Question

Find the value of k for which the following system of linear equations has no solutions:
$$\begin{gathered} kx+3y=k-3 \\ 12x+ky=k \end{gathered}$$

Answer 1

The given system of equations can be written as
$$\begin{gathered} kx+3y+3-k=0.....\left(i\right) \\ 12x+ky-k=0.....\left(ii\right) \end{gathered}$$
This system is of the form
$$\begin{gathered} a_{1}x+b_{1}y+c_1=0 \\ {a}_{2}x+b_{2}y+c_2=0 \end{gathered}$$
where $a_1=k,b_1=3,c_1=3-k\mathrm{and}{a}_2=12,b_2=k,c_2=-k$
For the given system of linear equations to have no solution, we must have
$$\begin{gathered} \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2} \\ \Rightarrow \frac{k}{12}=\frac{3}{k}\ne \frac{3-k}{-k} \\ \Rightarrow \frac{k}{12}=\frac{3}{k}\mathrm{and}\frac{3}{k}\ne \frac{3-k}{-k} \\ \Rightarrow k^2=36\mathrm{and}-3\ne 3-k \end{gathered}$$
$$\begin{gathered} \Rightarrow k=\pm 6\mathrm{and}k\ne 6 \\ \Rightarrow k=-6 \end{gathered}$$
Hence, $k=-6$.