The given system of equations:
x + 2y = 5
⇒ x + 2y - 5 = 0 ….(i)
3x + ky - 15 = 0 …(ii)
These equations are of the forms: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0
where, a1 = 1, b1= 2, c1 = -5 and a2 = 3, b2 = k, c2 = -15
For a unique solution, we must have:
Thus for all real values of k other than 6, the given system of equations will have a unique solution.