Question: Let's say we have

$$3x\_1+x\_2+x\_3+x\_4=0$$
$$5x\_1-x\_2+x\_3-x\_4=0$$

Then I would have matrix $\begin{bmatrix} 3 &1&1&1\cr8&0&2&0 \end{bmatrix}$ which simplifies to $\begin{bmatrix}-1&1&0&1\cr4&0&1&0\end{bmatrix}$

then $4x\_1+x\_3=0$ and $-x\_1+x\_2+x\_4=0$ after substitution we get $\frac{1}{4}x\_3+x\_2+x\_4=0$
and I'm not sure what to do from here.

Answer: 
\begin{bmatrix}-1&1&0&1\cr4&0&1&0\end{bmatrix}

Multiply 1st row by $4$ and add it to 2nd row to get :   

\begin{bmatrix}-1&1&0&1\cr 0&4&1&4\end{bmatrix}   

Let $x\_3=t$ and $x\_4=s$ and hence you have :   
$-x\_1+x\_2=-s$ and $4x\_2=-t-4s$ and hence $-x\_1=-s+t/4+s=t/4\implies x\_1=-t/4$  

$(x\_1,x\_2,x\_3,x\_4)^T=(-t/4,-t/4-s,t,s)^T=s(0, - 1,0,1)^T+t(-1/4,-1/4, 1,0)^T$

Basis of solution space (Null space in this particular case) is set $\{(0, -1,0,1)^T, (-1/4,-1/4, 1,0)^T\} $
You already have all the information, namely the echelon form, you need to write out the general solution.

The matrix $\begin{bmatrix}-1&1&0&1\cr4&0&1&0\end{bmatrix}$ you got tells you that
you can have $x\_1$ and $x\_2$ as the free variables and any solution can be written as
$$
x = a(1,0,-4,1)^T+b(0,1,0,-1)^T
$$

Let me know if you need more elaboration.