Solving the system $x^2+y^2+x+y=12$, $xy+x+y=-7$ I've been trying to solve this system for well over the past hour.$$x^2+y^2+x+y=12$$
$$xy+x+y=-7$$
I've tried declaring $x$ using $y$ ($x=\frac{-7-y}{y+1}$) and solving from there, but I've gotten to $$y^4+3y^3-9y^2-45y+30=0$$ and I don't see how we can get $y$ from here.
If anyone sees it, I'd appreciate the help. But if you see a simpler solution please do not hesitate to leave a comment and notify me of such.

$$(x+1)(y+1)=-7+1$$
Let $x+1=a,y+1=b\implies ab=-6$
$$12=x^2+y^2+x+y=(a-1)^2+(b-1)^2+a+b-2=a^2+b^2-(a+b)=(a+b)^2-2(-6)-(a+b)$$
$$\implies(a+b)(a+b-1)=0$$
So, we know $a+b$ and $a,b$
Can you take it home from here?
