Differential to Difference equation with two variables? For the following information : 
$$\frac{dx}{dt} = -10x+3y$$
$$\frac{dy}{dt}  = 2$$
How do I convert this to a difference equation ??
I want to use a simple discretisation technique (first order) rather than higher order (4th order Rung Kutta ). I prefer the usage of Euler's forward method.
I tried using Euler formula but I really can't get my head around solving this, im stuck at the first step. I basically want to convert the differential equation to a difference equation to plug in t values and interpolate future values to create a graph at continuous time. I would appreciate a sound reply with an informative step by step solution, I wish to learn this rather than just see the final outcome.
Regards.
EDIT
The equations here consider Newton's temperature rules. Where x is the room temperature and y is heater temperature. dy/dt is the rate of change of the heater's temperature with time when it is turned on. dx/dt is the rate of change of room's temperature according to the difference between room and heater temperatures. --  I will actually add this in question.
 A: The approach is just like a single equation.  For the forward Euler you have $$x_{n+1}=x_n+(-10x_n+3y_n)\Delta t \\ y_{n+1}=y_n +2\Delta t$$
A: Let $f: \mathbb{R}^2 \to \mathbb{R}^2$ be given by $f(x) = f(\begin{pmatrix}x_1\\x_2\end{pmatrix})= \begin{pmatrix}-10x_1+3x_2\\2\end{pmatrix}$. Then the differential equation can be written as $\dot{x} = f(x)$. Euler's forward method just approximates $\dot{x}$ on an interval $[t_0,t_0+h)$ by taking it to be the value $\dot{x}(t_0)$ on this interval. This gives the approximation $x(t) \approx x(t_0)+(t-t_0) f(x(t_0))$.
So, if we consider the time points $t_0 + kh$, $k=0,1,...$, then this gives the difference equation 
$$\tilde{x}(t_0+(k+1)h) = \tilde{x}(t_0+kh)+hf(\tilde{x}(t_0+kh)$$
subject to $\tilde{x}(t_0) = x_0$. This is typically written as 
$x_{k+1} = x_k+h f(x_k)$, where $x_k = \tilde{x}(t_0+kh)$. 
Euler's forward method is convergent and  conditionally stable. Loosely, convergent means that as the step size $h$ gets smaller, the solutions of the difference equations converge to the differential equation solution in some sense. Conditionally stable means that the step size needs to small enough to avoid the numerics 'going wild'.
(As an aside, the equations can be solved explicitly without any need for discretization.)
Addendum: To clarify, expanding the above for a step size $h$ gives (in the question's notation):
\begin{eqnarray}
x_{k+1} &=& x_k + h(-10 x_k + 3 y_k) \\
y_{k+1} &=& y_k +  h(2)
\end{eqnarray}
