Sum of two subspaces: representing it with equations 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!

Note that $(1,0,−1,−1,0,1)\in W_1$. Therefore, if$$W_3=\operatorname{span}\bigl\{(1,−1,1,−1,1,−1),(1,0,2,1,0,0),(2,1,0,0,0,0)\bigr\},$$then $W_1+W_2=W_1+W_3$.
Now, let us see what $W_1\cap W_3$ is. Asserting that$$\overbrace{\alpha(1,−1,1,−1,1,−1)+\beta(1,0,2,1,0,0)+\gamma(2,1,0,0,0,0)}^{\text{arbitrary element of }W_3}\in W_1$$means that $\alpha+3\beta+3\gamma=-\alpha+\beta=0$. This occurs if and only if $\beta=\alpha$ and $\gamma=-\frac{4\alpha}3$. Therefore,$$W_1\cap W_3=\left\{\left(-\frac{2 \alpha }{3},-\frac{7 \alpha }{3},3 \alpha,0,\alpha ,-\alpha \right)\,\middle|\,\alpha\in\Bbb R\right\};$$in particular, $\dim(W_1\cap W_3)=1$. So,\begin{align}\dim(W_1+W_3)&=\dim(W_1)+\dim(W_3)-\dim(W_1\cap W_3)\\&=4+3-1\\&=6.\end{align}Therefore, $W_1+W_3$ is the whole space $\Bbb R^6$.
