Show that the subset $S$ in $\mathbb{R}_3$ is a subspace. Show that the subset $S$ in $\mathbb{R}_3$ defined by $S=\{(a,b,c) \in \mathbb{R}_3 \text{ such that } a+b=c \}$ is a subspace.
I'm having trouble adapting the definition of subspace with the part $a+b=c$.

You want to show, that 
$$ S = \left\{ (a,b,c) \in \Bbb R^3 \; : \; a+b = c \right\} \subset \Bbb R^3$$
is a subspace of $\Bbb R^3$. 
First, note that $(0,0,0) \in S$, since $0 + 0 = 0$, so $S \neq \emptyset$. 
Next, let $v_1 := (a_1, b_1, c_1), \, v_2 := (a_2, b_2, c_2) \in S$. We have to show, that $v_1 + v_2 = (a_1 + b_1, a_2 + b_2, a_3 + b_3) \in S$. Since $v_1 \in S$, we have $a_1 + b_1 = c_1$, and since $v_2 \in S$, we have $a_2 + b_2 = c_2$. So we see, that 
$$(a_1 + a_2) + (b_1 + b_2) = c_1 + c_2 \; ,$$
which means that 
$$ v_1 + v_2 = (a_1 + a_2, b_1 + b_2, c_1 + c_2) \in S \; .$$
Finally, let $\alpha \in \Bbb R$ and $v := (a,b,c) \in S$. We need to show, that $\alpha v \in S$. Since $a+b = c$, we have $\alpha a + \alpha b = \alpha c$, which means that 
$$ \alpha v = (\alpha a, \alpha b, \alpha c) \in S \; .$$
This shows, that $S$ is a subspace of $\Bbb R^3$.
