Given: (k – 1)x – y = 5 – eq 1

(k + 1)x + (1 – k)y = (3k + 1) – eq 2

Here,

a₁ = (k - 1), b₁ = - 1, c₁ = - 5

a₂ = (k + 1) , b₂ = (1 - k), c₂ = - (3k + 1)

Given that system of equations has infinitely many solution

∴ = =

= =

Here,

=

(3k + 1) = - 5×(1 - k)

3k + 1 = - 5 + 5k

5K – 3k = 1 + 5

2k = 6

k =

k = 3

∴ k = 3