Solve the following systems of equations:
`3/(x + y) + 2/(x - y) = 2`
`9/(x + y) - 4/(x - y) = 1`
Solution
Let `1/(x + y) = u` and `1/(x - y) = v` Then, the given system of equation becomes
3u + 2v = 2 ....(i)
9u + 4v = 1 ...(ii)
Multiplying equation (i) by 3, and equation (ii) by 1, we get
6u + 4v = 4 ....(iii)
9u - 4v = 1 ...(iv)
Adding equation (iii) and equation (iv), we get
6u + 9u = 4 +1
=> 15u = 5
`=> u = 5/15 = 1/3`
Putting `u = 1/3` in equation (i) we get
`3 xx 1/3 + 2v = 2`
`=> 1 + 2v = 2`
=> 2v = 2 - 1
=> v = 1/2
Now, u = 1/(x + y)
`=> 1/(x + y) = 1/3`
=> x + y = 3 ....(v)
And `v = 1/(x - y)`
`=> 1/(x - y) = 1/2`
=> x - y = 2 .....(vi)
Adding equation (v) and equation (vi), we get
2x = 3 + 2
`=> x = 5/2`
Putting `x = 5/2` in equation (v) we get
`5/2 + y = 3`
`=> y = 3 - 5/2`
`=> y = (6-5)/2 = 1/2`
Hence, solution of the given system of equation is `x = 5/2, y = 1/2`