If $S$ is an affine independent subset of $R^n$ and $x \notin$ the affine span of $S$ then $S \cup \{x\}$ is affine independent. If $S$ is an affine independent subset of $R^n$ and $x \notin$ the affine span of $S$ then $S \cup \{x\}$ is affine independent.


I start by letting $S=\{a_0, \dots, a_m\}$ and therefore for by the definition of linear independence:
$$(\forall \{t_i\})[\sum_{i=1}^m t_i(a_i-a_0)=0 \Rightarrow t_i = 0, \forall i]$$  So then I need to show that $S \cup \{x\} = \{a_0, \dots , a_n, x\}$ is affine independent, or $$(\forall \{j_i\})[\sum_{i=1}^m j_i(a_i-a_0) + j_x(x-a_0)=0 \Rightarrow j_i = 0, \forall i]$$

But I'm a little confused as to how I'd continue from here after letting $\{j_i\}$ be arbitrary and assuming that $\sum_{i=1}^m j_i(a_i-a_0) + j_x(x-a_0)=0$.

Anyone have any ideas?
Let $x$ be outside the span of $S$. Suppose that $\alpha_1 v_1 + \dotsb + a_n v_n - \beta x = 0$ with $S = \{ v_1, \dotsc, v_n \}$ and $\alpha_1 + \dotsb + \alpha_n - \beta = 0$.

Suppose that $\beta \neq 0$, then $x = \frac{\alpha_1}{\beta} v_1 + \dotsb + \frac{\alpha_n}{\beta}$, and since we supposed $\alpha_1 + \dotsb + \alpha_n - \beta = 0$, we deduce that $\frac{\alpha_1}{\beta} + \dotsb + \frac{\alpha_n}{\beta} = 1$. This contradicts the hypothesis that $x$ is not in the span of $S$.

Therefore $\beta = 0$, we have $\alpha_1 v_1 + \dotsb + \alpha_n v_n = 0$ and $\alpha_1 + \dotsb + \alpha_n = 0$. The affine independence of $S$ implies that $\alpha_i = 0$ for all $i$. From this we deduce that $S \cup \{  x\}$ is affinely independent.
Starting from 

$ \sum_{i=1}^m j_i(a_i-a_0) + j_x(x-a_0)=0  $, you could write $ \sum_{i=1}^m j_i(a_i-a_0) = - j_x(x-a_0)  $

Now, the left-side term is a combination of elements in $S$ and therefore is an element of $span(S)$. But, $x$ being not in $span(S)$, the right-hand term is not in $span(S)$ unless $j_x = 0$. Given that the two terms are equal, they should both belong to the same vector spaces; in particular, they should belong to $span(S)$.

Hence, the right hand term is $0$, and you can conclude that all $j_i$ are $0$ from here.