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

  • 1
    $\begingroup$ It seems you can proceed as usual, for example eliminate $y$ from the second equation and you get an expression for what $x$ should be in terms of $z$. Then you can substitute that $x$ into the first equation to get a linear equation for $y,z$. Then substitute $y=C/z$ and you get a single nonlinear equation for $z$ that ends up being a quadratic equation. Then back-substitute for $y$ and $x$ once $z$ is obtained. $\endgroup$ – Michael Aug 21 '18 at 22:10
  • 1
    $\begingroup$ You can use the same idea from my answer to the other question. The difference here is that you'll get additional constant terms on the LHS of: $$ \begin{cases} (a_1c_2-a_2c_1)x \color{red}{- (d_1c_2-d_2c_1)} = -(b_1c_2-b_2c_1)y \\ (a_1b_2-a_2b_1)x\color{red}{- (d_1b_2-d_2b_1)} = -(b_2c_1-b_1c_2)z \end{cases} $$ After multiplying, you still get a quadratic in $x$, but this time it will have more than just the square term. $\endgroup$ – dxiv Aug 22 '18 at 1:00
  • $\begingroup$ @dxiv Please posy your comment is an answer $\endgroup$ – M a m a D Aug 22 '18 at 8:13

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.


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