$z(y^2-x^2)=16$, $y(z^2-x^2)=9$, $x(y^2-z^2)=5$ The best relationship I found is $16xy=9xz+5yz$, but that doesn't seem very useful either. With the help of Wolfram and I found out that the system has 2 real solutions, but I really don't know how to get them. This system seemed very simple, but I came up with terrible calculations. Can you help me please ?
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1Please show your calculations in your your question post, so we can see where you possibly went wrong. – amWhy May 18 '21 at 18:51
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Also, please post, in your question, how you derived $16xy=9xz+5z$, and how you think is shows a relationship. – amWhy May 18 '21 at 18:53
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The equations are correct. And one solution is $x=1, y=3, z=2$. The other solution is ugly. – Josie McLaren May 18 '21 at 18:53
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@Moo please don't answer in comments. – amWhy May 18 '21 at 18:57
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This site is not a do my work for me site. – amWhy May 18 '21 at 18:57
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I got $z^2=4/(x+y)+xy$ and $z=16xy/(9x+5y)$, but I don't think they are useful relationships because they are too complicated. – Josie McLaren May 18 '21 at 19:04
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1You can eliminate one variable first. – Dietrich Burde May 18 '21 at 19:05
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I also got similar relations for $x$ and $y$, but are ugly too. I really want to slove this system but I can't get more that this useless things. I will apreciate every hint or idea. – Josie McLaren May 18 '21 at 19:09
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Then look up $res(f,g)$, the resultant. It factors very nicely here, see my answer. – Dietrich Burde May 18 '21 at 19:11
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Yes, that is what I tried to do. I eliminate the variable z, but the calculations are too ugly. – Josie McLaren May 18 '21 at 19:11
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1@JosieMcLaren Please do not vandalize posted questions, including your own. Rolled back. – dxiv Jun 17 '21 at 17:32
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2Please do not deface the question. Changing a question invalidates the answers and comments already posted. – robjohn Jun 19 '21 at 06:59
1 Answers
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The computational way is to eliminate $z$ by $z=16/(y^2-x^2)$. This gives two equations $f(x,y)=0$ and $g(x,y)=0$. The resultant is a polynomial only in, say, $y$, and the roots are the common solutions. Here we have the product of three factors, namely $$(y-3)(y^2+3y+9)(3y^3+4)=0.$$ The first factor gives $(x,y,z)=(1,3,2)$, the second factor has no real solutions, but two complex solutions $y$ with $3z=2y$ and $3x=y$, and the third one gives another real solution, with the cube root of $-4/3$.
Dietrich Burde
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Thank you very much, it seems correct, you have the same solutions as Wolfram, but could you explain to me in a little more detail how you got the result? – Josie McLaren May 18 '21 at 19:30
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As I said above, use the resultant for these two polynomials $f$ and $g$. – Dietrich Burde May 18 '21 at 19:42
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Sorry, I meant the resultant, not the result. I didn't really understand how to calculate the resultant, but now I understand. The problem is that I obtained 2 polynomials in which the maximum degree of x is 6, respectively 5, so the determinant has the order 11 and not even Wolfram could calculate it for me. However, you managed to calculate it, so I assume that you obtained other polynomials, although I do not see how you could reach a lower degree. Could you please tell me how you proceeded to get this beautiful resultant? – Josie McLaren May 20 '21 at 16:40