Question

Find the value for k which each of the following systems of equations has a unique solution:

$kx+3y=(k-3),12x+ky=k.$

Answer 1

$kx+3y–(k–3)=0----\left(1\right)12x+ky–k=0----\left(2\right)a_1=k,b_1=3,c_1=-(k–3)a_2=12,b_2=k,c_2=-k$

For a unique solution, we must have,

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

Now,

$\frac{k}{12}\ne \frac{3}{k}{k}^2\ne 36k\ne \pm 6$

Thus, for all real values of k other than $\pm 6$, the given system of equations will have a unique solution.