Question

Find the value of K so that the pair of linear equations, (3 K + 1) x + 3 y – 2 = 0 and (K^2 + 1) x + (k–2)y – 5 = 0 is consistent​

Answer 1

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

(k²+1)x + (k-2)y - 5 = 0

3k+1/k²+1 = 3/k-2 ≠ -2/-5

Now, 3k + 1 / k² + 1 = 3/k - 2

⇒ ( 3k + 1 ) ( k-2 ) = 3 ( k² + 1 )

⇒ 3k² - 5k - 2 = 3k² + 3

⇒ -5k - 2 = 3

⇒ -5k = 5

⇒ k = -1

Hence, given system of equation will have no solution for

k = - 1.