Question

Determine the values of $k$ so that the following linear equations have no solutions

$(3k+1)x+3y-2=0;(k^2+1)x+(k-2)y-5=0$

Answer 1

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

Then,

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

Given,

$$\begin{gathered} (3k+1)x+3y-2=0..................................................\left(1\right) \\ (k^2+1)x+(k-2)y-5=0..............................................\left(2\right) \end{gathered}$$

Equations (1) and (2) are in standard forms

Step 2:Applying conditions

$$\begin{gathered} Here,a_1=3k+1b_1=3c_1=-2 \\ {a}_2=k^2+1b_2=k-2c_2=-5 \\ ie,\frac{3k+1}{k^2+1}=\frac{3}{k-2}\ne \frac{-2}{-5} \\ \Rightarrow \frac{3k+1}{k^2+1}=\frac{3}{k-2}\ne \frac{2}{5} \end{gathered}$$

Step 3:To find $k$ using cross multiplication method

$$\begin{gathered} Taking\frac{3k+1}{k^2+1}=\frac{3}{k-2} \\ \Rightarrow (3k+1)(k-2)=3\left(\right(k^2+1) \\ \Rightarrow 3k^2-6k+k-2=3k^2+3 \\ \Rightarrow 3k^2-5k-2-3k^2-3=0 \\ \Rightarrow -5k-5=0 \\ \Rightarrow -5k=5 \\ \Rightarrow k=-1 \end{gathered}$$

Checking

$$\begin{gathered} \frac{3}{k-2}=\frac{3}{-1-2}=\frac{3}{-3}=-1 \\ \frac{3k+1}{k^2+1}=\frac{3(-1)+1}{{(-1)}^2+1}=\frac{-3+1}{1+1}=\frac{-2}{2}=-1 \\ Therefore\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2} \end{gathered}$$

Therefore the value of $k=-1$