I have a system of equations shown as below:

$x_1x_3+x_5 = a$

$x_1x_4+x_5 =b$

$x_2 x_3 +x_5 =c$

$x_2x_4 +x_5 =d$,

where $x_1,x_2,x_3,x_4,x_5$ are variables and $a,b,c,d$ are constant.

Given any $(a,b,c,d)$, is it always possible to find a valid solution?

For example, given $a=1,b=c=d=0$, I can find a solution as $x_1=x_3=1,x_2=x_4=x_5=0$.

This is not a homework problem that I'm trying to find an answer. I just don't know where to start to tackle this problem. Can someone provide some hints?


Please check if this statement is right: If $b-a = d-c \neq 0$ and ($a\neq c$ or $b\neq d$), then there is no solution; otherwise, there is always a solution.


No, it is not always possible. The solution is given, for arbitrary $x_4\neq 0$ by $$ x_1=\frac{ab - ad - b^2 + bd}{x_4(a - b - c + d)}, $$

$$ x_2=\frac{bc - bd - cd + d^2}{x_4(a - b - c + d)}, $$

$$ x_3=\frac{x_4(a - c)}{b - d}, $$

$$ x_5=\frac{ad - bc}{a - b - c + d}, $$

provided not one of the denominators equals zero. If $$ a-b-c+d=0, $$ then there is not always a solution. Take $(a,b,c,d)=(1,2,3,4)$. This can be found by using Buchberger's algorithm, or by substitution.

  • $\begingroup$ Thank you i will check the algorithm and try to give a proof. $\endgroup$ – Paradox Sep 27 '17 at 19:10
  • $\begingroup$ I am still computing the case $a-b-c+d=0$ - hope to finish the answer soon. $\endgroup$ – Dietrich Burde Sep 27 '17 at 19:10
  • $\begingroup$ I just checked this case. Actually you can find a valid solution for $a=1,b=0,c=1,d=0$ $\endgroup$ – Paradox Sep 27 '17 at 19:12
  • $\begingroup$ But not for $a=1,b=2,c=3,d=4$. You'll need, that two of them are equal. $\endgroup$ – Dietrich Burde Sep 27 '17 at 19:13
  • $\begingroup$ I think the condition is when $a\neq c \neq 0$ or $b\neq d \neq 0$, the system does not have a solution; otherwise, it always have a solution. $\endgroup$ – Paradox Sep 27 '17 at 19:23

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