The pair of equations $3 x + 4 y = k, 9 x + 12 y = 6$ has infinitely many solutions if

k = 2

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k = 6

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k \neq 6

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k = 3

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Solution

The correct option is A $k = 2$
The equations are
$3 x + 4 y - k = 0$
$9 x + 12 y - 6 = 0$
Here, $a_{1} = 3, b_{1} = 4, c_{1} = - k$
and $a_{2} = 9, b_{2} = 12, c_{2} = - 6$
For the system to have infinite solutions condition is
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
$\Rightarrow \frac{3}{9} = \frac{4}{12} = \frac{- k}{- 6}$
$\frac{4}{12} = \frac{- k}{- 6}$
$\Rightarrow \frac{1}{3} = \frac{- k}{- 6}$
$\Rightarrow 3 k = 6$
$\Rightarrow k = 2$

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