\[x2y+kz=1\] |
\[2x+y+z=2\] |
\[3xykz=3\] |
has a solution \[(x,y,z),z\ne 0,\]then (x,y) lies on the straight line whose equation is : |
A) \[3x4y1=0\]
B) \[3x4y4=0\]
C) \[4x3y4=0\]
D) \[4x3y1=0\]
Correct Answer: C
Solution :
\[x2y+kz=1\] ...(1) \[2x+y+z=2\] ...(2) \[3xykz=3\] ...(3) (1) + (3) \[\Rightarrow 4x3y=4\]You need to login to perform this action.
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