Correct option is (A) x = 3, y = 2
\(\frac{10}{x+y}+\frac{2}{x-y}= 4\) ________(1)
and \(\frac{15}{x+y}-\frac{5}{x-y}=-2\) ________(2)
Put \(\frac1{x+y}=X\;\&\;\frac{1}{x-y}=Y\)
Then equations (1) & (2) converts to
10X + 2Y = 4 ________(3)
and 15X - 5Y = -2 ________(4)
Multiply equation (3) by \(\frac52,\) we get
25X + 5Y = 10 ________(5)
By adding equation (4) & (5), we get
(25X + 5Y) + (15X - 5Y) = 10 - 2
\(\Rightarrow\) 40X = 8
\(\Rightarrow\) X \(=\frac8{40}=\frac15\)
Then from (3), we get
\(10\times\frac15+2Y=4\)
\(\Rightarrow\) 2 + 2Y = 4
\(\Rightarrow\) 2Y = 4 - 2 = 2
\(\Rightarrow\) Y = \(\frac22\) = 1
\(\because\) \(X=\frac15\)
\(\Rightarrow\) \(\frac1{x+y}=\frac{1}5\)
\(\Rightarrow\) x+y = 5 ________(6)
and Y = 1
\(\Rightarrow\) \(\frac1{x-y}=1\)
\(\Rightarrow\) x - y = 1 ________(7)
By adding equations (6) & (7), we get
2x = 5+1 = 6
\(\Rightarrow\) x = \(\frac62\) = 3
\(\therefore\) y = 5 - x
= 5 - 3 = 2
