Question

Question 2
For which value(s) of k will the pair of equations
kx + 3y = k – 3,
12x + ky = k
has no solution?

Answer 1

Given pair of linear equations is
$kx+3y=k–3$ and $12x+ky=k$

On comparing with $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, we get

$a_1=k,b_1=3and{c}_1=-(k–3)a_2=12,b_2=kand{c}_2=-k$

For the pair of linear equations to have no solution,

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

$\Rightarrow \frac{k}{12}=\frac{3}{k}\ne \frac{-(k-3)}{-k}$

Now,
$\frac{k}{12}=\frac{3}{k}$

$\Rightarrow k^2=36\Rightarrow k=\pm 6$

And,

$\frac{3}{k}\ne \frac{k-3}{k}\Rightarrow 3k\ne k(k-3)\Rightarrow 3k-k(k-3)\ne 0\Rightarrow k(3-k+3)\ne 0\Rightarrow k\ne 0andk\ne 6$
Hence, required value of k for which the given pair of linear equations has no solutions is – 6.