System of linear recurrences; finding an explicit description for both sequences involved. $$\left\{\begin{aligned} a_n &&= &&2a_{n-1} + b_{n-1} + a_{n-2} - b_{n-2} && n \ge 2 && (1)\\ b_n &&=&& b_{n-1} + b_{n-2} - a_{n-2} && n \ge 2&& (2)\end{aligned}\right.$$
with $a_0 = 5, a_1 = 3, b_0 = 0, b_1 = 3$.
From (2) I get: $a_{n-2} = b_{n-1}+b_{n-2} - b_n$. Substituting in (1): $a_n = 2a_{n-1} + 2b_{n-1} - b_n$.  
Now I'm stuck. I don't see what I can do next... 

Hint:
Adding the two equations gives
$$
a_n + b_n = 2(a_{n-1} + b_{n-1}) \quad {\rm{for}} \quad n \ge 2 
$$ 
so 
$$
a_n + b_n = 2^{n-1}(a_{1} + b_{1}) = 6 \cdot 2^{n-1}
$$ 
Plugging this into the second recursion gives 
$$
b_n = b_{n-1} + 2 b_{n-2} - (a_{n-2} + b_{n-2}) = b_{n-1} + 2 b_{n-2} -6 \cdot 2^{n-3}
$$
Likewise,the first equation gives 
$$
a_n = a_{n-1} + (a_{n-1}+ b_{n-1}) + 2 a_{n-2} - (a_{n-2} + b_{n-2}) = \\
= a_{n-1} + 6 \cdot 2^{n-2} + 2 a_{n-2} - 6 \cdot 2^{n-3} = \\
= a_{n-1}  + 2 a_{n-2}+ 3 \cdot 2^{n-2} 
$$
So there are two single-variable recursions which can be solved.
