For which value (s) of k will the pair of equations

kx + 3y = k - 3


12x + ky = k


has no solution?


The given pair of linear equations is

kx + 3y = k - 3 …(i)


and 12x + ky = k …(ii)


On comparing with ax + by = c = 0, we get


a1 = , b1 = 3, c1 = -(k - 3)


And a2 = 12, b2 = , c2 = - k


a1 /a2 = /12


b1 /b2 = 3/


c1 /c2 =


For no solution of the pair of linear equations,


a1/a2 = b1/b2 c1/c2


So =


Taking first two parts, we get


/12 = 3/


k2 = 36


k = 6


Taking last two parts, we get



3k k(k - 3)


k2 - 6k 0


so, k 0,6


Hence, required value of k for which the given pair of linear equations has no solution is k = - 6.


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