Question

The value of k for which the pair of linear equations 3kx + 6y= $\sqrt{50}$and $\sqrt{18} x+\sqrt{24} y$=$\sqrt{75}$ have a unique solution is

(a) k ≠ -6
(b) k ≠ 6
(c) k ≠ $\sqrt{3}$
(d) k ≠ -3

Answer 1

Answer:

k=

$k \: is \: not \: equal \: to \: \sqrt{3} \: \: option \: c$