If the system $6 x - 2 y = 3, k x - y = 2$ has a unique solution, then.

k = 3

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k \neq 3

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k = 4

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k \neq 4

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Solution

The correct option is B $k \neq 3$
$6 x - 2 y = 3, k x - y = 2$
$y = k x - 2$
$6 x - 2 (k x - 2) = 3$
$6 x - 2 k x + 4 = 3$
$x (6 - 2 k) = - 1$
$x = \frac{- 1}{6 - 2 k}$
$k \neq 3$ Since denominator

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Similar questions

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

k \neq 0

k \neq 3

k x - y = 2

6 x - 2 y = 3

\begin{matrix} k x + 2 y - z = 1 (k - 1) y - 2 z = 2 (k + 2) z = 3 \end{matrix}

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

$k x + 3 y = (k - 3), 12 x + k y = k .$

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) $k \neq 3$

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