The pair of linear equations $3 x + 7 y = k, 12 x + 2 k y = 4 k + 1$ do not have any solution if

k = 7

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k = 14

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k = 21

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k = 28

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Solution

The correct option is B $k = 14$
The pair of linear equations
$3 x + 7 y - k = 0$
$12 x + 2 k y - (4 k + 1) = 0$
Here, $a_{1} = 3, b_{1} = 7, c_{1} = - k$
and $a_{2} = 12, b_{2} = 2 k, c_{2} = - (4 k + 1)$
The system has no solution if
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
$\Rightarrow \frac{3}{12} = \frac{7}{2 k} \neq \frac{- k}{- (4 k + 1)}$
$\Rightarrow \frac{1}{4} = \frac{7}{2 k} \neq \frac{k}{4 k + 1}$
$\Rightarrow 2 k = 28$
$\Rightarrow k = 14$

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