Question

Solve the following systems of equations:

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

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

Answer 1

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`