The pair of linear equations $13 x + k y = k, 39 x + 6 y = k + 4$ has infinitely many solutions if

k = 1

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k = 2

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k = 4

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k = 6

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Solution

The correct option is B $k = 2$
The pair of linear equations
$13 x + k y - k = 0$
$39 x + 6 y - (k + 4) = 0$
Here, $a_{1} = 13, b_{1} = k, c_{1} = - k$
$a_{2} = 39, b_{2} = 6, c_{2} = - (k + 4)$

The system has infinitely many solutions if
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
$\Rightarrow \frac{13}{39} = \frac{k}{6} = \frac{- k}{- (k + 4)}$
Taking,
$\frac{13}{39} = \frac{k}{6}$
$\Rightarrow \frac{1}{3} = \frac{k}{6}$
$\Rightarrow 3 k = 6$
$\Rightarrow k = 2$
Taking,
$\frac{k}{6} = \frac{- k}{- (k + 4)}$
$\Rightarrow \frac{1}{6} = \frac{1}{(k + 4)}$
$\Rightarrow k + 4 = 6 \Rightarrow k = 2$
Thus, $k = 2$ is the correct answer

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