Question

for what value of k will the following equation have no solution in the following equation (3k+1)x+3y=2 and (k2+1)x+(k-2)y=5​

Answer 1

ans is k=-1 plzzz give me thanks and brainliest ans

Answer 2

Heyaa☺

_____________________

The lines which have no solution are parallel lines.

i.e, a1/a2 = b1/b2 is not equals to c1/c2.

Here,

a1 = 3k + 1 ; a2 = k^2 + 1

b1 = x + 3 ; b2 = k - 2

c1 = -2 ; c2 = -5.

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

3(k^2 + 1) = (k-2)3k + 1

=》 k = -1.

_______________________

Hope you understand...✌