Find the value of $k$ for which the following system of equations has no solution:
$k x + 3 y = 3$
$12 x + k y = 6$

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Solution

Given,

$k x + 3 y = 3$

$12 x + k y = 6$

Condition for equations to have no solutions,

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

$∴ \frac{k}{12} = \frac{3}{k} \neq \frac{- 3}{- 6}$

$\Rightarrow \frac{k}{12} = \frac{3}{k} \Rightarrow k = \pm 6$

$\Rightarrow \frac{3}{k} \neq \frac{- 3}{- 6} \Rightarrow k \neq 6$

the given equations have no solutions when $k = - 6$

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Similar questions

Find the value of k for which each of the following systems of equations has no solution:

$k x + 3 y = 3, 12 x + k y = 6.$

k x + 3 y = k - 3

12 x + k y = k

k x + 3 y = k - 3; 12 x + k y = k; k \neq 0

(k - 3) x + 3 y = k; k x + k y = 12

k x + 3 y - (k - 3) = 0, 12 x + k y - k =

k =

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