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Find the value of x and y in the following equations:
\(\dfrac{10}{x+y}+\dfrac{2}{x-y}=4\)
\(\dfrac{15}{x+y}-\dfrac{5}{x-y}=-2\)
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Find the value of x and y in the following equations:
\(\dfrac{10}{x+y}+\dfrac{2}{x-y}=4\)
\(\dfrac{15}{x+y}-\dfrac{5}{x-y}=-2\)

Answer

\(x=3, y=2\)
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Solution

Given:
\(\dfrac{10}{x+y}+\dfrac{2}{x-y}=4\)
\(\dfrac{15}{x+y}-\dfrac{5}{x-y}=-2\)
Substituting \(\frac{1}{x+y} = m\) and \(\frac{1}{x-y} = n\) in the given equations, we get, 
\(10m + 2n = 4 \implies 10m + 2n – 4 = 0\)....  (i)
\(15m – 5n = -2 \implies 15m – 5n + 2 = 0\)...  (ii)
Using cross-multiplication method, we get,
\(\dfrac{m}{4-20}=\dfrac{n}{-60-(20)}=\dfrac{1}{-50-30}\)
\(\dfrac{m}{-16}=\dfrac{n}{-80}=\dfrac{1}{-80}\)
\(\dfrac{m}{-16}=\dfrac{1}{-80}\)and\(\dfrac{n}{-80}=\dfrac{1}{-80}\)
\(m=\dfrac{1}{5} \)and \(n=1\)
\(m=\dfrac{1}{x+y}=\dfrac{1}{5}\) and \(n=\dfrac{1}{x-y}=1\)
\(x + y = 5\) .... (iii)
and \(x – y = 1\) .....  (iv)
Adding equation (iii) and (iv), we get
\(2x = 6 \implies x = 3\) .....  (v)
Putting the value of \(x = 3\) in equation (3), we get
\(y = 2 \)
Hence, \(x = 3\) and \(y = 2 \)
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