# 30/x-y+4y/x+y=10  40/x-y+55/x+y=13

Dear Student,

I assume your question as below and have provided the solution, as it appears to be faulty.

Given : $\left(\frac{44}{x+y}\right)$  + $\left(\frac{30}{x-y}\right)$ = 10
$\left(\frac{55}{x+y}\right)$  - $\left(\frac{40}{x-y}\right)$ = 13

To find : the values of x and y

Let $\left(\frac{1}{x+y}\right)$ be a and  $\left(\frac{1}{x-y}\right)$ be b
then,
=>   44a +30b =10 .................(1)
and 55a +40b=13..................(2)

Multiply eq (1) by 4 and eq (2) by 3
=>   176a +120b =40 .................(3)
and 165a +120b=39..................(4)

Now subtract eq (3) by eq 4 , we get
11a = 1
=> a = 1/ 11
applying this value of a  in eq 1 we get

=> 44$\left(\frac{1}{11}\right)$ + 30 b = 10
=> 30 b = 6
=> b = 1/5

Now as a = $\left(\frac{1}{x+y}\right)$ and b = $\left(\frac{1}{x-y}\right)$
=> x+y = 11............(5)
and x-y = 5.............(6)

adding eq (5) and eq (6)

=>2x = 16
=> x= 8

similarily applying this value in eq (5) we get
=> 8+ y = 11
=> y = 3

Regards.

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