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

Answer 1

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.

$$\begin{gathered} \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2} \\ \Rightarrow \frac{3k+1}{k^2+1}=\frac{3}{k-2}\ne \frac{-2}{-5} \\ \Rightarrow \frac{3k+1}{k^2+1}=\frac{3}{k-2}\mathrm{and}\frac{3}{k-2}\ne \frac{2}{5} \end{gathered}$$

Now,

$$\begin{gathered} \frac{3k+1}{k^2+1}=\frac{3}{k-2} \\ \Rightarrow \left(3k+1\right)\left(k-2\right)=3\left(k^2+1\right) \\ \Rightarrow 3k^2-5k-2=3k^2+3 \\ \Rightarrow -5k=5 \end{gathered}$$
$\Rightarrow k=-1$

When k = −1,

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

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

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