Answer:Except -1 for all real values of k given the equations has a unique solution.
Step-by-step explanation:
Except -1 for all real values of k given the equations has a unique solution.
Step-by-step explanation:
Compare given linear pair of equations
(3k+1)x+3y-2 = 0 ----(1)
(k²+1)x+(k-2)y-5=0 ----(2)
with
----(2)
----(3)
we get ,
and
\* Given linear equations have unique solution *\
=> 3k(k-2)+1(k-2)≠3k²+3
=> 3k²-6k+k-2≠3k²+3
=> 3k²-5k-3k²≠3+2
=> -5k ≠ 5
Divide each term by (-5) , we get
=> k ≠-1
Therefore,
Except -1 for all real values of k given the equations has a unique solution.