How to determine if the following function subsets from $\mathbb{R}$ are vector subspaces (beginner)

Question

As per the title, I wish to determine which of the following subsets from $\mathbb{R}$ are vector subspaces.

I have consulted this related question, and I understand that three criteria need to be fulfilled. (namely, the 0 element is contained, the set is closed und addition and the set is closed under multiplication). Nonetheless, the notation below isn't intuitive to me. Below I have provided two subsets, as I think that the two examples are complimentary for forming an understanding.

$U_1 := \{ (x,y,z) | 3x - 4y + 2z = 0, x + 2y - z = 0\}$

...

$U_4 := \{ (2t + s, s, t-s) | s,t \in \mathbb{R}\}$

Any help in understanding the notation would be appreciated.

Thanks

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EDIT: It has been pointed out that the following proposed solution is wrong.

Proposed solution for $U_4$:

I) Subset contains the 0 element:

If $s = 0, s \in \mathbb{R} = 0.$

$s = 0 \land 2t + s = 0 \implies t \in \mathbb{R} = 0.$

II) Subset is closed under addition:

If $s = 0 \land t = 0$, then $s + t = 0$.

III) Subset is closed under multiplication:

If $s = 0 \land \lambda \in \mathbb{R}$ then $\lambda s = 0$

If $t = 0 \land \lambda \in \mathbb{R}$ then $\lambda t = 0$

Answer 1

What's preventing you from checking the three conditions? In the case of $U_1$:

1. Since $3\times0-4\times0+2\times0=0+2\times0-0=0$, $(0,0,0)\in U_1$.
2. If $3x_1-4y_1+2z_1=3x_2-4y_2+2z_2=0$, then $3(x_1+x_2)-4(y_1+y_2)+2(z_1+z_2)=0$. Also, if $x_1+2y_1-z_1=x_2+2y_2-z_2=0$, then $x_1+x_2+2(y_1+y_2)-(z_1+z_2)=0$
3. If $3x_1-4y_1+2z_1=x_1+2y_1-z_1=0$ and $\lambda\in\mathbb R$, $3\lambda x_1-4\lambda y_1+2\lambda z_1=\lambda(3x_1-4y_1+2z_1)=0$ and $\lambda x_1+2\lambda y_1-\lambda z_1=\lambda(x_1+2y_1-z_1)=0$.

Can you do it now for $U_4$?