If $k x + 2 y = 5$ and $3 x + y = 1$ has a unique solution, if

k = 6

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k \neq 6

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k = 0

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k has any value

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Solution

The correct option is B $k \neq 6$
$k x + 2 y = 5$
$3 x + y = 1$
Condition for unique solution is
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
$\Rightarrow \frac{k}{3} \neq \frac{2}{1} \Rightarrow k \neq 6$
Hence, the correct option is (B).

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The system x-2y=3 and 3x+ky=1 has a unique solution only when

(a) k=-6 (b) $k \neq - 6$ (c) k=0 (d) $k \neq 0$

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