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Solve for $x,y$: \begin{align} x^3 + y^3&=2\\ x^2 +x + 9y - 3y^2&=8 \\ \end{align}

I can find $x=y=1$ by guessing. Please help me solve it without using computer.

Thanks

Edited, sorry, I have to change $x^2$ to $x^3$ to have a nice solution, but still can't done it by hand.

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The solutions definite doesn't look great. –  Inceptio May 22 '13 at 15:11
    
Perhaps you can resort to graphing software to see how many solutions there are, at least then you know how many to expect when going for the algebraic approach. Let me see what I can find... –  imranfat May 22 '13 at 15:13
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$y$ is a root of the polynomial equation $(y-1)(y^5+7y^4-2y^3-43y^2+74y-34)=0$, and apart from $y=1$, I don't think there is a nice expression for the other roots... –  Ivan Loh May 22 '13 at 15:21
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Did you want to get approximate values for the solutions instead? –  Ivan Loh May 22 '13 at 15:22
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Why do you think it is solvable without a computer? Of course, you can do numeric approximation, but it will be a lot of work. –  Ross Millikan May 22 '13 at 15:56

2 Answers 2

Substracting equation $2$ from $1$ we obtain $y^3 - x - 9y + 3y^2 = -6 \iff y^3 - \sqrt{2-y^3} -9y + 3y^2 = -6$ Now you have an expression that only consists of $y$'s. Try to see if you can manipulate this such that you get your desired result. I am not sure though if you cand find it algebraically. If you can, chances are that long division may be needed.

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x=y=1 is in essence a double root (I know, it's a poor choice of wording)since the curves are tangent to each other here. There are two more solutions, symmetric about the y-axis, but they are "ugly" –  imranfat May 22 '13 at 15:27
    
Sorry, please see my edited post. –  Xeing May 23 '13 at 2:05

I think you can use Newton-Raphson or any other approximation methods for multivariate equations (this is an example)

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