Algebraic Identity with Real Variables

Rearranging the equation: $x^2 + (x - 2y - 1)^2 = -4xy - 4y^2$ $(x^2 + 4xy + 4y^2) + (x - 2y - 1)^2 = 0$ $(x + 2y)^2 + (x - 2y - 1)^2 = 0$

Since $x$ and $y$ are real numbers, the sum of two squares is zero if and only if each term is zero:

1. $x + 2y = 0 \implies x = -2y$
2. $x - 2y - 1 = 0$

Substituting $x = -2y$ into the second equation: $-2y - 2y - 1 = 0 \implies -4y = 1 \implies y = -\frac{1}{4}$ Hence, $x = \frac{1}{2}$.

Then $x - 2y = \frac{1}{2} - 2\left(-\frac{1}{4}\right) = \frac{1}{2} + \frac{1}{2} = 1$.