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Consider the following system of equations where $x,y,z$ are variables and for a constant $\mathrm C$, $y \times z = \mathrm C \neq 0$

\begin{equation} \left\{ \begin{array}{lcl} a_1x + b_1y + c_1z &= 0\\ a_2x + b_2y + c_2z &= 0\\ \end{array} \right. \end{equation}

The most straightforward solution is to replace $z$ by $\dfrac{\mathrm C}{y}$ and convert this to a system of two nonlinear equations with two variables. I would like to know

  1. Is there another solution to this system?
  2. Is there any necessary and sufficient conditions for solvability of this system?

Thanks in advance

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    $\begingroup$ Is there another solution Use the linear equations to express $y$ and $z$ in terms of $x$, then substitute those into $y \cdot z = C$ which becomes an equation in $x$. $\endgroup$ – dxiv Aug 21 '18 at 18:12
  • $\begingroup$ How can I express $y$ in terms of $x$? $\endgroup$ – No one Aug 21 '18 at 18:15
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Hint:   eliminating $z$ and $y\,$, respectively, between the linear equations:

$$ \begin{cases} (a_1c_2-a_2c_1)x+(b_1c_2-b_2c_1)y = 0 \\ (a_1b_2-a_2b_1)x+(b_2c_1-b_1c_2)z = 0 \end{cases} \;\; \iff \;\; \begin{cases} (a_1c_2-a_2c_1)x = -(b_1c_2-b_2c_1)y \\ (a_1b_2-a_2b_1)x = -(b_2c_1-b_1c_2)z \end{cases} $$

Multiplying the latter equations:

$$ \begin{align} (a_1c_2-a_2c_1)(a_1b_2-a_2b_1)x^2 &= (b_1c_2-b_2c_1)(b_2c_1-b_1c_2)yz \\ &= (b_1c_2-b_2c_1)(b_2c_1-b_1c_2)C \end{align} $$

Solvability depends on the signs of the coefficients above, and whether any of them is $0$.

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Multiply your system by z $$\begin{equation} \left\{ \begin{array}{lcl} a_1xz + b_1yz + c_1z^2 &= 0\\ a_2xz + b_2yz + c_2z^2 &= 0\\ \end{array} \right. \end{equation}$$

Substitute $C$ for $yz$ and you get a linear system in $xz$ and $z^2$ which is easy to solve. $$\begin{equation} \left\{ \begin{array}{lcl} a_1xz + c_1z^2 = -b_1C\\ a_2xz + c_2z^2 =-b_2C \\ \end{array} \right. \end{equation}$$ Once you have your $xz$ and $z^2$ you can solve for $z$ and $x$ and find your $y$ as well.

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  • $\begingroup$ kheili az shoma mamnoonam :) $\endgroup$ – No one Aug 21 '18 at 18:32
  • $\begingroup$ lotf dareed. movaffagh basheed. $\endgroup$ – Mohammad Riazi-Kermani Aug 21 '18 at 18:34
  • $\begingroup$ What if the right hand side is non zero? I think your solution will create a new variable $z$ and will not be linear again. Is there a solution for that case? $\endgroup$ – No one Aug 21 '18 at 18:57
  • $\begingroup$ @Drupalist this is a new question and should not be postrd as comments but as new question. $\endgroup$ – miracle173 Aug 21 '18 at 19:38

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