Solve the following pairs of equation for `x and y`, `15/(x-y)+22/(x+y)=5` and `40/(x-y) + 55/(x+y)=13`

Explanation

Putting ` (1)/(x - y ) = uand (1)/( x + y ) = v `, the given equations become
` 15u + 22v = 5 " " `… (i)
` 40 u + 55v = 13" " `… (ii)
Multiplying (ii)by 2 and (i) by 5 and subtracting the results, we get
`80 u - 75 u = 26 - 25`
`rArr 5u = 1 `
` rArr u = (1)/(5)`
Putting ` u = (1)/(5)` in (i), we get
` ( 15 xx (1)/(5) ) + 22 v = 5`
` rArr 3 + 22 v = 5 `
` rArr 22 v = 2 rArr v = (2)/(22) = (1)/(11)`
Now, ` u = (1)/(5) rArr (1)/(x- y ) = (1)/(5) rArr x - y = 5 " " `... (iii)
And, ` v = (1)/(11) rArr (1)/( x+ y ) = (1)/(11) rArr x +y = 11 " " `... (iv)
On adding (iii) and (iv), we get ` 2x= 16 rArr x = 8`.
On subtracting (iii) from (iv), we get ` 2y = 6 rArr y = 3.`
Hence, ` x = 8 and y = 3`.