Two-dimensional Vectors Problem (a) Show that any two-dimensional vector can be expressed in the form
$s \begin{pmatrix} 3 \\ -1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 7 \end{pmatrix},$
where $s$ and $t$ are real numbers.
(b) Let $\vec{u}$ and $\vec{v}$ be non-zero vectors. Show that any two-dimensional vector can be expressed in the form
$s \vec{u} + t \vec{v},$
where $s$ and $t$ are real numbers, if and only if of the vectors $\vec{u}$ and $\vec{v}$, one vector is not a scalar multiple of the other vector.
I've already gotten part (a) but (b) is a challenge. I have solved part a hough:
Let's say that $ \begin{pmatrix} 3 \\ -1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 7 \end{pmatrix}$ is equal to $\binom{a}{b}$.
\begin{align*}
 \begin{pmatrix} 3 \\ -1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 7 \end{pmatrix} &= \binom{a}{b}\\
\binom{3s}{-s}+\binom{2t}{7t}&=\binom{a}{b}\\
\binom{3s+2t}{-s+7t}&=\binom{a}{b}\\
\Rightarrow 3s+2t&=a\\
-s+7t&=b
\end{align*}
 A: Hint: Finding $s$ and $t$ for a given vector can be done in a similar manner to what you did in part (a). Thinking of $x$, $u$, and $v$ as constant vectors and $s$ and $t$ as real variables, the equation $x = su+tv$ is equivalent to the following system of equations in the variables $s$ and $t$, which you're probably more comfortable working with:
\begin{align*}
u_1s+v_1t &= x_1 \\
u_2s+v_2t &= x_2
\end{align*}
where
$$
    u = \begin{pmatrix}u_1 \\ u_2\end{pmatrix},\qquad
    v = \begin{pmatrix}v_1 \\ v_2\end{pmatrix},
    \qquad\text{and}\qquad
    x = \begin{pmatrix}x_1\\x_2\end{pmatrix}.
$$
Try to solve this for $s$ and $t$ like you normally would, and see what assumptions you have to make about $u_1,u_2,v_1,v_2$ to ensure that a solution $(s,t)$ exists for all values of $x_1,x_2$.
Edit: I can elaborate if necessary, but you're on the right track. I think you can figure this out using your approach to part (a).
A: $$su_1+tv_1=a$$
$$su_2+tv_2=b$$
set up for elimination ...
$$su_1 u_2+tv_1 u_2=a u_2$$
$$su_2 u_1+tv_2 u_1=b u_1$$
subtract (1) - (2)
$$t(v_1 u_2-v_2 u_1)= a u_2 - b u_1 $$
a solution can exist only if $v_1 u_2 \ne v_2 u_1$ ( $ \vec u $ and $\vec v $ not parallel) 
or if  $a u_2 = b u_1 $ ( all three vectors parallel )
