Question

Solve the following systems of equations:

`6/(x + y) = 7/(x - y) + 3`

`1/(2(x + y)) = 1/(3(x - y))`, where x + y ≠ 0 and x – y ≠ 0

Answer 1

Let `1/(x + y) = u and 1/(x - y) = v` Then, the given system of equation becomes

6u = 7v + 3

=> 6u - 7v = 3 .....(i)

And `u/2 = v/3`

=> 3u = 2v

=> 3u - 2v = 0 ......(ii)

Multiplying equation   ii by 2, and equation (i) by 1, we get

6u - 7v = 3 ....(iii)

6u - 4v = 0 ......(iv)

Subtracting equation (iv) from equation (iii), we get

-7 + 4v = 3

=> -3v = 3

=> v = -1

Puttting v = -1 in equation (ii), we get

`3u - 2xx (-1) = 0`

=> 3u + 2 = 0

=> 3u = -2

`=> u = (-2)/3`

Now `u = (-2)/3`

=> `1/(x + 2) = (-2)/3`

`=> x + y = (-3)/2`   ...(v)

And v= -1

`=> 1/(x - y) = -1`

=> x - y = -1 ...(vi)

Adding equation (v) and equation (vi), we get

`2x = (-3)/2 - 1`

`=> 2x = (-3-2)/2`

`=> 2x = (-5)/2`

`=> x = (-5)/4`

Putting x = (-5)/4  in equation (vi), we get

`(-5)/4 - y = -1`

`=> (-5)/4 + 1 = y`

`=> (-5+4)/4 = y`

`=> (-1)/4 = y`

`=> y = (-1)/4`

Hence, solution of the system of equation is `x = (-5)/4, y = (-1)/4`