Show that $u, v, w$ are in the span of $\{u+v, 2u+3v, 4v+6w\}?$ I know this has to do with linear combinations, namely that you would set out to solve the following set of equations to show that $c_{1}, c_{2}$, and  $c_{3}$ exist and are not all 0, but I'm unclear as to how I actually solve for those in this case. 
That is, I know I should have these equations:
$u = c_{1}(u+v) + c_{2}(2u+3v) + c_{3}(4v+6w)$
$v = c_{1}(u+v) + c_{2}(2u+3v) + c_{3}(4v+6w)$
$w = c_{1}(u+v) + c_{2}(2u+3v) + c_{3}(4v+6w)$
Do you not need to solve for c to do this proof?

The solution I got is as follows:
$u = 3(u+v) + -1(2u+3v) + 0(4v+6w)$
$v = -2(u+v) + 1(2u+3v) + 0(4v+6w)$
$w = \frac{4}3(u+v) + \frac{-2}3(2u+3v) + \frac{1}6(4v+6w)$
