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Solve the system of equations :\

$x+y^2=7$
$x^2+y=11$
my attempt :
$$x=7-y^2 \implies (7-y^2)^2 +y = 11 \text{ Hence } y^4-14y^2+y+38=0 $$
i'm stuck here any help

calipaw
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  • The formula for biquadratic equations give only complicated complex solutions, sorry. – vonbrand Sep 17 '21 at 22:49
  • @vonbrand It's not biquadratic as there is a $y$ term. – Alan Abraham Sep 17 '21 at 22:50
  • @vonbrand but for example $(3,2)$ is a solution of the system – calipaw Sep 17 '21 at 22:53
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    Unfortunately, $ \ (3 \ , \ 2) \ $ is the only "nice" solution. The "horizontal" and "vertical" parabolas so described do have four intersections, but three of them have irrational coordinates. –  Sep 17 '21 at 23:00
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    Does this answer your question? System of equations: $x^2+y=7, y^2+x=11$. If you search your question on Approach0, you will find this one. – Toby Mak Sep 18 '21 at 00:05
  • There's a slightly earlier version of the problem here: https://math.stackexchange.com/questions/108962/steps-to-solve-this-system-of-equations-sqrtxy-7-sqrtyx-11 . After about ten years, it looks like the "beautiful" way to solve this system has yet to be posted... –  Sep 18 '21 at 04:25

2 Answers2

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$$x+y^2=7\Rightarrow x=7-y^2$$ sub in: $$(7-y^2)^2+y=11$$ solve this for values of $y$ (pretty hard to do by hand) then sub back in to find values of $x$. As you may be able to tell, there should be $4$ solutions, only one of which has integer values.

This is just an extension of what you said really, I'm not sure of a method that doesn't yield a quartic

Henry Lee
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$$x=7-y^2 \implies (7-y^2)^2 +y = 11 \text{ Hence } y^4-14y^2+y+38=0$$

With $y=2$ we get $$16-56+2+38=0$$

Thus we have one solution, $ x=3, y=2 $ finding other solutions involves solving a cubic polynomial.$$y^3+2y^2-10y-19=0$$

Mohammad Riazi-Kermani
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