If $x$ and $y$ are real numbers satisfying \[x^2+y=10=y^2+3x,\] find all possible values for $x+y$.
Enter all the possible values, separated by commas.
We arrange the equation $(y-5)^2=20+3x$ for easier comparison:
\[(y-5)^2 - 3x = 20.\tag{1}\]We try expanding the left hand side and see if anything nice happens:
$y^2-10y+25-3x=20,$ so $y^2-10y-3x+5=0.$
Recall that we want factors of the form $py+q$ to appear, for some $p,q.$ Therefore we try to put $y$ into the form $p(y-5)$:
Rewriting: \begin{align*}
y^2-10y-3x+5 &= y^2-10y+25-3x-20\\
&=(y-5)^2-3x-20.
\end{align*}This time, let's compare to $(1):$
$(y-5)^2=20+3x.$
Thus \[(y-5)^2-3x = (y-5)^2-3x-20+20 = 0.\]Now we have a quadratic in the form $(p+q)^2 - (p+q)+20=0,$ which factors as $(p+q)(p+q-1)+20=(p+q-1)(p+q)=0.$
Thus $(y-5)+(y-5+1)=2y-4$ must be one of the constant terms in the factorization. Solving $2y-4=0,$ we find $y=2.$ Plugging this into either of our original equations gives us \[(x,y) = \boxed{(7,2)}.\]