Question

determine k
(3k=1)x + 3y - 2 =0
(k^2+1)x + (k-2)y - 5 = 0

Answer 1

Answer:For having no solution

a1/ a2 = b1/ b2 not= c1/ c2

3k + 1 / k2 + 1 = 3 / k - 2

3k2 - 6k + k - 2 = 3 k2 + 3

-5k -2 = 3

-5k = 5

K=-1