# Seemingly simple system of equations

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).

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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. – Srivatsan Sep 2 '11 at 20:55
If you know that 6 is a root then divide x-6 into your cubic (en.wikipedia.org/wiki/Polynomial_long_division) You'll be left with a quadratic that's easy to solve. – Dan Piponi Oct 8 '11 at 17:44
@user80: $6$ is the $y$ correspoinding to $x=5$, not a separate root. – Ross Millikan Oct 17 '11 at 15:48

It appears that your roots are not rational, nor are they square roots of rationals. So the remaining $y$-coordinates will not be rational either.