c + ba = 30
a + bc = 18
SOLVE FOR a, b, c
Adding and substracting the two equations we have
$\displaystyle \displaystyle \left\{\begin{array}{ll}(a-c)(b-1)=12\\(a+c)(b+1)=48\end{array}\right.$
So, $\displaystyle b-1$ must divide 12 and $\displaystyle b+1$ must divide 48.
For example:
$\displaystyle b-1=1\Rightarrow b=2\Rightarrow b+1=3|48$.
Replacing $\displaystyle b=2$ in system we have
$\displaystyle \displaystyle \left\{\begin{array}{ll}a-c=12\\a=c=16\end{array}\right.\Rightarrow a=14,c=2$.
And so on...