If $f(1-x) + 2f(x) = 3x$, what is $f(0)?$ I did the following:
$$f(1-0) + 2f(0) = 3\cdot 0$$
$$f(1) + 2f(0) = 0$$
This reminds me of the equation of the straight line in the plane, then:
$$\left< \begin{pmatrix}
{1}\\ 
{2}
\end{pmatrix}
,
\begin{pmatrix}
{f(1)}\\ 
{f(0)}
\end{pmatrix} \right> =0$$
$(1,2)$ is a normal vector to $(f(1),f(0))$, then It's possible that: $(f(1)=-2,f(0)=1)$ because:
$$\left< \begin{pmatrix}
{1}\\ 
{2}
\end{pmatrix}
,
\begin{pmatrix}
{-2}\\ 
{1}
\end{pmatrix}\right> =-2+2=0$$
With generality, It is possible that for all $\alpha$:
$$\left< \begin{pmatrix}
{1}\\ 
{2}
\end{pmatrix}
,
\begin{pmatrix}
{\alpha\cdot (-2)}\\ 
{\alpha \cdot 1}
\end{pmatrix}\right> = 
\left< \begin{pmatrix}
{1}\\ 
{2}
\end{pmatrix}
,
\alpha \begin{pmatrix}
{ -2}\\ 
{ 1}
\end{pmatrix}\right> =1\cdot -2\alpha+2\alpha \cdot 1=-2\alpha + 2 \alpha =0$$
I tagged with "functional equations" because it seems to be related. I don't know if my solution is correct, what I'm saying with this is that there are infinite solutions but I don't know if assuming arbitrary values for $f(1),f(0)$ can be made because of $3x$.
EDIT: I know that It is possible to solve: 
$$f(1) + 2f(0) = 0$$
$$f(0) + 2f(1) = 3$$
And get the solution. But why do I have to use the latter instead of the former?

Just in case if you want to solve for the actual function next time:Substituting $x$ with $1-x$
$$f(1-(1-x))+2f(1-x)=3(1-x)$$$$f(x)+2f(1-x)=3-3x\tag{1}$$Now multiply the original equation by 2:$$2f(1-x)+4f(x)=6x\tag{2}$$
Subtract: $(2)-(1)$ We get
$$3f(x)=9x-3$$Therefore$$f(x)=3x-1$$$$f(0)=-1$$
