Question

24. Solve by eliminating method 44/x+y +30/x-y =10.​

Answer 1

Answer:

where is your second equation your question is incomplete

Answer 2

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