Question

Find the value of k for which the following system of linear equations has an infinite number of solutions:
$$\begin{gathered} \left(k-3\right)x+3y=k \\ kx+ky=12 \end{gathered}$$

Answer 1

The given system of equations can be written as
$$\begin{gathered} \left(k-3\right)x+3y-k=0.....\left(i\right) \\ kx+ky-12=0.....\left(ii\right) \end{gathered}$$
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 $a_1=k-3,b_1=3,c_1=-k\mathrm{and}{a}_2=k,b_2=k,c_2=-12$
For the given system of linear equations to have an infinite number of solutions
, we must have
$$\begin{gathered} \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \\ \Rightarrow \frac{k-3}{k}=\frac{3}{k}=\frac{-k}{-12} \\ \Rightarrow \frac{k-3}{k}=\frac{3}{k}\mathrm{and}\frac{3}{k}=\frac{-k}{-12} \\ \Rightarrow k-3=3\mathrm{and}{k}^2=36 \end{gathered}$$
$$\begin{gathered} \Rightarrow k=6\mathrm{and}k=\pm 6 \\ \Rightarrow k=6 \end{gathered}$$
Hence, k = 6.