Find the value of k for which the system of equations has a unique solution: $x-k y=2$ $3 x+2 y+5=0$

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Find the value of k for which the system of equations has a unique solution:
$x-k y=2$
$3 x+2 y+5=0$

Find the value of k for which the system of equations has a unique solution:
$x-k y=2$
$3 x+2 y+5=0$

Solution:

The system of equations given are
$\begin{array}{l} x-k y-2=0 \\ 3 x+2 y+5=0 \end{array}$
This system of equations given is of the form:
$a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$
where, $a_{1}=1, b_{1}=-k, c_{1}=-2$ and $a_{2}=3, b_{2}=2, c_{2}=5$
So now, for the system of equations to have a unique solution, we must have:
$\begin{array}{l} \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \\ \Rightarrow \frac{1}{3} \neq \frac{-k}{2} \\ \Rightarrow \mathrm{k} \neq-\frac{2}{3} \end{array}$
As a result, $\mathrm{k} \neq-\frac{2}{3}$ .