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Question

If the system of linear equations
x2y+kz=12x+y+z=23xykz=3
has a solution (x,y,z),z0, then (x,y) lies on the straight line whose equation is :
  1. 4x3y1=0 
  2. 3x4y1=0  
  3. 3x4y4=0  
  4. 4x3y4=0   


Solution

The correct option is D 4x3y4=0   
x2y+kz=1     (1)
2x+y+z=2     (2)
3xykz=3     (3)

Adding (1) and (3), we get
4x3y=4
4x3y4=0

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