Let the system of linear equations $x+y+kz=2$; $2x+3y-z=1$; $3x+4y+2z=k$ have infinitely many solutions. Then the system $(k+1)x+(2k-1)y=7$; $(2k+1)x+(k+5)y=10$ has:

  • A
    infinitely many solutions
  • B
    unique solution satisfying $x-y=1$
  • C
    no solution
  • D
    unique solution satisfying $x+y=1$

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