Solution:

We note that $G = \dfrac{x}{1 - y}$ and $G' = \dfrac{y}{1 - x}$. Setting them equal gives
$$\frac{x}{1 - y} = \frac{y}{1 - x}$$
Cross-multiplying, we get
$$x(1 - x) = y(1 - y)$$
Expanding both sides:
$$x - x^2 = y - y^2$$
Bringing all terms to one side:
$$x - x^2 - y + y^2 = 0$$
$$(x - y) + (y^2 - x^2) = 0$$
$$(x - y) + (y - x)(y + x) = 0$$
$$(x - y) - (x - y)(x + y) = 0$$
$$(x - y)[1 - (x + y)] = 0$$
Since $x < y$, $x - y \neq 0$, so $1 - (x + y) = 0$, i.e., $x + y = 1$.