Question

find the values of k for which the following pair of linear equation is 6x +3y k 3: 12x +ky = k

Answer 1

::Only applicable if it has got a unique solution!
If  a1 /  a2 ≠  b1 /  b2 then the pair of linear equations a1x + b1y + c1 = 0

a2x+ b2y + c2 = 0  has a unique solution.

Given pair of  equations K x + 3y = K-3, 12x + Ky = K.

a1 = k ,b1 = 3 , a2 = 12 , b2 = k .

k / 12 ≠ 3 / k

k^2≠ 36.

k  ± 6.

Answer 2

Answer:

6x+3y=k-3

12x+ky=k

Step-by-step explanation:

a1/a2 = b1/b2 = c1/c2

6/12 = 3/k = k-3/k

so,

6/12=3/k => 1/2=3/k => k=6

hence, we get,

          a1/a2=b1/b2=c1/c2