Determine the value of k so that the following linear equations have no solution: (...

The given equational system is
(3k + 1) x + 3y − 2 = 0
$(k^2+1)x+(k-2)y-5=0$
Here, a1 = 3k + 1, b1 = 3, c1 = −2
a2 = k2 + 1, b2 = k − 2, c2 = −5
There is no answer to the given system of equations.
$\therefore \frac{a}{1}=\frac{b}{1}\ne \frac{c}{1}\Rightarrow \frac{3k+1}{k^2+1}=\frac{3}{k-2}\ne \frac{-2}{-5}\Rightarrow \frac{3k+1}{k^2+1}=\frac{3}{k-2}\mathrm{and}\frac{3}{k-2}\ne \frac{2}{5}$
Now,
$\frac{3k+1}{k^2+1}=\frac{3}{k-2}\Rightarrow \left(3k+1\right)\left(k-2\right)=3\left(k^2+1\right)\Rightarrow 3k^2-5k-2=3k^2+3\Rightarrow -5k=5$
$\Rightarrow k=-1$
When k = −1,
$\frac{3}{k-2}=\frac{3}{-1-2}=\frac{3}{-3}=-1$
Thus, for k = −1, $\frac{3}{k-2}\ne \frac{2}{5}$
Therefore, the given system of equations has no solution when k = −1.