Find the sum.

Given equations are, $x^{2}-4y=-7$ ....(i) and $y^{2}-2x=2$ ....(ii)

Adding (i) and (ii), we get---

$x^{2}-4y+y^{2}-2x=-5$

$\implies x^{2}-4y+y^{2}-2x+5=0$

$\implies (x^{2}-2x+1)+(y^{2}-4y+4)=0$

$\implies (x-1)^{2}+(y-2)^{2}=0$

Now, since square of any real no. cannot be (-ve), so we can see that the above equation is satisfied only if $(x-1)^{2}$ and $(y-2)^{2}$ are 0, i.e, $(x-1)$ and $(y-2)$ . are 0.

Thus, $x-1=y-2=0 \implies x=1 , y=2 \implies x+y=1+2=\boxed{3}$

Adding two equations,

$(x^{2}-4y)+(y^{2}-2x)=-7+2$

$x^{2}-2x+y^{2}-4y=-5$

Completing the squares,

$(x-1)^{2}+(y-2)^{2}=0\implies x=1, y=2$

Thus, $\boxed {x+y=3}$

If you add the two equations, you find that $x^2-2x+y^2-4y=-5.$ If you complete the square, you find that $x^2-2x+1+y^2-4y+4=(x-1)^2+(y-2)^2=0.$ This is only satisfied when $(x-1)^2=(y-2)^2=0\text{,}$ so $x=1$ and $y=2.$ $x+y=\boxed{3}$