# 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.

• Check your coefficient on $y$. I get (that is to say, Mathematica gets) $-17$, which allows the quartic to decompose into two linear factors and one quadratic.
– Blue
Dec 26, 2021 at 9:54
• Add the second equation to the first one twice. You'll get $x^2+2xy+y^2+3x+3y=-2$, which you can simplify to $(x+y)^2+3(x+y)=-2$, which you can solve for $x+y$. Dec 26, 2021 at 9:54

Let $$S = x+y$$ and $$P=xy$$. You have $$S^2 - 2P +S = 12$$ and $$P+S=-7$$. Therefore

$$S^2 - 2P +S = S^2-2(-7-S) + S=S^2+3S+14=12$$ or $$S^2+3S+2=0.$$ The roots of this last equation are $$-1,-2$$. Therefore $$(S,P) \in \{(-1, -6) , (-2,-5)\}$$. Which implies that $$x,y$$ are the roots of either $$u^2 +u -6 = 0$$ or of $$u^2 +2u -5=0$$. Which are quadratic equations that you can solve... You'll get

$$(x,y) \in \{(2,-3),(-3,2),(-1- \sqrt 6, -1 + \sqrt 6),(-1+ \sqrt 6, -1 - \sqrt 6)\}$$

• I'm assuming you made a typo and replaced $S$ with $u$ Dec 26, 2021 at 10:33
• @Cookie I may be missing something as I don't understand your point. $u$ is just a free variable that I used for the quadratic equations. Dec 26, 2021 at 10:35
• Oops, my bad, I got confused, thanks for your help. Dec 26, 2021 at 10:35

Alternatively, set $$u=x+y$$ and $$v=x-y$$. Then first equation plus twice of second equation gives $$(x^2+y^2+2xy)+3(x+y)=-2$$ which is $$u^2+3u+2=0$$

Also, first equation minus second gives $$x^2+y^2-xy=19$$ which can be rewritten as $$\frac{1}{4}(x+y)^2+\frac{3}{4}(x-y)^2=19 \Rightarrow u^2+3v^2=76$$

The first quadratic has easy roots, $$u \in \{-1,-2\}$$. From above two values of $$v$$ are obtained. Finally you have to solve for $$x+y= \ldots$$ and $$x-y=\ldots$$

The original equations are of a circle and a hyperbola, which on Geogebra can be seen to have four intersections (solutions). Hence both values of $$u,v$$ give valid solutions.

$$(x+1)(y+1)=-7+1$$

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?