Find the value of k for which each of the following systems of equations has no solution:
$k x + 3 y = k - 3, 12 x + k y = k .$

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Solution

Equations are written as
(kx + 3y) - (k - 3) = 0,

12x+ky-k=0

For given equations to have no solution
a1/a2 = b1/b2 is not equal to c1/c2
k/12 = 3/k
k^2 = 36
k = 6 or k = -6
On putting value of k in the given equation
we get a1/a2 = b1/b2 is not equal to c1/c2
hence the given equation will have no solution both for k=6 & k = -6

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