Question

Find the value of k for which the system of equations has a unique solution:
kx + 3y = (k – 3),
12x + ky = k

Answer 1

The given system of equations:
kx + 3y = (k – 3)
⇒ kx + 3y – (k - 3) = 0             ….(i)
And, 12x + ky = k
⇒12x + ky - k = 0                        …(ii)
These equations are of the following form:
${\displaystyle a_{1}x+b_{1}y+c_1=0,a_{2}x+b_{2}y+c_2=0}$
Here, ${\displaystyle a_1=k,b_1=3,c_1=-(k–3)\text{and}{a}_2=12,b_2=k,c_2=-k}$
For a unique solution, we must have:
${\displaystyle \frac{a_1}{a_2}\ne \frac{b_1}{b_2}}$
i.e., ${\displaystyle \frac{k}{12}\ne \frac{3}{k}}$
⇒ ${\displaystyle k^2\ne 36\Rightarrow k\ne \pm 6}$
Thus, for all real values of k, other than ±6, the given system of equations will have a unique solution.