kx – y + 3 – k = 0; 4x – ky + k = 0

(ii) kx – y + 3 – k = 0; 4x – ky + k = 0


Sol. kx – y + 3 – k = 0
∴  kx – y = k – 3
Comparing with a1x + b1y = c1 we get, a1 = k, b1 = – 1, c1 = k – 3

4x – ky + k = 0
∴  4x – ky = – k
Comparing with a2x + b2y = c2 we get, a2 = 4, b2 = – k, c2 = – k

The simultaneous equations have infinitely many solutions.

∴ a1/a2 = b1/b2 = c1/c2

k/4 = -1/-k = k-3/-k

∴ k/4 = 1/k     OR    1/k = k-3/-k
∴ k2 = 4          OR    -k/k = k – 3
∴ k = ±√4       OR   -1 = k – 3
∴ k = ± 2       OR     k = 2


k = 2 satisfies both conditions hence k = 2 is the value for which the given simultaneous equations have infinitely many solutions.

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