Question

For which value(s) of k will the pair of equations have no solution?
$kx+3y=k-3;12x+ky=k;k\ne 0$

Answer 1

The given pair of linear equation can be written as

$kx+3y-k+3=0$ and $12x+ky-k=0$

Here, $a_1=k,b_1=3,c_1=-k+3$ and $a_2=12,b_2=k,c_2=-k$

Now, the given pair of linear equation has no solution if

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}\Rightarrow \frac{k}{12}=\frac{3}{k}\ne \frac{-k+3}{-k}$

$\frac{k}{12}=\frac{3}{k}$ and $\frac{3}{k}\ne \frac{k-3}{k}$

$k^2=36$ and $3\ne k-3$

$k=\pm 6$ and $k\ne 6$

$k=-6$

Hence, the given pair of linear equation has no solution if $k=-6$.