Question

answer for this question​

Answer 1

Answer:

Step-by-step explanation:

(1)

$Let \;\;\frac{1}{x+y}=a\\and\\\frac{1}{x-y}=b$

Now, the equations become

$5a+b=2....(i)\;\;\;\;\;;\;;\ 10a+3b=5.........(ii)$

Multiplying $(i)$ by 2, we get:

$2(5a+b)=2(2)\\=>10a+2b=4......(iii)$

Now subtracting:

Taking  $(iii) \;\;\;from \;\;\;(ii)$

$=>10a+3b-10a-2b=5-4\\=>b=1.....(iv)\\Now, \;using(iv)\;in\;(i),\;we\;get\\ 5a+b=2\\=>5a+1=2\\=>a=\frac{1}{5}$

but

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

so,

$\frac{1}{x+y}=\frac{1}{5}\\=>x+y=5....(v)\\\\\frac{1}{x-y}=1\\=>x-y=1....(vi)\\\\\\adding(v)\;and\;(vi),\;we\;get\\x+y+x-y=5+1\\=>2x=6\\=>x=3....(vi)\\using\;(vi)\;in(v)\\we\;get\\3+y=5\\=>y=2\\\\\\\\Hence\\x=3\\\\y=2\\\\\\\\Learning\;\;together\;:)$