For which value (s) of k will the pair of equation kx + 3y =k-3 , 12x + kv =k has no solution ?
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For no solution
a1/a2=b1/b2 not=c1/c2
k/12=3/k not eq to( k-3)/k
k^2=36
k=6 or k=-6
for k=6
the eq become 6x+3y=3 or 2x+y=1
& 12x+6y=6 or 2x+y=1
which gives infinite solutions
so k is not 6, k=-6
a1/a2=b1/b2 not=c1/c2
k/12=3/k not eq to( k-3)/k
k^2=36
k=6 or k=-6
for k=6
the eq become 6x+3y=3 or 2x+y=1
& 12x+6y=6 or 2x+y=1
which gives infinite solutions
so k is not 6, k=-6
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