Determine $x+y$ where $x$ and $y$ are real numbers such that $(2x+1)^2+y^2+(y-2x)^2=\frac{1}{3}$ Determine $x+y$ where $x$ and $y$ are real numbers such that $(2x+1)^2+y^2+(y-2x)^2=\frac{1}{3}$
I used the quadratic equation to get $$x=\frac{y-1\pm\sqrt{-2y-3y^2-\frac{5}{3}}}{4}$$
But I don’t see how that helps, hints and solutions would be appreciated 
Taken from the 2006 IWYMIC 

Multiply out and collect terms $$8x^2+4x-4xy+2y^2+\frac{2}{3}=0.$$
Multiply by $6$ and complete the square for $x$ terms $$3(4x+1-y)^2+9y^2+6y+1=0$$
$$3(4x+1-y)^2+(3y+1)^2=0$$
Therefore $x=y=-\frac{1}{3}.$
