Question

for which value of k will pair of equations kx + 3y =k-3 and 12x + ky =k have no solution

Answer 1

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Answer 2

For no solution,
$\frac{ a_{1} }{ a_{2} } = \frac{ b_{1} }{ b_{2} } \neq \frac{ c_{1} }{ c_{2} }$
$\frac{k}{12} = \frac{3}{k} \neq \frac{k-3}{k}$
Considering,
$\frac{k}{12} = \frac{3}{k}$
$k^{2} =36$
$k= \sqrt{36}$
k=6

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