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Having this system:

$$\begin{cases} 2 x(x^2 + y - 11) + x + y^2 - 7 = 0,\\ x^2 + y - 11 + 2 y (x + y^2 - 7) = 0 . \end{cases}$$

How can I solve it without using a computer?.

Thank you!

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I would start by multiplying both sides of the second equation by $2x$, subtracting the second equation from the first, and factoring. You get $(1-4xy)(x+y^2-7)=0$. This is not a complete solution, but it is progress. –  Stefan Smith Apr 7 '13 at 16:26
    
Actually, that helps me a lot!, by studying both cases, thanks! –  Alfonso Pérez Apr 7 '13 at 16:28
    
I did a couple more steps and it led to a fourth-degree polynomial, which are always possible to solve exactly, but sometimes extremely messy. Do you know if the answer(s) are supposed to work out nice? If not, can you use a computer algebra system to check if the answer(s) will work out nice? –  Stefan Smith Apr 7 '13 at 16:31
    
I checked it in Wolfram Alpha and yes, it had 3 solutions, being one of them a beautiful one, (3,2) I think. The thing is that I wanted to know how to start working with it without a computer, and you helped me a lot. Thank you! –  Alfonso Pérez Apr 7 '13 at 16:37
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2 Answers 2

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Note first that a substitution of $x=0$ yields the system $$\begin{cases}y^2-7=0\\y-11+2y(y^2-7)=0,\end{cases}$$ which you can show has no solutions in $y$. Thus, we needn't worry about any problems that might arise from having $x=0$.


Now, multiplying the second equation in our original system by $2x$ gives us the equivalent system $$\begin{cases}2x(x^2+y-11)+x+y^2-7=0\\2x(x^2+y-11)+4xy(x+y^2-7)=0,\end{cases}$$ from which it follows that $$(4xy-1)(x+y^2-7)=0,$$ meaning $$y=\frac1{4x}\quad\text{or}\quad x=7-y^2.$$ Now, by $x=7-y^2$ and $x^2+y-11+2y(x+y^2-7)=0,$ we find $$0=(7-y^2)^2+y-11=y^4-14y^2+y+38.$$ Using the rational root test and a little trial and error, we find that $2$ is a root of that equation, and we can factor it as $$0=(y-2)(y^3+2y^2-10y-19).$$ Thus, $(3,2)$ is one solution to the system. We can explicitly calculate the roots of that cubic factor using Cardano's Method--be warned, they are real, even though they won't look like it, and they aren't very nice in any case--and from substituting those $y$-values back into $x=7-y^2$ find the corresponding $x$-values.

From $y=\frac1{4x}$ and $2x(x^2+y-11)+4xy(x+y^2-7)=0$ we find $$0=2x^3+\frac12-22x+x+\left(\frac1{4x}\right)^2-7=2x^3-21x-\frac{13}2+\frac1{16x^2},$$ whence multiplication by $16x^2$ yields $$0=32x^5-336x^3-104x^2+1.$$ This is messier, still. No rational roots exist, and quintics don't have a nice general method of solution. I wouldn't hold out much hope....

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You have some flaws, not in the approach but in the steps that lead you to the coefficients of the equation, for instance, it is not $y=-\frac1{4x}$ but $y=\frac1{4x}$. –  user67878 Apr 7 '13 at 17:56
    
@Thus: Thanks! I think it's all fixed, now. –  Cameron Buie Apr 7 '13 at 19:43
    
You guys are awesome!, Thanks! –  Alfonso Pérez Apr 7 '13 at 21:45
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Let $a = x^2+y-11$ and $b = x+y^2-7$. The equations we get are as follows: $$2ax+b=0$$ $$a+2by=0$$ Replacing $a$ in the first equation yields: $$-4bxy+b = 0$$ which gives us two options: $1-4xy = 0$ ($x = 1/4y$) or $b = 0$ ($x = 7-y^2)$. In both cases, we have expressed $x$ explicitly, we can insert it back the second original equation and solve it.

1.) $(1/4y)^2 + y - 11 + 2y(1/4y + y^2 -7) = 0$ that is (hopefully) equivalent to $$ 32y^5 -14y^4+16y^3-21\cdot 18y^2+1 = 0$$ which looks painful.

2.) $(7-y)^2 + y - 11 = 0$ which is easily solvable.

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