Find all pairs of real numbers (x,y) such that

(x-2)/y + 5/xy = (4-y)/x - |y-2x|/xy

End result:  (1,2) is the only solution I found

Case 1:  y>2x
(x-2)/y + 5/xy = (4-y)/x - (y-2x)/xy   mult by xy
x^2 - 2x + 5 = 4y - y^2 - y + 2x
x^2 - 4x + 5 = -(y^2 - 3y  )
x^2 - 4x + 5 + y^2 - 3y  = 0
x^2 - 4x + 3 + y^2 - 3y + 2  = 0
(x-1)(x-3) + (y-1)(y-2) = 0       
(x,y) = {(1,1), (1,2), (3,1), (3,2)}
except all are rejected
because none of these have y>2x

Case 2:  y<2x
(x-2)/y + 5/xy = (4-y)/x - (2x-y)/xy   mult by xy
x^2 - 2x + 5 = 4y - y^2  - 2x + y
x^2   + 5 - 5y + y^2 = 0
x^2 + y^2 - 5y  + 5  = 0
x^2 + (y-1)(y-5) = 0
(x,y) = {(0,1), (0,5)}
again all are rejected since y is not less than 2x

Case 3:  y=2x
(x-2)/y + 5/xy = (4-y)/x
(x-2)/2x + 5/2x^2 = (4-2x)/x  mult by 2x^2
x(x-2) + 5 = 2x(4-2x)
x^2 - 2x + 5 = 8x - 4x^2
5x^2 - 10x + 5 = 0
(x-1)^2 = 0      (x,y) = (1,2)

So the only solution is (1,2), and this does solve the original equation.

Posted by Larry on 2023-07-22 12:22:03