Pestañeishon

$\begin{aligned} (2x+2y-1)(x+y) &= -\frac{1}{2} \\ \\ 2x^2-2xy+2y^2-x-y+\frac{1}{2} &= 0 \\ \\ 4x^2-4xy+4y^2-2x-2y+1&= 0 \\ \\ (x^2+2xy+y^2-2x-2y+1) + (3x^2-6xy+3y^2)&= 0 \\ \\ (x+y-1)^2 + 3(x-y)^2&= 0 \end{aligned}$

Since $x$ and $y$ are real numbers, the only possible solution needs to fulfill the following equations:

$x+y-1 = 0$ and $x-y = 0$

Adding both equations results in

$x=\frac1{2}$ and $y=\frac1{2}$ .

Therefore

$a=1$ and $b=2$ $\Rightarrow a+b=\boxed{3}$ .

Expand:

$2y^2-y(2x+1)+2x^2-x+1/2=0$

Solve after $y$ :

$y=\frac14\left(2x+1\pm\sqrt{12x-12x^2-3}\right)=\frac14\left(2x+1\pm\sqrt{3}\sqrt{-(2x-1)^2}\right)$

Since we're interested in real solutions $2x-1$ must be zero, hence $x=1/2$