Given any 12 real constants $a_1,a_2,a_3,b_1,b_2,b_3,c_1,c_2,c_3,d_1,d_2,d_3$ all $>0$ I'm asking myself if is it always possible to find positive reals $x,y,z,t,u,v$ such that
$$a_1x+a_2y+a_3z=b_1t+b_2u+b_3x=c_1t+c_2y+c_3v=d_1v+d_2u+d_3z$$
But I can not find an answer since I don't know how to work with the condition of $x,y,z,t,u,v$ positive.
What do you think? If the answer is "no" is there a condition on $a_1,a_2,a_3,b_1,b_2,b_3,c_1,c_2,c_3,d_1,d_2,d_3$ to guarantee that there exist such $x,y,z,t,u,v$?
As pointed out by user Marty Cohen one could substitute the variables $x,y,\dots$ with the variables $x^2,y^2,\dots$ and get quadratic equation. At this point one could introduce another variable $k\in \mathbb{R}$ and ask if the following system
$$a_1x^2+a_2y^2+a_3z^2=k\\ b_1t^2+b_2u^2+b_3x^2=k\\ c_1t^2+c_2y^2+c_3v^2=k\\ d_1v^2+d_2u^2+d_3z^2=k$$
has a solution such that all $x,y,z,t,u,v$ are $\neq 0$. Note that this is the intersection of 4 quadrics of $\mathbb{R}^6$
Edit: I'm searching conditions on the $a_1,a_2,a_3,b_1,b_2,b_3,c_1,c_2,c_3,d_1,d_2,d_3$ which guarantee the existence of $x,y,zt,u,v>0$, I don't want to solve the system with a linear program