kx + y = k – 2; 9x + ky = k

(i) kx + y = k – 2; 9x + ky = k

Sol. kx + y = k – 2

Comparing with a1x + b1y = c1 we get, a1 = k, b1 = 1, c1 = k – 2

9x + ky = k

Comparing with a2x + b2y = c2 we get, a2 = 9, b2 = k, c2 = k

∵ The simultaneous equations have infinitely many solutions.

a₁/a₂ = b₁/b₂ = c₁/c₂

∴ k/9 = 1/k =  k-2/k   [From eq.

∴ k/9 = 1/k    OR   1/k =  k-2/k

∴ k² = 9         OR  k/k = k-2

∴ k = ±√9       OR  1 = k – 2

∴ k =  ± 3       OR   3 = k

k = 3 satisfies both conditions hence k = 3 is the value for which the given simultaneous equations have infinitely many solutions.