Question

find the value of k such that the pair of linear equation kx + 3 y = 2 k - 3 and 12 x + k y = 2 k is consistent dependent​

Answer 1

Given pair of linear equations:

kx+3y-(k-3)=0—-(1)

12x+ky-k=0—-(2)

Compare above equations with,

a1x+b1y+c1 =0 and a2x+b2y+c2=0, we get

a1=k , b1 = 3, c1 = -(k-3);

a2=12 , b2 = k , c2 = -k,

Now,

a1/a2 ≠ b1/b2

/* Given, equations have unique

solution*/

k/12 ≠ 3/k

=> k² ≠ 36

=>k ≠ √36

=> k ≠ ± 6

Therefore,

k should be all real values other than k ≠ ±6