44/x+y +30/x-y =10,55/x+y +40/x-y=13,x+y and x-y are not equal to zero

Asked by samridhi arora | 22nd Jun, 2013, 03:45: PM

Expert Answer:

Answer : Given : (44/(x+y) ) +(30/(x-y)) =10
(55/(x+y)) +(40/(x-y)) =13
 
To find : the values of x and y
 
Let 1/(x+y) be a and 1/(x-y) 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(1/11) + 30 b = 10
=> 30 b = 6
=> b = 1/5 
 
Now as a= 1/( x+ y) and b = 1/(x-y)
=> x+y = 11............(5)
 and x-y = 5.............(6)
 
adding eq 5 and eq 6 
 
=>2x = 16 
=> x= 8 Answer
 
similarily applying this value in eq 5 we get
=> 8+ y = 11
=> y = 3 Answer
 


Answered by  | 22nd Jun, 2013, 06:05: PM

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