# Solving a homogeneous system of two equations and three variables where product of two of the variables are constant

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

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

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?

• 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 $x$. – dxiv Aug 21 '18 at 18:12
• How can I express $y$ in terms of $x$? – No one Aug 21 '18 at 18:15

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 $0$.

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

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

• kheili az shoma mamnoonam :) – No one Aug 21 '18 at 18:32
• lotf dareed. movaffagh basheed. – Mohammad Riazi-Kermani Aug 21 '18 at 18:34
• 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? – No one Aug 21 '18 at 18:57
• @Drupalist this is a new question and should not be postrd as comments but as new question. – miracle173 Aug 21 '18 at 19:38