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.

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