Question

Find the value of k for which the following equations has a unique solution:
$$\begin{gathered} 2x+3y-5=0 \\ kx-6y-8=0 \end{gathered}$$

Answer 1

The given system of equations are
$$\begin{gathered} 2x+3y-5=0.....\left(i\right) \\ kx-6y-8=0.....\left(ii\right) \end{gathered}$$
This system is of the form
$a_{1}x+b_{1}y+c_1=0\mathrm{and}{a}_{2}x+b_{2}y+c_2=0$
where $a_1=2,b_1=3,c_1=-5\mathrm{and}{a}_2=k,b_2=-6,c_2=-8$
Now, for the given system of equations to have a unique solution, we must have
$$\begin{gathered} \frac{a_1}{a_2}\ne \frac{b_1}{b_2} \\ \Rightarrow \frac{2}{k}\ne \frac{3}{-6} \\ \Rightarrow k\ne -4 \end{gathered}$$
Hence, $k\ne -4$.