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

a_{1} = , b_{1} = 3, c_{1} = -(k - 3)

And a_{2} = 12, b_{2} = , c_{2} = - k

a_{1} /a_{2} = /12

b_{1} /b_{2} = 3/

c_{1} /c_{2} =

For no solution of the pair of linear equations,

a_{1}/a_{2} = b_{1}/b_{2} c_{1}/c_{2}

So =

Taking first two parts, we get

/12 = 3/

k^{2} = 36

k = 6

Taking last two parts, we get

3k k(k - 3)

k^{2} - 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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