Two numbers x and y (x > y) such that their sum is equal to three times their difference. Then value of \( \Large \frac{3xy}{2 \left(x^{2}-y^{2}\right) } \) will be;

A) \( \Large \frac{2}{3} \)

B) 1

C) \( \Large 1\frac{1}{2} \)

D) \( \Large 1\frac{1}{3} \)

Correct answer:
B) 1

Description for Correct answer:
x > y

given :-

\( \Large x+y = 3 \left(x-y\right) \)

=> \( \Large x + y = 3x - 3y \)

=> \( \Large x - 3x = -3y - y \)

=> \( \Large -2x = -4y => x = 2y \)

Therefore, \( \Large \frac{3xy}{2 \left(x^{2}-y^{2}\right) }=\frac{3 \times 2y \times y}{2 \left( \left(2y\right)^{2}-y^{2} \right) } \)

= \( \Large \frac{6y^{2}}{2 \times \left(4y^{2}-y^{2}\right) } \)

= \( \Large \frac{6y^{2}}{6y^{2}} = 1 \)


Please provide the error details in above question