Determine the value of k so that the following system of equation have no solution

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

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

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Solution

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

(k^2+1)x+(k−2)y−5=0

For a system which has no solution,

a1/a2 = b1/b2 ≠ c1/c2

∴3k+1/(k^2+1) = 3/(k−2) ≠ 2/5

∴3k+1/(k^2+1) = 3/(k−2)

⟹ (3k+1)(k−2) = 3(k^2+1)

⟹ 3k^2− 6k + k − 2 = 3k^2 + 3

-5k = 5

k = -1

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