How to balance ionic equations

Balance this ionic equation: $SiF_6^{2-}+OH^-\to SiO_3^{2-}+F^-+H_2O$.

SOLUTION:
Write down the stoichiometric coefficients in letters:
${\color{red} a}SiF_6^{2-}+{\color{red} b}OH^-\to {\color{red} c}SiO_3^{2-}+{\color{red} d}F^-+{\color{red} e}H_2O$

Now balance considering the principle of conservation of mass, and the principle of conservation of charge (the total charge in reactants will be equal to the total charge in products):
$\begin{cases}Si:a=c\\ F:6a=d\\ O:b=3c+e\\ H:b=2e\\ Charge:2a+b=2c+d\end{cases}$

Try now to express every variable in terms of $a$. From the silicon equation we obtain $c=a$; from the fluorine equation we obtain $d=6a$. Now look at the oxygen and hydrogen equations:
$\begin{cases}O:b=3c+e\\ H:b=2e\\ c=a\end{cases}\Rightarrow 3c+e=2e\Rightarrow e=3c\Rightarrow e=3a$
And finally $b=2e\Rightarrow b=6a$. To verify, we can try to substitute the value of $b$, $c$, $d$ and $e$ (in terms of $a$) in the charge equation (if correct, we will obtain $a=a$).
So we have:
$\begin{cases}a=a\\ b=6a\\ c=a\\ d=6a\\ e=3a\end{cases}$

Now assign $a=1$. The balanced equation is:
$SiF_6^{2-}+{\color{red} 6}OH^-\to SiO_3^{2-}+{\color{red} 6}F^-+{\color{red} 3}H_2O$

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PROBLEM 9:
When dissolved beryllium chloride reacts with dissolved silver nitrate in water, acqueous beryllium nitrate and silver chloride powder are made. Write the balanced equation and put it in the ionic form.