Question

Answer.... tis....,...............

Answer 1

We have equations :

( k - 3 )x + 3y = k

So,

( k - 3 )x + 3y - k = 0

And

kx + ky = 12

So,

kx + ky - 12 = 0

And for given equations to have infinitely many solutions , So

a1a2 = b1b2 =c1c2So, a1a2 = b1b2⇒k − 3k = 3k⇒k − 3= 3 ⇒k = 6

Answer 2

(k-3)x+3y =k

k(x+y)=12

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

a2=k, b2=0 and c2= 12

For infinite number of solutions

a1/a2 =b1/b2

k-3/k = 3/0

k= 1/3

a1/a2 = c1/c2

k-3/k = k/12

k^2= 12k-36

k^2-12k+36

k^2-6k-6k+36

k(k-6)-6(k-6)

(k-6)(k-6)

k-6=0

k= 6

.................