x+y-2z=1;x-y+z=0;2x+3y-4z=2

2x+y-z=1 x-y+z=2 3x+y-2z=-1

The values of x, y, z for the equations x-y+z=1, 2x-y=1, 3x+3y-4z=2 are

Solve the system: x+3y-2z=0,2x-y+4z=0,x-11y+14z=0 .

The intersection of the spheres x^(2)+y^(2)+z^(2)+7x-2y-z=13andx^(2)+y^(2)+z^(2)-3x+3y+4z=8 is the same as the intersection of one of the spheres and the plane a.x-y-z=1 b.x-2y-z=1 c.x-y-2z=1 d.2x-y-z=1

x+y+z=1 x-2y+3z=2 5x-3y+z=3

2x+y+z=1x-y+2z=-1,3x+2y-z=4

The system of equations x +2y+3z=1, x-y+4z=0, 2x+y+7z=1 has

The solution of 2x+y+z=1,x-2y-3z=1,3x+2y+4z=5 is

x+y+z=4 2x-y+z=-1 2x+y-3z=-9