Question

If y(3x-y):x(4x+y)= 5:12, then find the ratio (x^2+y^2):(x+y)^2

Answer 1

Answer:

Step-by-step explanation:

Given  

If y(3x-y):x(4x+y)= 5:12, then find the ratio (x^2+y^2):(x+y)^2

So we need to find the ratio. So product of means is equal to product of extremes.

So 12 y(3 x – y) = 5 x(4 x + y)

36xy – 12y^2 = 20x^2 + 5xy

36xy – 5xy = 20x^2 + 12 y^2

31xy = 20x^2 + 12y^2

20x^2 – 31xy + 12 y^2 = 0

20x^2 – 16 xy – 15 xy + 12y^2 = 0

4x(5x – 4y) – 3y(5x – 4y) = 0

5x – 4y = 0        4x – 3y = 0

5x = 4y                 4x = 3y

x = 4y / 5               x = 3y / 4

Now we need to find  

x ^2 + y^2 / (x + y)^2

Substituting for x we get

x = 4 y/5

16 y^2 / 25 + y^2 / (9 y / 5)^2

41 y^2 / 25 / 81 y^2 / 25

= 41 / 81

When x =3 y/4 we get

9 y^2 / 16 + y^2 / (7y / 4)^2

= 25 y^2 / 16 / 49 y^2 / 16

= 25 / 49