3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Blue
    Dec 26, 2021 at 9:54
  • 7
    $\begingroup$ 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 $x+y$. $\endgroup$
    – Vercassivelaunos
    Dec 26, 2021 at 9:54

3 Answers 3

Reset to default
4
$\begingroup$

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

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

$\endgroup$
3
  • $\begingroup$ I'm assuming you made a typo and replaced $S$ with $u$ $\endgroup$
    – Cookie
    Dec 26, 2021 at 10:33
  • $\begingroup$ @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. $\endgroup$
    – mathcounterexamples.net
    Dec 26, 2021 at 10:35
  • $\begingroup$ Oops, my bad, I got confused, thanks for your help. $\endgroup$
    – Cookie
    Dec 26, 2021 at 10:35
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

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?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .