The pair of linear equations $k x + 4 y = 5, 3 x + 2 y = 5$ is consistent only when

k \neq 6

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k = 6

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k \neq 3

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k = 3

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Solution

The correct option is A $k \neq 6$
A pair of linear equations is consistent if it has a solution either a unique or
infinitely many.
Given equations are:
$k x + 4 y = 5$ .....(1)
$3 x + 2 y = 5$ ......(2)
Here, $a_{1} = k, b_{1} = 4, c_{1} = - 5$
and $a_{2} = 3, b_{2} = 2, c_{2} = - 5$
For given system,
$\frac{a_{1}}{a_{2}} = \frac{k}{3}$
$\frac{b_{1}}{b_{2}} = \frac{4}{2} = 2$
$\frac{c_{1}}{c_{2}} = \frac{- 5}{- 5} = 1$
Clearly, $\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
Hence, the system does not have infinitely many solution.
So, the equation has a unique solution.
$\Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
$\Rightarrow \frac{k}{3} \neq 2$
$\Rightarrow k \neq 6$

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