z=\dfrac{1}{2}\\4y=3x-z-4=\dfrac{3}{2}-\dfrac{1}{2}-4=1-4=-3\end{matrix}\right.\)

=>x=1/2;z=1/2;y=-3/4