If $(2 k - 1, k)$ is a solution of the equation $10 x - 9 x y = 12$ then find the value of $k$ .

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Solution

$a s (2 k - 1, k) i s a s o l u t i o n o f t h e e q u a t i o n h e n c e i t m u s t s a t i s f y t h e g i v e n e q u a t i o n p u t x = 2 k - 1 a n d y = k t h e n 10 (2 k - 1) - 9 (2 k - 1) (k) = 12 o r 20 k - 10 - 9 (2 k^{2} - k) = 12$

$o r 20 k - 10 - 18 k^{2} + 9 k = 12 o r - 18 K^{2} + 29 k - 22 = 0$

$18 k^{2} - 29 k + 22 = 0 a s i t s D = (- 29)^{2} - 4.18.22 = 841 - 1584 < 0$

$∴ n o v a l u e o f k i s e q u a t i o n^{'} s s o l u t i o n$

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