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2014-12-03T14:17:00+05:30

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       2 x + ( k - 1) y = 6
       3 x + (2 k + 1) y = 9

Multiply the equation 1 by 3 and multiply equation 2 by 2, and subtract,

       (k + 5) y = 0      , so y = 0  or  k = -5

Possibility 1 :  y = 0  =>  x = 3        => unique solution
                     =>  k can be any real value

Possibility 2 :  k = -5,      2 x -6 y = 6    and  3 x - 9 y = 9
    =>  x - 3 y = 3      and  there is no unique solution.  There are infinite solutions.

Hence, the system has unique solution only if y = 0 and x = 3 and it is true for any real value of k.

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