Not Egyptian Fraction

$\frac{x+y}{xy}=1\\ x+y=1\times xy=xy\\ x-xy=-y \\ x(1-y)=-y\\ x=\frac{-y}{1-y}\\ x=\frac{-y\times(-1)}{(1-y)\times(-1)} \\ x=\frac{y}{-1+y} \\ \boxed{x=\dfrac{y}{y-1}}$

we have $\frac {x+y}{xy}=1$ so $x+y=xy$ hence $x(y-1)=y$ thus $x=\frac{y}{y-1}$

x+y/xy=1
x+y/x=y
x/x+y/x=y
1+y/x=y
y/x=y-1
1/x=y-1/y
x=y/y-1
as in the option.

(x+y)/xy=1

-> x+y=xy

-> y=xy-x

-> y= x(y-1)

->x=y/(y-1)

(x+y)/(xy) = 1 (1/y)+(1/x) = 1 1 - (1/y) = (1/x) (y-1)/(y) = 1/x x = y/(y-1)

It may not be as algebraic, but since 2+2 and 2*2 both equal 4, dividing them would equal 1. Which means x & y both equal 2, and 2 divided by 2 - 1 = 2.