Question

For what value of k will the following system of linear equations has no solution:

3x + y = 1

(2k − 1)x + (k − 1)y = 2k + 1

Answer 1

The given system of equations is

3x + y = 1

(2k − 1)x + (k − 1)y = 2k + 1

Here, a₁ = 3, b₁ = 1, c₁ = 1

a₂ = 2k − 1, b₂ = k − 1, c₂ = 2k + 1

The given system of linear equations has no solution.

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

Now,

$$\begin{gathered} \frac{3}{2k-1}=\frac{1}{k-1} \\ \Rightarrow 3k-3=2k-1 \\ \Rightarrow 3k-2k=-1+3 \\ \Rightarrow k=2 \end{gathered}$$

When k = 2,

$\frac{1}{k-1}=\frac{1}{2-1}=1$ and $\frac{1}{2k+1}=\frac{1}{2\times 2+1}=\frac{1}{5}$

Thus, for k = 2, $\frac{1}{k-1}\ne \frac{1}{2k+1}$

Hence, the given system of linear equations will have no solution when k = 2.