5
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I have the following system:

$x^{2} + y = 31$

$x + y^{2} = 41$

As I try to solve it via simple substitution, I get into 4-th power equations, which I can simplify to $(x-5)(x^{3}+5x^{2}-37x-184)$ (and I am not sure how to get the cubic here). Is there a simpler way to solve this? There are 4 pairs of answers, I have got one (5 and 6).

  • 0
    [Rational root theorem](http://en.wikipedia.org/wiki/Rational_root_theorem) will be helpful in finding if this cubic has a rational root. Wolfram|Alpha seems to think that there are 3 real, but irrational, roots. [link](http://www.wolframalpha.com/input/?i=x^3%2B5x^2%E2%88%9237x%E2%88%92184).2011-09-02
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    If you know that 6 is a root then divide x-6 into your cubic (http://en.wikipedia.org/wiki/Polynomial_long_division) You'll be left with a quadratic that's easy to solve.2011-10-08
  • 0
    @user80: $6$ is the $y$ correspoinding to $x=5$, not a separate root.2011-10-17

2 Answers 2