Question

Question 2 (ii)
For which value of k will the following pair of linear equations have no solution?
3x + y = 1
(2k -1)x + (k -1)y = 2k + 1

Answer 1

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

Compare the given equations with

$a_{1}x+b_{1}y+c_1=0and{a}_{2}x+b_{2}y+c_2=0$.

$\frac{a_1}{a_2}=\frac{3}{2k-1}$
$\frac{b_1}{b_2}=\frac{1}{k-1}$
$\frac{c_1}{c_2}=\frac{-1}{-2k-1}=\frac{1}{2k+1}$
For no solutions,
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
$\frac{3}{2k-1}=\frac{1}{k-1}\ne \frac{1}{2k+1}$
$\frac{3}{2k-1}=\frac{1}{k-1}$
$k=2$
Hence, for k = 2, the given equation has no solution.