Question

For what value of $\mathrm{k}$ the following system of linear equations will have no solution?

$$\begin{gathered} 3\mathrm{x}+\mathrm{y}=1 \\ (2\mathrm{k}-1)\mathrm{x}+(\mathrm{k}-1)\mathrm{y}=2\mathrm{k}+1 \end{gathered}$$

Answer 1

Find the value of $\mathrm{k}$ for the given condition.

Given: System of linear equation is

$$\begin{gathered} 3\mathrm{x}+\mathrm{y}=1 \\ (2\mathrm{k}-1)\mathrm{x}+(\mathrm{k}-1)\mathrm{y}=2\mathrm{k}+1 \end{gathered}$$

A system of linear equation

$$\begin{gathered} {\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0 \\ {\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2=0 \end{gathered}$$

will have no solution if $\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}\ne \frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}$

On comparing given system of linear equations with general system of linear equations, we get;

${\mathrm{a}}_1=3,{\mathrm{b}}_1=1,{\mathrm{c}}_1=-1$ and ${\mathrm{a}}_2=2\mathrm{k}-1,{\mathrm{b}}_2=\mathrm{k}-1,{\mathrm{c}}_2=-2\mathrm{k}-1$

Here, $\frac{3}{2\mathrm{k}-1}=\frac{1}{\mathrm{k}-1}\ne \frac{-1}{-\left(2\mathrm{k}+1\right)}$

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

$\Rightarrow$ $3\mathrm{k}-3=2\mathrm{k}-1$

$\Rightarrow$ $3\mathrm{k}-2\mathrm{k}=3-1$

$\Rightarrow$ $\mathrm{k}=2$

Hence, the given system of equation has no solution for $\mathrm{k}=2$