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For the following system of equation determine the value of $k$ for which the given system has no solution:.$2x-ky+3=0$ and $3x+2y-1=0$.
Given: The system of equations: $2x-ky+3=0$ and $3x+2y-1=0$.
To do: To determine the value of $k$ for which the given system has no solution.
Solution:
The given system of equation: $2x-ky+3=0$ and $3x+2y-1=0$.
Here, $a_1=2,\ b_1=-k,\ c_1=3$ and $a_2=3,\ b_2=2,\ c_2=-1$
For no solution, $\frac{a_1}{a_2}=\frac{b_1}{b_2}≠\frac{c_1}{c_2}$
$\Rightarrow \frac{2}{3}=\frac{-k}{2}$
$\Rightarrow -3k=4$
$\Rightarrow k=-\frac{4}{3}$
Thus, for $k=-\frac{4}{3}$, the given system of equations has no solution.
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