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I would really appreciate help with this system of equations:

$$ \left\{ \begin{array}{c} ax+by=c \\ cx+az=b \\ bz+cy=a \end{array} \right. $$

Here $a, b$ and $c$ are given real numbers, such that $abc\neq 0$ and $x, y$ and $z$ are the unknowns.

I am kind of stuck with it and can't think of anything that I could do with it. I think that to solve it I should express $x, y$ and $z$ in terms of $a, b$ and $c$ but I can't figure out just how should I do it.

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Well, there are various approaches. You could use the first equation to express $x$ in terms of other things, including $y$. And the last equation to express $z$ in terms of things including $y$ (possible because $abc \neq 0$). Substitute both into the middle equation to get one linear equation in $y$. You then already have $x$ and $z$ in terms of $y$ and can find those. – Mark Bennet Oct 20 '12 at 18:47
up vote 2 down vote accepted

$c\cdot(1)-a\cdot(2)$ produces $bcy-a^2z=c^2-ab$. Subtract $b\cdot(3)$ to obtain $ -(a^2+b^2)z=c^2-2ab$, hence $$z = \frac{2ab-c^2}{a^2+b^2}.$$ (provided you are allowed to divide by $a^2+b^2$ - why are you?)

The other values can be obtained the same way or just make use of the cyclic symmetry of the system.

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Should we warn the OP consider $c\neq 0$? I don't know if he just wanted to have solutions or he also considered the number of solutions? – Babak S. Oct 20 '12 at 19:00

$$ \begin{array}{c} ax+by=c \\ cx+az=b \\ bz+cy=a \end{array}$$

multiply the first two up to get

$$ \begin{array}{c} acx+bcy=c^2 \\ acx+a^2z=ab \\ \end{array}$$

eliminate $acx$ to get


multiply up the third equation $b^2z+bcy=ab$ and add to to eliminate $bcy$:


which is a simple quadratic equation for $z$ you can solve that, then use that $z$ to find $y$ and use that to find $x$.

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Your equations are linear in the variables $x,y,z$. So the method of Gaussian Elimination applies. Even easier is to use Cramer's rule, which gives $x,y,z$ as ratios of determinants formed from the coefficients. For your equations, the denominator for each of $x,y,z$ is $c(a^2+b^2)$, and the numerators are respectively

For $x$ : $-a^2b+ac^2+b^3,$

For $y$ : $a^3-ab^2+bc^2,$

For $z$ : $2abc-c^3.$

When given equations in which none of $x,y,z$ are squared, there are always simple solutions in terms of the coefficients.

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