Question

for what value of k will the system of equation has no solution (3x+1)x + 3y - 2=0 & (k^2+1)x + k-2y-5=0

Answer 1

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.

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