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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).

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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
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2 Answers

One equation represents a parabola with a vertical axis opening downward, and the other represents a parabola with a horizontal axis opening leftward. You can see from a sketch that there should be four intersection points.

The roots of your cubic can be solved for exactly using the cubic formula: http://en.wikipedia.org/wiki/Cubic_function#General_formula_of_roots.

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.

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OP will probably want to use the trigonometric form of the cubic formula, given that this is the casus irreduciblis. –  J. M. Sep 3 '11 at 1:55
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In general, your method seems to work great for two parabolas. The reason why is because we have Cardano's Method to solve the cubic and General Solutions to the Quartic for cases when it does not reduce. These aren't so bad to use, right? But what's great is that this will always work for nondegenerate cases.

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