The system $k x - y = 2 and 6 x - 2 y = 3$ has a unique solution only when
(a) k = 0 (b) $k \neq 0$ (c) k = 3 (d) $k \neq 3$

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Solution

The given equations are
$k x - y - 2 = 0 . . . . . (i) 6 x - 2 y - 3 = 0 . . . . . (i i)$
Here, $a_{1} = k, b_{1} = - 1, c_{1} = - 2, a_{2} = 6, b_{2} = - 2 and c_{2} = - 3 .$
For the given system to have a unique solution, we must have
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \Rightarrow \frac{k}{6} \neq \frac{- 1}{- 2} \Rightarrow k \neq 3$
Hence, the correct answer is option (d).

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Similar questions

x - 2 y = 3 and 3 x + k y = 1

The system kx-y=2 and 6x-2y=3 has a unique solution only when

(a) k=0 (b) $k \neq 0$ (c) k=3 (d) $k \neq 3$

k x - y = 2

6 x - 2 y = 3

Find the value for k which each of the following systems of equations has a unique solution:

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