Recurrence Substitution Method with multiple givens Solve by using the substitution method $T(n)=T(n-1)+2T(n-2)+3$
Given $T(0)=3$ and $T(1)=5$  

I kind of understand it with only one given and one recurrence call by expanding the call using what is inside.
ex. if it was $T(n)=T(n-1)+3$
Given $T(1)=5$
I would take the $n-1$ and plug it into the the first equation $T(n)=T(n-1)+3$
and repeat until I can find a general case. After that simplify it to get my answer.  

My problem is when I have two givens and two recurrences. I first though about solving each one separately using the first recurrence and then the second but it wasn't making any sense to me.
Let $S(n) = T(n) + 1.5$. Then
$$
S(n) = T(n) + 1.5 = T(n-1) + 2T(n-2) + 4.5 = S(n-1) + 2S(n-2).
$$
The recurrence $S(n) = S(n-1) + 2S(n-2)$ is homogeneous with constant coefficients, so we know how to solve it. We first solve the equation $\lambda^2 = \lambda + 2$, finding that the solutions are $2,-1$. This implies that there exist constants $A,B$ such that
$$
S(n) = A 2^n + B (-1)^n.
$$
Since $S(0) = 4.5$ and $S(1) = 6.5$, we see that $A + B = 4.5$ and $2A - B = 6.5$, from which we find that $A = 11/3$ and $B = 5/6$. Therefore
$$
T(n) = \frac{11}{3} 2^n + \frac{5}{6} (-1)^n - \frac{3}{2}.
$$