Question

Find the value of k for which the following system of equations has no solution:
kx+3y=312x+ky=6

Answer 1

k = +-6

Step-by-step explanation:

Given:  

kx + 3y = 3

12x +ky = 6

The system of equations has no solution.

a1 = k, b1 = 3, c1 = 3

a2 = 12, b2 = k, c2 = 6

 So  a1/a2 = b1/b2 but not equal to c1/c2

 k/ 12 = 3/ k

 k*k = 12 * 3

 k = root of 36

Therefore k = +-6

Answer 2

$\Large{\underline{\underline{\bf{Solution :}}}}$

As, it is given equations has no solution.

We know the case of no solution.

$\Large{\star{\boxed{\rm{\frac{a_{1}} {a_{2}} = \frac{b_1}{b_2} ≠ \frac{c_1}{c_2}}}}}$

Where,

a1 = k, a2 = 12

b1 = 3, b2 = k

c1 = 3, c2 = 6

________________[Put Values]

$\sf{→\frac{k}{12} = \frac{3}{k} ≠ \frac{3}{6}} \\ \\ \sf{→\frac{k}{12} = \frac{3}{k}} \\ \\ \sf{→k \times k = 12 \times 3} \\ \\ \sf{→k^2 = 36} \\ \\ \sf{→k = \sqrt{36}} \\ \\ \sf{→k = (\sqrt{6})^2} \\ \\ \sf{k = \pm 6} \\ \\ \Large{\star{\boxed{\sf{K = \pm 6}}}}$