I found the following excercise:
Let $W_1 = \{(x_1, ..., x_6) : x_1 + x_2 + x_3 = 0, x_4 + x_5 + x_6 = 0 \}$. Let $W_2$ be the span of $S := \{(1, -1, 1, -1, 1, -1), (1, 0, 2, 1, 0, 0), (1, 0, -1, -1, 0, 1), (2, 1, 0, 0, 0, 0)\}$. Give a base, a dimension and an equation representation of $W_1 + W_2$
I'm new to the concept of sum of subspaces. But as I understand it, the first step would be to note that any $\textbf{x} = (x_1, ..., x_6) \in W_1$ satisfies
\begin{equation*} \begin{cases} x_1 = -x_2 - x_3 \\ x_4 = -x_5 - x_6 \end{cases} \end{equation*}
so that its general form is
\begin{equation*} \textbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{pmatrix} = \begin{pmatrix} -x_2 - x_3 \\ x_2 \\ x_3 \\ -x_5 - x_6 \\ x_5 \\ x_6 \end{pmatrix} \end{equation*}
We also know any $\textbf{y} \in W_1$ is of the general form
\begin{align*} \textbf{y} = \begin{pmatrix} y_1 \\ y_2 \\ y_3 \\ y_4 \\ y_5 \\ y_6 \end{pmatrix} = \begin{pmatrix}x + y +z + 2w \\ -x + w \\ x + 2y - z \\ -x + y - z \\ x\\ -x + z\end{pmatrix} \end{align*}
Then for generals $\mathbf{x}, \mathbf{y}$ we have
\begin{align*} \textbf{x} + \textbf{y} &= \begin{pmatrix} x + y +z + 2w + (-x_2 - x_3)\\ -x + w + x_2\\ x + 2y - z + x_3\\ -x + y - z + (-x_5 - x_6)\\ x + x_5\\ -x + z + x_6 \end{pmatrix} \end{align*}
One can then conclude
$$W_1 + W_2 = \Big\{\big(x + y + z + 2w - x_2 - x_3\big), \big(-x + w + x_2 \big), \big(x +2y - z + x_3 \big), \big(-x + y - z - x_5 - x_6 \big), \big(x + x_5 \big), \big(-x + z + x_6 \big) \mid x, y, z, x_2, x_3 \in \mathbb{R} \Big\}$$
But what would be an representation via equations of this system? I'm very new to linear algebra so go easy on me!