Question

Solve the following systems of equations:

`1/(3x + y) + 1/(3x - y) = 3/4`

`1/(2(3x + y)) - 1/(2(3x - y)) = -1/8`

Answer 1

The given equation is

`1/(3x + y) + 1/(3x - y) = 3/4`

`1/(2(3x + y)) - 1/(2(3x - y)) = -1/8`

Let `1/(3x + y) = u and 1/(3x - y) = v` then equation are

`u + v = 3/4` ...(i)

`u/2 - v/2 = 1/8` ....(ii)

Multiply equation (ii) by 2 and add both equations, we get

`u + v = 3/4`

`u - v = -1/4`

`2u = 1/2`

`u = 1/4`

Put the value of u  in equation (i) we get

`1 xx 1/4 + v = 3/4`

`v= 1/2`

Then

`1/(3x + y) = 1/4`   ....(iii)

3x  + y = 4

`1/(3x -y) = 1/2`  ...(iv)

3x - y = 2

Add both equation we get

3x + y = 4

3x - y = 2

_________

6x = 6

x = 1

Put the value of x in equation (iii) we get

3 x 1 + y = 4

y = 1

Hence value of x = 1 and  y = 1