Question

Find the value of k for which each of the following system of equations has no solution:
$(3k+1)x+3y-2=0,(k^2+1)x+(k-2)y-5=0.$

Answer 1

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:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
i.e. $\frac{3k+1}{k^2+1}=\frac{3}{\left(k-2\right)}\ne \frac{-2}{-5}$
$\frac{3k+1}{k^2+1}=\frac{3}{\left(k-2\right)}$ and $\frac{3}{\left(k-2\right)}\ne \frac{2}{5}$
⇒ (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 $k\ne \frac{19}{2}$
Thus, $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$ holds when k is equal to −1.
Hence, the given system of equations has no solution when k is equal to −1.