The given system of equations:
(3k + 1)x + 3y − 2 = 0 ...(i)
And, (k2 + 1)x + (k − 2)y − 5 = 0 ...(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = (3k + 1), b1= 3, c1 = −2 and a2 = (k2 + 1), b2 = (k − 2), c2 = −5
In order that the given system has no solution, we must have:
i.e.
and
⇒ (k − 2) (3k + 1) = 3(k2 + 1) and 15 ≠ 2(k − 2)
⇒ 3k2 + k − 6k − 2 = 3k2 + 3 and 15 ≠ 2k − 4
⇒ −5k = 5 and 2k ≠ 19
⇒ k = −1 and
Thus, holds when k is equal to −1.
Hence, the given system of equations has no solution when k is equal to −1.