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

Question:

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

(3k + 1) x + 3y − 2 = 0

(k2 + 1) x + (k − 2) y − 5 = 0

Solution:

The given system of equations is

(3k + 1) x + 3y − 2 = 0

(k2 + 1) x + (k − 2) y − 5 = 0

Here, a1 = 3k + 1, b1 = 3, c1 = −2

a2 = k2 + 1, b2 = k − 2, c2 = −5

The given system of equations has no solution.

$\therefore \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

$\Rightarrow \frac{3 k+1}{k^{2}+1}=\frac{3}{k-2} \neq \frac{-2}{-5}$

$\Rightarrow \frac{3 k+1}{k^{2}+1}=\frac{3}{k-2}$ and $\frac{3}{k-2} \neq \frac{2}{5}$

Now,

$\frac{3 k+1}{k^{2}+1}=\frac{3}{k-2}$

$\Rightarrow(3 k+1)(k-2)=3\left(k^{2}+1\right)$

$\Rightarrow 3 k^{2}-5 k-2=3 k^{2}+3$

$\Rightarrow-5 k=5$

$\Rightarrow k=-1$

When k = −1,

$\frac{3}{k-2}=\frac{3}{-1-2}=\frac{3}{-3}=-1$

Thus, for $k=-1, \frac{3}{k-2} \neq \frac{2}{5}$

Hence, the given system of equations has no solution when k = −1.