The pair of linear equations kx + 2y = 5 and 3x + y = 1 has a unique solution if:

$k\neq 6$

The given equations are

kx + 2y = 5

3x + y = 1

We know that,

If (a1/a2) ≠ (b1/b2), then the system of equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 has a unique solution.

kx + 2y - 5 = 0      ----(1)

a1 = k, b1 = 2, c1 = -5

3x + y - 1 = 0      ----(2)

a2 = 3, b2 = 1, c2 = -1

According to the question,

(a1/a2) ≠ (b1/b2)

⇒ (k/3) ≠ (2/1)

⇒ 6 ≠ k