Question

Solve for x and y:
`1/(3x+y) + 1/(3x−y) = 3/4, 1/(2(3x+y)) - 1/(2(3x−y)) = −1/8`

Answer 1

The given equations are
`1/(3x+y)+ 1/(3x−y) = 3/4`      ……(i)
`1/(2(3x+y)) - 1/(2(3x−y)) = −1/8`
`1/(3x+y) - 1/(3x−y) = −1/4` (Multiplying by 2)         ……(ii)
Substituting `1/(3x+y) = u and 1/(3x−y)` = v in (i) and (ii), we get:
u + v = `3/4`                             ……..(iii)
u – v = −`1/4`                                  …….(iv)
Adding (iii) and (iv), we get:
2u = `1/2`

⇒ u = `1/4`
` ⇒3x + y = 4   (∵1/(3x+y) =u)`                …..(v)
Now, substituting u = `1/4` in (iii), we get:
`1/4 + v = 3/4`
v = `3/4 - 1/4`
⇒v = `1/2`
⇒ 3x – y = 2              (∵`1/(3x−y` = v)       …..(vi)
Adding (v) and (vi), we get
6x = 6 ⇒x = 1
Substituting x = 1 in (v), we have
3 + y = 4 ⇒y = 1
Hence, x = 1 and y = 1.