Can choosing a different value for a free variable yield a different solution to a Row Reduced Matrix? Let $ U = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 4 \end{bmatrix} $ and
$ c = \begin{bmatrix} 5 \\ 8 \end{bmatrix}$
I have been asked to reduce the matrix $ \begin{bmatrix} U & c \end{bmatrix} $ to $ \begin{bmatrix} R & d \end{bmatrix} $ and then solve $ Rx = d$
$ \begin{bmatrix} U & c \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 4 & 8 \end{bmatrix}$
Dividing row 2 by 2 and then subtracting it thrice from row 1 gives: $ \begin{bmatrix} 1 & 2 & 0 & -1 \\ 0 & 0 & 1 & 2 \end{bmatrix}$
This means that I can treat column 2 as a free column and thus $ x_2 $ is a free variable.
Writing out $ Rx = d $:
$$x_1 + 2x_2 = -1$$
$$x_3 = 2$$
If I take $x_2 = 0$ then the solution becomes $x = \begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix}$
If I take $x_2 = 1$ then the solution becomes $x = \begin{bmatrix} -3 \\ 1 \\ 2 \end{bmatrix}$
Both solutions seem to work. However, the website on which I found this question only shows the second option. Did I do something wrong when I set the free variable equal to 0?
Edit: The original question is number 8.2 on this pdf: https://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/ax-b-and-the-four-subspaces/solving-ax-b-row-reduced-form-r/MIT18_06SCF11_Ses1.8sol.pdf

Since $x_2$ is a free variable, every value you assign to $x_2$ will yield a solution to the linear system. They are all valid. There is nothing strange about only one of the solutions being in the alternatives... I imagine they asked which of the alternatives were solutions of the system.
