Question

- For what value of 'K', the system of equations
kx + 3y = 1, 12x + ky = 2 has no solution.
[Board Term-1, 2011, Set-A2 NCERT]​

Answer 1

Step-by-step explanation:

the value of k will be 6.

Answer 2

Answer:

since

kx+3y=1,*2

=2kx+6y=2...eq1

12x-ky=2....eq2

so by sub both eq..

ans=2kx-12x+6y-ky=0.

so, 2kx-12x=ky-6y.

so take 2 common in lhs

2(kx-6x)=ky-6y

take xandycommon

2x(k-6).=y(k-6)

by cancelling k-6

so, 2x=y

so, kx+3y=1

=kx+3(2x)=1

=kx+6x=1

=x=1/k+6

so eq 2

12x-ky=2

12x-2kx=2

12(1/k+6)-2k(1/k+6)=2

12/k+6-2k/k+6=2

12-2k/k+6=2

12-2k/2k+12=0

6-k/6+k=0

6-k=6+k

2k=12

since k=6

Step-by-step explanation:

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