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.