Given that: $$2c-3bt=-358,\\2c+3b+4t=-102,\\-2ct+b=-318$$ find the value of $2(c+bt)$: is it required to find $c,b,$ and $t$ individually? If we must find them separately, then how? By WA, this system has 3 real solutions. I do not know how to solve it.
Any help/hint would be really appreciated. THANKS!
One can use the system by using resultants, but I still believe that a solution by substitution is so easy. First we have $b=2ct-318$, and then we have $t=-\frac{1}{3c+2}(c - 426)$. This gives $$ (3c^2 + 1332c + 14540)(c + 14)=0 $$ and we are done.
By the way, $2(c+tb)$ is not always equal for the three solutions. For the case $c=-14$ we have $$ (b,c,t)=(-10,-14,-11) $$ so that $2(c+tb)=192$.