Show that any 2D vectors can be expressed in the form... 
(a) Show that any 2D 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 $u$ and $v$ be non-zero vectors. Show that any 2D vector can be expressed in the form
  $s u + t v$
  where $s$ and $t$ are real numbers, if and only if of the vectors $u$ and $v$, one vector is not a scalar multiple of the other vector.

No idea what to do to start these problems. Hints appreciated.
 A: Hint
You can ignore (a), because (b) gives (a) as a byproduct.
For (b) start by letting $x = \pmatrix{x_1\\x_2} \in \mathbb R^2$ arbitrary and $u, v$ linearly independent. Then try to find a general solution to the system
$$\pmatrix{u_1 & v_1\\u_2 & v_2} \pmatrix{s\\t} = \pmatrix{x_1\\x_2}$$
(again, assuming independence - independence is equivalent to $u_1 v_2 - u_2 v_1 \ne 0$)
For the reverse, you can argue by contradiction, since not every vector in $\mathbb R^2$ will be a scalar multiple of a given $u$ and if $v = \lambda u$ then $s u + tv = (s+\lambda t)u$ always is a scalar multiple of $u$.
A: (a) Let $\begin{pmatrix} x\\ y\end{pmatrix}$ a 2D vector. Finding $s$ et $t$ such that $$s \begin{pmatrix} 3 \\ -1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 7 \end{pmatrix}=\begin{pmatrix} x\\ y\end{pmatrix}$$
is the same thing as solving:
$$\begin{cases}3s+2t=x\\
-s+7t=y\end{cases}$$
where $s$ and $t$ are the unknowns.
(b) Now if you replace $\begin{pmatrix} 3 \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 2 \\ 7 \end{pmatrix}$ by $u$ and $v$, what are the conditions on the previous system for it to admit a (unique) solution?
