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

Question

Consider the following non-homogeneous 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 &= d_1\\ a_2x + b_2y + c_2z &= d_2\\ \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?

The homogeneous model of this problem already asked in Solving a system of two equations and three variables where product of two of the variables are constant and very good answers are given.

Thanks in advance

Answer 1

Hint (borrowing from my answer to the homogeneous problem):   eliminating $z$ and $y\,$, respectively, between the linear equations:

$$ \begin{cases} (a_1c_2-a_2c_1)x+(b_1c_2-b_2c_1)y = d_1c_2−d_2c_1 \\ (a_1b_2-a_2b_1)x+(b_2c_1-b_1c_2)z = d_1b_2−d_2b_1 \end{cases} \\ \iff \quad \begin{cases} (a_1c_2-a_2c_1)x - (d_1c_2−d_2c_1) = -(b_1c_2-b_2c_1)y \\ (a_1b_2-a_2b_1)x - (d_1b_2−d_2b_1) = -(b_2c_1-b_1c_2)z \end{cases} $$

Multiplying the latter equations:

$$ \begin{align} \big((a_1c_2-a_2c_1)x - (d_1c_2−d_2c_1)\big)\big((a_1b_2-a_2b_1)x - (d_1b_2−d_2b_1)\big) &= (b_1c_2-b_2c_1)(b_2c_1-b_1c_2)yz \\ &= (b_1c_2-b_2c_1)(b_2c_1-b_1c_2)C \end{align} $$

The above is a quadratic in $x$ (in the general case, unless some of the coefficients are $\,0$), and solvability in reals depends on the sign of the discriminant thereof.