**Question:**

For which value (s) of k will the pair of equations

kx+3y = k – 3,

12 x + ky =k

has no solution?

**Solution:**

Given pair of linear equations is

kx + 3y = k – 3

and 12x + ky = k

On comparing with ax + by + c = 0, we get

a1 = k, b1 = 3 and c1 = -(k-3)

a2 = 12,b2 = k and c2 = -k

For no solution of the pair of linear equations,

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

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

Taking first two parts, we get

$\Rightarrow \quad \frac{k}{12}=\frac{3}{k}$

$\Rightarrow \quad k^{2}=36$

$\Rightarrow \quad k=\pm 6$

Taking last two parts, we get

$\frac{3}{k} \neq \frac{k-3}{k}$

$\Rightarrow \quad 3 k \neq k(k-3)$

$\Rightarrow \quad 3 k-k(k-3) \neq 0$

$\Rightarrow \quad k(3-k+3) \neq 0$

$\Rightarrow \quad k(6-k) \neq 0$

$\Rightarrow \quad k \neq 0$ and $k \neq 6$

Hence, required value of k for which the given pair of linear equations has no solution is -6.