Q:

recurrence relations and closed form

Let $a_n = 5a_{n-1} - 6a_{n-2}$. $a_0 = 12, a_1 = 27$. What is the closed form for this recurrence? What is $a_{10}$?
I have figured out what $a_n$ is up to $a_{10}$ Which is:
$a_0 = 12$,
$a_1 = 27$,
$a_2 = 63$,
$a_3 = 153$,
$a_4 = 387$,
$a_5 = 1017$,
$a_6 = 2763$,
$a_7 = 7713$,
$a_8 = 21987$,
$a_9 = 63657$,
$a_{10} = 186363$
However I am stuck on getting the closed form equation. I came up with this equation based on the example Image 2.4.6 below.
$x^2 - 5x + 6 = 0$
solving for x:
$(x-3) (x-2) = 0$
$x=3,2$
recurrence relation form:
$a_n = a3^n + b2^n$
plug in $n = 0, n = 1$
$12 = a3^0 + b2^0 = a + b$
$27 = a3^1 + b2^1 = 3a + 2b = 27$
This is the part that I am stuck on. In the example 2.4.6 I am not sure how they solved for this system and got $a = \frac{7}{3}$ and $b = - \frac{1}{3}$
Any insight will be much appreciated 

A:

You got $a+b=12$ and $3a+2b=27.$
Subtract twice the first equation from the second to get $a=3$. 
Then solve for $b$ from the first equation.  

