# Thread: Why some solutions of system of linear equations create contradictions?

1. ## Why some solutions of system of linear equations create contradictions?

Lets take a simple system of two equations:
$\begin{cases} a_{1}*q^{2}-a_{1}=15 \\ a_{1}*q^{3}-a_{1}*q=6 \end{cases}$

after evaluating a_{1} from the first equation and putting it into the second we get
$15q^3-6q^2-15q+6=0$

this gives
$(15q-6)(q-1)(q+1)=0$
so, solutions are: q=1, q=-1 and q= $\frac{2}{5}$

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

thanks

2. You're first evaluation is incorrect... If you solve the first equation for $a_1$ you get:

$a_1(q^2 - 1) = 15$

$a_1 = \frac{15}{q^2 - 1}$

Plug this into the second equation to get:

$\frac{15}{q^2 - 1}q^3 - \frac{15}{q^2 -1}q = 6$

$\frac{15q}{q^2 - 1}(q^2 - 1) = 6$

$15q = 6$

$q = \frac{2}{5}$

That's why the other answers don't work, because they are not solutions to the system.

3. <blushing> Thanks

4. No problem.