Solution to the equation please.....?

Question

If 12(x^2+y^2)=25xy then show that x^2-y^2 is either 3/4 xy or 4/3xy.

Answer 1

12(x^2 + y^2) = 25xy
12x^2 + 12y^2 = 25xy
12x^2 - 25xy + 12y^2 = 0
12x^2 + 12y^2 = 25xy
(4x - 3y) (3x - 4y) = 0
x = 3/4y
y - 4/3x

x^2 - y^2 = 0
(x + y) (x - y) = 0

I just realized you have a negative sign in the second equation; is that supposed to be negative?
I hope this is correct

Answer 2

Continue from above after the line:
12x^2 - 25xy + 12y^2 = 0 .....(A)
Note that x and y are interchangeable.

Since x = (3/4) y or y = (4/3) x
Multiply by x, or y, respectively:
x^2 = (3/4) xy or y^2 = (4/3) xy, vice-versa because we can also factorize it as (4y - 3x) (3y - 4x) = 0

Alternatively, by dividing throughout eq. (A) by either y^2 or x^2, it can also be solved to get x/y = 3/4 or 4/3, and vice versa.

Thus, using the vice-versa case: x^2 - y^2 = (4/3 - 3/4) xy
= ((16-9)/12) xy
= (7/12) xy
So, it cannot be shown that x^2-y^2 is either 3/4 xy or 4/3 xy.