Question

Solve the following system of equations:

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

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

Solve: `3/(x + y) + 2/(x - y) = 2` and `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`

⇒ `y = 1/2`

Hence, solution of the given system of equation is `x = 5/2, y = 1/2`.