Question

Solve the following systems of equations:

`44/(x + y) + 30/(x - y) = 10`

`55/(x + y) + 40/(x - y) = 13`

Answer 1

Let `1/(x + y) = u and 1/(x - y) = v`

Then, the system of the given equations becomes

44u + 30v = 10 ....(i)

55u + 40v = 13 ....(ii)

Multiplying equation (i) by 4 and equation (ii) by 3, we get

176u + 120v = 40 ...(iii)

165u + 120v  = 39 ...(iv)

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

176 - 165u = 40 - 39

=> 11u = 1

`=> u = 1/11`

Putting u = 1/11 in equation (i) we get

`44 xx 1/11 + 30v = 10`

4 + 30v = 10

=> 30v = 10 - 4

=> 30v = 6

`=> v = 6/30 = 1/5`

Now `u = 1/(x + y)`

`=> 1/(x + y) = 1/11`

=> x + y = 11 ...(v)

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

2x = 11 + 5

=> 2x = 16

`=> x = 16/2 = 8`

Putting x = 8 in equation (v) we get

8 + y = 11

=> y = 11 - 8 - 3

Hence, solution of the given system of equations is x = 8, y = 3