Determine whether or not the function f it is bijective 2f(3-2x)+f(3/2-x/2)=x, where x is a real number
On the second equivalent equation you found,
$2f(y)+f(\frac{3-y}{4})=\frac{3-y}{2}$, (1)
use twice the map $x\mapsto (3-x)/2$, ie, let
$z=\frac{3-(\frac{3-y}{2})}{2}=\frac{y+3}{4}$,
$\Rightarrow y=4z-3$.
Plugging this in (1) yields
$2f(4z-3)+f(-z)=3-2z$
and by subtracting what we found from your first equation we can conclude
$f(z)=f(-z)$.
Hence $f$ is not injective, as it's an even function.