How does one solve this recurrence relation? We have the following recursive system:
$$
\begin{cases}
& a_{n+1}=-2a_n -4b_n\\
& b_{n+1}=4a_n +6b_n\\
& a_0=1, b_0=0
\end{cases}
$$
and the 2005 mid-exam wants me to calculate answer of $ \frac{a_{20}}{a_{20}+b_{20}} $. 
Do you have any idea how to solve this recursive equation to reach a numerical value?

Observe that
$$a_{n+1}+b_{n+1}=2a_n+2b_n=2(a_n+b_n)\;,$$
and $a_0+b_0=1$, so in general $a_n+b_n=2^n$. 
Quickly calculating a few values, we see that the numbers $b_n$ are a little nicer than the numbers $a_n$:
$$\begin{array}{rcc}
n:&0&1&2&3&4\\
a_n:&1&-2&-12&-40&-112\\
b_n:&0&4&16&48&128\\
\end{array}$$
Concentrating on the $b_n$, we see that
$$b_{n+1}=4(a_n+b_n)+2b_n=2^{n+2}+2b_n\;,$$
so that
$$\begin{align*}
b_n&=2b_{n-1}+2^{n+1}\\
&=2(2b_{n-2}+2^n)+2^{n+1}\\
&=2^2b_{n-2}+2\cdot2^{n+1}\\
&=2^2(2b_{n-3}+2^{n-1})+2\cdot 2^{n+1}\\
&=2^3b_{n-3}+3\cdot 2^{n+1}\\
&\;\;\vdots\\
&=2^kb_{n-k}+k2^{n+1}\\
&\;\;\vdots\\
&=2^nb_0+n2^{n+1}\\
&=n2^{n+1}\;,
\end{align*}$$
so $a_n=2^n-n2^{n+1}=2^n(1-2n)$, and
$$\frac{a_n}{a_n+b_n}=\frac{2^n(1-2n)}{2^n}=1-2n\;.$$
(There are other ways to solve that first-order recurrence for $b_n$; I just picked the most elementary one.)
