Lets split this into two halves, y-2x>=0 and y-2x<=0.
First half, when y-2x>=0, then makes the equation into:
(x-2)/y + 5/xy = (4-y)/x - (y-2x)/xy
This can be simplified to an equation of a circle:
(x-2)^2 + (y-3/2)^2 = 5/4
But the line y-2x=0 is actually tangent to the circle at (1,2)
It happens that the circle lies on the side of the tangent with y-2x<=0.
Then the only point on the circle satisfying y-2x>=0 is the tangent point.
Second half, when y-2x<=0, is nearly identical:
This makes the equation into:
(x-2)/y + 5/xy = (4-y)/x + (y-2x)/xy
This can be simplified to an equation of a circle:
x^2 + (y-5/2)^2 = 5/4
But the line y-2x=0 is actually tangent to the circle at (1,2)
It happens that the circle lies on the side of the tangent with y-2x>=0.
Then the only point on the circle satisfying y-2x<=0 is the tangent point.
So the solution set to the original equation is the single point (1,2).
Addendem: What I find more interesting is to change the sign on preceding absolute value term. Then the solution set is a union of a pair of tangent circles.
Solution
