Find k for which the system kx − y = 2 and 6x − 2y = 3 has a unique solution.

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Solution

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

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Similar questions

Find k for which the system kx-y=2 and 6x-2y=3 has a unique solution.

k x - y = 2

6 x - 2 y = 3

k \neq m

k x - y = 2

6 x - 2 y = 3

k x - y = 2 and 6 x - 2 y = 3

k \neq 0

k \neq 3

6 x - 2 y = 3, k x - y = 2

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