Lets take a simple system of two equations:

$\displaystyle \begin{cases} a_{1}*q^{2}-a_{1}=15 \\ a_{1}*q^{3}-a_{1}*q=6 \end{cases}$

after evaluating a_{1} from the first equation and putting it into the second we get

$\displaystyle 15q^3-6q^2-15q+6=0$

this gives

$\displaystyle (15q-6)(q-1)(q+1)=0$

so, solutions are: q=1, q=-1 and q=$\displaystyle \frac{2}{5}$

but if you try to put this into the original system of equations, only q=$\displaystyle \frac{2}{5}$ works, why?

thanks