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I have a system of polynomials, where the first one is a multivariate linear polynomial, but the rest are univariate quadratic polynomials. How would I solve such a system (finding one or all solutions, or showing there are no solutions)? For example,
$$17x+16y-5z-67=0 \\ x^2+3x-5=0 \\ 4y^2-7y-4=0 \\ z^2-6z-3=0$$
The system has no solution. This can be seen by computing a Groebner basis, for example. But also a direct approach is possible. We can compute $x,y,z$ from the second, third and last equation (two solutions each) and then substitute it into the first one. Even if we would replace the first equation by $$ 17x+16y-5z-a=0 $$ where $a$ is an integer, there is no solution.
Well, one way is to just throw it into Wolfram. But, the procedure in this case is quite simple. Start by finding all solutions of the bottom three single variable equation. Then see if any combinations of solutions from these equations produce a solution to the first equation. In this case no such combinations exist.
The discriminants of the three quadratics are $29$, $113$ and $48 = 4^2 \cdot 3$ respectively. A linear combination of $x$, $y$, $z$ and $1$ over the rationals with some nonzero coefficients is a linear combination of $\sqrt{29}$, $\sqrt{113}$, $\sqrt{3}$ and $1$ over the rationals with some nonzero coefficients. But in fact, since $29$, $113$ and $3$ are distinct primes, $\sqrt{29}$, $\sqrt{113}$, $\sqrt{3}$ and $1$ are linearly independent over the rationals: see e.g. this.
We have $$0=4(x^2+3x-5)+(4y^2-7y-4)+(z^2-6z-3)-(17x+6y-5z-67)=$$ $$=4x^2-5x+4y^2-13y+z^2-z+40=$$ $$=\left(2x-\frac{5}{4}\right)^2+\left(2y-\frac{13}{4}\right)^2+\left(z-\frac{1}{2}\right)^2+\frac{221}{8}>0,$$ which says that our system has no real solutions.
