Question

For what value of $k$ will the following system of linear equations has no solutions

$3x+y=1;(2k-1)x+(k-1)y=2k+1$

Answer 1

Step 1: If the pair of linear equations has no solutions

Applying condition,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Given,

$$\begin{gathered} 3x+y=1 \\ \Rightarrow 3x+y-1=0........................................\left(1\right) \\ (2k-1)x+(k-1)y=2k+1 \\ \Rightarrow (2k-1)x+(k-1)y-(2k+1)=0.......................................\left(2\right) \end{gathered}$$

Equation (1) and (2) are in standard form

$$\begin{gathered} Here{a}_1=3b_1=1c_1=-1 \\ {a}_2=2k-1b_2=k-1c_2=-(2k+1) \\ here\frac{3}{2k-1}=\frac{1}{k-1}\ne \frac{-1}{-(2k+1)} \\ \Rightarrow \frac{3}{2k-1}=\frac{1}{k-1}\ne \frac{1}{2k+1} \end{gathered}$$

Step 2: To find the values of $\mathit{k}$ by cross multiplication method

$$\begin{gathered} taking\frac{3}{2k-1}=\frac{1}{k-1} \\ \Rightarrow 3(k-1)=2k-1 \\ \Rightarrow 3k-3=2k-1 \\ \Rightarrow 3k-2k=3-1 \\ \Rightarrow k=2 \end{gathered}$$

Checking,

$$\begin{gathered} \frac{3}{2k-1}=\frac{3}{4-1}=\frac{3}{3}=1 \\ \frac{1}{k-1}=\frac{1}{2-1}=1 \\ \frac{1}{2k+1}=\frac{1}{5} \\ here,\frac{3}{2k-1}=\frac{1}{k-1}\ne \frac{1}{2k+1} \end{gathered}$$

Therefore the value of $k=2$