Vector equations, solve for what and how to I got this question as a task for second grade high school math, and I can't really figure out what to solve for and how to solve it. 
$$2(k\vec a+\vec x)+m\vec x=5\vec a-k\vec x-m\vec a$$
I think this should be an easy question, as we just started working with vectors and this was at the start of the test. 
I've tried isolating $x$ and $a$, I got 
$\vec a(5-m-2k)$ and 
$\vec{x}(k+2+m)$
but this didn't get me any further. I don't know what i'm supposed to solve for / find when solving, could anyone point me in the right direction? 
 A: How about gathering like terms (like vectors), with the objective of isolating one of the variables (a vector) to express it as a function of the other variable (a vector):
$$\begin{align}2(k\vec a + \vec x) +m\vec x = 5\vec a - k\vec x -m\vec a & \iff 2k\vec a + 2\vec x+m\vec x = 5\vec a - k\vec x -m\vec a\\ \\ & \iff  2k\vec a - 5 \vec a + m \vec a   = -k\vec x - m\vec x \\ \\ & \iff (2k - 5+ m)\vec a = -(k + 2 + m)\vec x\\ \\ & \;\;\;\text{or}\quad(k + 2 + m)\vec x = (5 - 2k - m)\vec a\end{align}$$ 
Now you can express $\vec x$ as a function of $\vec a$, or vice versa, dividing by the respective parenthetical coefficient, provided $k + 2 + m \neq 0$ when solving for $\vec x$, or provided $(5 - 2k - m)\neq 0$ in the case of solving for $\vec a$.
A: Note that, as when working with scalars, you just perform the same operations on each side of the equality. This means that what you have done so far can be expressed like this:
$$ \vec{x}(k+2+m)=\vec{a}(5-m-2k) $$ 
This should make it easier to identify the next step necessary to isolate $x$.
