Determine basis for solution space when there are more unknowns than equations 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.

Take it to reduced row-echelon form. You have
$$\begin{bmatrix}-1&1&0&1\\4&0&1&0\end{bmatrix} \sim \begin{bmatrix}\color{red}1&0&\frac{1}{4}&0\\0&\color{red}1&\frac{1}{4}&1\end{bmatrix}.$$
From here, identify the pivots, coloured in red, and the columns without pivots in them. In this case, we have the third and fourth columns. This means it is safe to consider the third and fourth variables (i.e. $x_3$ and $x_4$) as free variables. The other two variables, $x_1$ and $x_2$, will have a unique solution given fixed values of $x_3$ and $x_4$.
Our equations are
\begin{align*}
x_1 + \frac{1}{4} x_3 &= 0 \\
x_2 + \frac{1}{4} x_3 + x_4 &= 0,
\end{align*}
or equivalently,
\begin{align*}
x_1 &= -\frac{1}{4} x_3 \\
x_2 &= -\frac{1}{4} x_3 - x_4 \\
x_3 &= x_3 \\
x_4 &= x_4.
\end{align*}
This gives us
\begin{align*}
(x_1, x_2, x_3, x_4) &= \left(-\frac{1}{4}x_3, -\frac{1}{4}x_3 - x_4, x_3, x_4\right) \\
&= \left(-\frac{1}{4}x_3, -\frac{1}{4}x_3, x_3, 0\right) + (0, -x_4, 0, x_4) \\
&= x_3\left(-\frac{1}{4}, -\frac{1}{4}, 1, 0\right) + x_4(0, -1, 0, 1).
\end{align*}
If you're up to bases yet, this tells you that
$$\left\{\left(-\frac{1}{4}, -\frac{1}{4}, 1, 0\right) + (0, -1, 0, 1)\right\}$$
forms a basis for your solution space.
