Question

44÷x+y + 30÷x-y =10; 55÷x+y + 40÷x-y=13

Answer 1

heya,
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Sol:
Given that   (44/x+y) + (30/x-y) = 10 and (55/x+y) + (40/x--y) = 13
 
Let 1/(x + y) =a, 1/(x - y) =b
 
⇒ 44a + 30b = 10 and 55a + 40b = 13
 
Solving the equations:
 
    4 [44a + 30b = 10]
    3 [55a + 40b = 13]
 
    176a +120 b = 40 ----------- (1)
  -165a+ 120 b = 39 ----------- (2)
 
Solving (1) and (2), we get a = 1/11
 
Substitute the value of a in 44a + 30b = 10
 
    44/11 + 30b = 10
    ⇒ 4 + 30b = 10
    ⇒ 6 = 30b
    ⇒ b = 6/30 = 1/5
 
But a = 1/(x + y), and b = 1/(x - y)
 
    a = 1/11 = 1/x + y
 
Then, x + y = 11 --------------- (3)
 
    b = 1/5 = 1/x - y
 
Then x - y = 5 --------------- (4)
 
Solving the equations (3) and (4), we get x = 8 and y = 3.
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hope this helped u,
cheers

Answer 2

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