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Find the value of $k$ for which the system of equations $kx-y=2$ and $6x-2y=3$ has a unique solution.
Given: The system of equations $kx-y=2$ and $6x-2y=3$.
To do: To find the value of $k$ for which the system of equations has a unique solution.
Solution:
The given system is
$kx-y-2=0\ .....( i)$
$6x-2y-3=0\ ....( ii)$
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:
$\frac{a_1}{a_2}≠\frac{b_1}{b_2}$
$\Rightarrow \frac{k}{6}≠\frac{-1}{-2}$
$\Rightarrow -2k≠-6$
$\Rightarrow k≠\frac{-6}{-2}$
$\Rightarrow k≠3$
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