The solution of `2x + y + z = 1 , x-2y-3z = 1 , 3x +2y +4z=5` is (A) 1,2,3 (B) 1,2,-3 (C) 1,-3,2 (D) 1,3,2

The solution of the system of equations is x-y+2z=1,2y-3z=1 and 3x-2y+4y=2 is

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

The number of solutions of the system of equations: 2x+y-z=7,\ \ x-3y+2z=1,\ \ x+4y-3z=5 is (a) 3 (b) 2 (c) 1 (d) 0

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

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

The equations x+2y+3z=1 2x+y+3z=1 5x+5y+9z=4

Solve : x+y+z =1, 2x+2y+2z=3, 3x+3y+3z=4

If ( x - 1 )^3 + ( y - 2 )^3 + ( z - 3 )^3 = 3 ( x - 1 ) ( y - 2 ) ( z - 3 ) and x - 1 != y -2 != z-3 , then x + y + z is equal to : ( A ) 2 ( B ) -6 ( C ) 6 ( D ) 4

The values of x,y,z in order of the system of equations 3x+y+2z=3,2x-3y-z=-3,x+2y+z=4, are (A)2,1,5(B)1,1,1(C)1,-2,-1(D)1,2,-1