System of 3 differential equations I'm trying to solve this system
$$
\begin{align}
x'&=x-3y+3z\\
y'&=-2x-6y+13z\\
z'&=-x-4y+8z
\end{align}
$$
must be reduced to a single equation
I tried to express the x 3 and substitute in the other two
but then I have not reduced y or z
I can not understand how to solve it
Euler method can not be solved because it is written in the job to reduce to a single equation
 A: Matrix approach: name the equations as (1), (2), (3). To reduce them into one, you have to do
$$a\cdot (1) +b\cdot (2) +c\cdot (3)$$
such that the right hand side is exactly a multiple of the right hand side. That is
$$(ax+by+cz)'=(a-2b-c)x+(-3a-6b-4c)y+(3a+13b+8c)z\\
=\lambda(ax+by+cz)$$
This makes it an eigenvalue problem. You need to solve for eigenvalue $\lambda$, then find the eigenvectors. The eigenvector will make a good set of $a,b,c$.
Another approach: (without using matrix)
$$2\cdot (3)-(2): 2z'-y'=-2y+3z\tag{4}\\
(1)+(3): x'+z'=-7y+11z\implies x'=-z'-7y+11z$$
Now equate this with (1):
$$-z'-7y+11z=x-3y+3z\implies x=-z'-4y+8z \tag{5}$$
This equation can be used to eliminate the variable $x$. 
Taking the derivative of (5) gives
$$x'=-z''-4y'+8z' \tag{6}$$
Plug (5) and (6) into (1):
$$-z''+9z'-11z=4y'-7y\tag{7}$$
$$(7)-4\cdot (4): y=-z''+z'+z\tag{8}$$
Take derivative of (8):
$$y'=-z'''+z''+z'\tag{9}$$
Now plug (8) and (9) into (7):
$$2z'-3z=-z'''+z''+z'-2(-z''+z'+z)$$
This is the single ode for $z$.
A: I would look at the problem this way:
consider the vector $v = (x,y,z)^T$ and $ v' = (x',y',z')^T$.
Then you can rewrite the problem like such:
$v'(t) = Av(t)$ 
where A is the matrix containing the coefficients of your functions. In your case:
$A :=\left( \begin{matrix}
  1 &-3 & 3 \\
  -2 & -6 & 13 \\
  -1 & -4 & 8
 \end{matrix} \right)$
Now you know that the solution must have the form $ v(t) = e^{At} \cdot c$ where $c \in \mathbb{R}^3$ is the vector with the constant values you could determine with given information about the functions. Assuming you know how the Jordan normal form works, you can rewrite this matrix as $ A = SJS^{-1}$.
Then we have $  v(t) = e^{SJS^{-1}t} \cdot c$ and by using the definition of the exp function we get $ v(t) = S\cdot  e^{Jt}\cdot S^{-1}\cdot c$.
I will leave the matrix multiplication to you but I always thought that the $e^{Jt}$ was kinda tricky:
In your case:
$e^{Jt} = \exp\left( \begin{matrix}
  t & t & 0 \\
  0 & t & t \\
  0 & 0 & t
 \end{matrix} \right)$
You can divide this matrix into two separate ones:
$I_n := \left( \begin{matrix}
  t & 0 & 0 \\
  0 & t & 0 \\
  0 & 0 & t
 \end{matrix} \right)$
and $N :=\left( \begin{matrix}
  0 & t & 0 \\
  0 & 0 & t \\
  0 & 0 & 0
 \end{matrix} \right)$
Then using $\exp(x+y) = \exp(x) \cdot \exp(y)$ you can now rewrite as such:
$e^{Jt} = e^{I_n} \cdot e^{N}$
and by taking a close look see that
$e^{I_n} = \left( \begin{matrix}
  e^t & 0 & 0 \\
  0 & e^t & 0 \\
  0 & 0 & e^t
 \end{matrix} \right)$
Almost done! But you must see that $N^k = 0  \forall k \geq 3$
That means you can calculate $e^N$ quite quickly and see that (if I calculated it correctly that is :) )
$e^N = \left( \begin{matrix}
  1 & t & 0.5\cdot t^2 \\
  0 & 1 & t \\
  0 & 0 & 1
 \end{matrix} \right)$
And therefore
$e^{Jt} = e^{I_n} \cdot e^{N} = \left( \begin{matrix}
  e^t & t\cdot e^t & 0.5\cdot t^2 \cdot e^t \\
  0 & e^t & t \cdot e^t \\
  0 & 0 & e^t
 \end{matrix} \right)$
And then you can finally see what your solutions must be by 'simple' matrix multiplication
$v(t) = S \cdot \left( \begin{matrix}
  e^t & t\cdot e^t & 0.5\cdot t^2 \cdot e^t \\
  0 & e^t & t \cdot e^t \\
  0 & 0 & e^t
 \end{matrix} \right) \cdot S^{-1} \cdot c $ 
I hope this could help you! If you are familiar with the Jordan normal form I think its a cool way to solve problems like this! 
Best wishes, 
Gustav
