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what is the best way to solve this system of equations:

$$ax^2 +by^2-2y=0$$ $$axy+byz-z=0$$ $$ay^2+bz^2-c=0$$ Solve for x,y,z where a,b,c are constants.

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    $\begingroup$ Are you sure of the second equation ? $\endgroup$ Commented May 2, 2016 at 8:10
  • $\begingroup$ "best way" in what sense ? $\endgroup$
    – user65203
    Commented May 2, 2016 at 8:11
  • $\begingroup$ @ClaudeLeibovici sorry made a typo. Edited. $\endgroup$
    – saak
    Commented May 2, 2016 at 8:11
  • $\begingroup$ @YvesDaoust Easiest. Also would best great to know if there is procedure to approach such problems. $\endgroup$
    – saak
    Commented May 2, 2016 at 8:12

2 Answers 2

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Extract $y$ from the second equation to get $$y=\frac{z}{a x+b z}$$ Plug in the third equation which is a quadratic in $x$; the solutions are $$x_{\pm}=\frac{\pm\sqrt{a^3 c z^2-a^3 b z^4}-a b^2 z^3+a b c z}{a^2( b z^2- c)}$$ You are left with a nasty equation in $z$ using the first equation.

I suppose that you have four sets of solutions.

Have fun !

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From the second equation,

$$x=\frac{1-by}{ay}z.$$

Plug this in the first,

$$\frac{(1-by)^2}{ay^2}z^2+by^2-2y=0.$$

Then you can eliminate $z^2$ with the third, giving

$$\frac{ay^2(2y-by^2)}{(1-by)^2}=\frac{c-ay^2}b,$$ and by magic the equation simplifies to $$(a-cb^2)y^2+2bcy-c=0.$$

Now the problem has turned to the resolution of a quadratic equation in a single unknown. There are at most two real solutions in $y$. $z$ and $x$ easily follow from $y$, for a total of four distinct solution sets.

Generally speaking, you can address such problems by putting the system in a triangular form, leading to higher degree univariate polynomials. (The degree reflects the number of possible solutions.) Good root-finding numerical methods are available.

There are systematic approaches (involving Gröbner bases), https://en.wikipedia.org/wiki/System_of_polynomial_equations, but this remains a difficult topic.

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  • $\begingroup$ Shouldn't it be (c-ay^2)/b when we substitute z^2? $\endgroup$
    – saak
    Commented May 2, 2016 at 8:52
  • $\begingroup$ @Akhilesh1990: quite right. I am fixing the typo. $\endgroup$
    – user65203
    Commented May 2, 2016 at 8:55
  • $\begingroup$ Much better than my junk ! $\endgroup$ Commented May 2, 2016 at 8:55
  • $\begingroup$ Thank you so much! this was very helpful. @YvesDaoust $\endgroup$
    – saak
    Commented May 2, 2016 at 9:01
  • $\begingroup$ @ClaudeLeibovici: yep, leaving radicals called for trouble. And then it turns out that the problem is quite tractable :) $\endgroup$
    – user65203
    Commented May 2, 2016 at 9:02

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