Question

Find the value of a so that the following linear equations have no solution: (3a+1)x 3y-
2=0, (a+1)X + (a-2)y-5 = 0.​

Answer 1

Given :- Equation will have no solution.

(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.