System of recursive equation. Let's  consider:
$$u_o = -1, v_0 = 3$$
$$\begin{cases} u_{n+1} = u_n + v_n \\ v_{n+1} = -u_n + 3v_n \end{cases}$$
I tried:
$$x^n = u_n , y^n = v_n$$
$$\begin{cases} x^{n+1} = x^n + y^n \\ y^{n+1} = -x^n + 3y^n \end{cases}$$
$$\begin{cases} x = 1 + \frac{y^n}{x^n}  \\ y = -\frac{x^n}{y^n} + 3 \end{cases}$$
And I don't know how to continue. And it is the basic problem.

The second problem:
(*)The question which occured while attemption solving it, and basically it is not connected with main problem. 
We have $y = 4-x$ and now $f =\frac{y^n}{x^n}= (\frac{4-x}{x})^n $
Lets observe that $n \to \infty$ so the $\lim_{n \to \infty}  f = 0$
So we can ignore it in our system of equation? It is my doubt. I'm nearly convinced it is not correct but I can't convince of myself why.
Help me, please!
 A: $$\begin{cases}
u_o = -1 \\
v_0 = 3 \\
u_{n+1} = u_n + v_n \\
v_{n+1} = -u_n + 3v_n
\end{cases}$$
So:
$$\begin{bmatrix}
u_{n+1} \\
v_{n+1} \\
\end{bmatrix} = 
\begin{bmatrix}
1 & 1 \\
-1 & 3
\end{bmatrix}
\begin{bmatrix}
u_{n} \\
v_{n} \\
\end{bmatrix}$$
In order to make it nonrecursive, convert the sequence of matrix multiplications to a $\text{matrix}^\text{integer}$ exponent:
$$\begin{bmatrix} u_{n} \\ v_{n} \\ \end{bmatrix} = \begin{bmatrix} 1 & 1 \\
-1 & 3 \end{bmatrix} ^ n \begin{bmatrix} u_{0} \\ v_{0} \\ \end{bmatrix}$$ And if you want to write it without matrices, a Jordan decomposition works:
$$
\begin{align}\begin{bmatrix} u_{n} \\ v_{n} \\ \end{bmatrix} &= \left(\begin{bmatrix} -1 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} -1 & 1 \\ -1 & 0 \end{bmatrix}^{-1}\right)^n\begin{bmatrix} u_{0} \\ v_{0} \\ \end{bmatrix} \\
&= \begin{bmatrix} -1 & 1 \\ -1 & 0 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}^n\begin{bmatrix} 1 & 1 \\ -1 & 3 \end{bmatrix}\begin{bmatrix} u_{0} \\ v_{0} \\ \end{bmatrix} \\
&= \begin{bmatrix} -1 & 1 \\ -1 & 0 \end{bmatrix}\begin{bmatrix} 2^n & n2^{n-1} \\ 0 & 2^n \end{bmatrix}\begin{bmatrix} 1 & 1 \\ -1 & 3 \end{bmatrix}\begin{bmatrix} u_{0} \\ v_{0} \\ \end{bmatrix} \\
&= \begin{bmatrix} u_0\left(2^n + n2^{n-1}\right) - v_0 n2^{n-1} \\ v_0\left(2^n + n2^{n-1}\right) - u_0 n2^{n-1} \end{bmatrix}
\end{align}$$
A: I don’t necessarily recommend it, but in this case the original problem is easily solved by rather elementary ad hoc methods. Adding the two recurrences, we see that $u_{n+1}+v_{n+1}=4v_n$. From the first recurrence we know that $u_{n+1}+v_{n+1}=u_{n+2}$, so $u_{n+2}=4v_n$, and $u_n=4v_{n-2}$ for $n\ge 2$. Substituting this into the second recurrence, we find that
$$v_{n+1}=3v_n-4v_{n-2}\;.\tag{1}$$
The recurrence $(1)$ has the characteristic equation $x^3-3x^2+4=0$. A little work with the rational root test produces the root $2$, so we can divide out a linear factor $x-2$ and then factor the resulting quadratic:
$$x^3-3x^2+4=(x-2)(x^2-x-2)=(x+1)(x-2)^2\;.$$
Thus, we expect $(1)$ to have the general solution
$$v_n=(A+Bn)2^n+C(-1)^n\;.\tag{2}$$
We have the initial condition $v_0=3$, and we can compute $v_1=10$, $u_1=2$, and $v_2=28$; substituting $n=0$, $n=1$, and $n=2$ into $(2)$ gives us
$$\left\{\begin{align*}
&A+C=3\\
&2A+2B-C=10\\
&4A+8B+C=28\;,
\end{align*}\right.$$
which is easily solved: $A=3$, $B=2$, and $C=0$, so
$$v_n=(3+2n)2^n=3\cdot2^n+n2^{n+1}=(n+1)2^{n+1}+2^n\;.$$
(You can use whichever form seems most convenient.) Finally, 
$$u_n=4v_{n-2}=4\left((n-1)2^{n-1}+2^{n-2}\right)=(n-1)2^{n+1}+2^n\;.$$
