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 $\mathrm C$, $y \times z = \mathrm C \neq 0$
\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 


*

*Is there another solution to this system? 

*Is there any necessary and sufficient conditions for solvability of this system?


Thanks in advance
 A: 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. 
A: 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 $0$.
