How can I determine general formula of this sequence? I am trying to find general formula of the sequence $(x_n)$  defined by
$$x_1=1, \quad x_{n+1}=\dfrac{7x_n + 5}{x_n + 3}, \quad \forall n>1.$$
I tried
put $y_n = x_n + 3$, then $y_1=4$ and
$$\quad y_{n+1}=\dfrac{7(y_n-3) + 5}{y_n }=7 - \dfrac{16}{y_n}, \quad \forall n>1.$$
From here, I can't solve it. How can I determine general formula of above sequence?
With Mathematica, I found $x_n = \dfrac{5\cdot 4^n-8}{4^n+8}$. I want to know a method to solve problem, than have a given formula.

The chracteristic equation of the given sequence is
$$y=\dfrac{7y+5}{y+3} \Leftrightarrow y_1 = 5 \lor y_2 = -1.$$
Let us consider the sequence
$$b_n = \dfrac{x_n-y_1}{x_n - y_2}=\dfrac{x_n-5}{x_n+1}.$$
We note that
$$b_{n+1}=\dfrac{x_{n+1}-5}{x_{n+1}+1}=\dfrac{\dfrac{7x_n+5}{x_n+3}-5}{\dfrac{7x_n+5}{x_n+3} + 1}=\dfrac{2x_n-10}{8x_n + 8} = \frac{1}{4} \cdot \dfrac{x_n-5}{x_n +1 }=\frac{1}{4} b_n.$$
Therefore $(b_n)$ is a geometric progression, with ratio $\dfrac{1}{4}$ and the first term
$$b_1 = \frac{x_1-5}{x_1+1}=\frac{1-5}{1+1}=\frac{-4}{2}=-2.$$
Therefore
$$b_{n+1}=b_1 \cdot \left (\dfrac{1}{4}\right)^n= - 2 \cdot \left (\dfrac{1}{4}\right)^n,$$
or equaivalently,
$$\dfrac{x_{n+1}-5}{x_{n+1}+1}=- 2 \cdot \left (\dfrac{1}{4}\right)^n \Leftrightarrow x_{n+1} = \frac{5\cdot 4^n - 2}{4^n +2}\Leftrightarrow x_n = \frac{5\cdot 4^n - 8}{4^n + 8}.$$
