Question

Q31 if you solve it you area genius

Answer 1

Hey friend, Harish here.

Here is your answer:

Let:

$\frac{1}{x+y} = a \ \ \ \ \& \ \ \ \ \frac{1}{x-y} = b$

Then,

$\frac{3}{x+y} + \frac{2}{x-y} = 2 ; \\ \\ \to 3a + 2b = 2 \ \ -\ (i) \\ \\ \frac{9}{x+y} - \frac{4}{x-y} = 1 ; \\ \\ \to 9a - 4b = 1 \ \ - (ii) \\ \\ Now \ multiply \ (i) \ with \ 3, and \ subtract\ (ii) \ from \ (i): \\ \\ Then, \\ \\ 10b = 5 \\ \\ \to b = \frac{1}{2} \\ \\ Now \ substituting \ b \ in \ (i) \ we \ get\ a = \frac{1}{3} . \\ \\ \boxed{\bold{Therefore \ a = \frac{1}{3} ; \ b = \frac{1}{2} }}$
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Hope my answer is helpful to you.