For what value of 'k' will the pair of linear equations given below have infinite solutions? Kx+ 3y-(k-3)=0 12x+ky-k=0

Question

For what values of K , the pair of equations Kx+3y= K-3, 12x+Ky=K has unique solution?

Answer 1

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.

Here a1 = k ,b1 = 3 , a2 = 12 , b2 = k .

k / 12 ≠ 3 / k

k² ≠ 36.

k ≠ ± 6.