Answer
$f(x)=-\dfrac{1}{2}x-\dfrac{1}{2}$
Work Step by Step
Using the properties of equality, the given equation, $
2y+x=3
,$ is equivalent to
\begin{array}{l}
2y=-x+3
\\\\
y=-\dfrac{1}{2}x+\dfrac{3}{2}
.\end{array}
Using $y=mx+b$, where $m$ is the slope, the slope of the given line is
\begin{array}{l}
m=-\dfrac{1}{2}
.\end{array}
Using $m=
-\dfrac{1}{2}
$ (parallel lines have same slope) and the given point $(
3,-2
),$ then the equation of the line is
\begin{array}{l}
y-(-2)=-\dfrac{1}{2}(x-3)
\\\\
y+2=-\dfrac{1}{2}x+\dfrac{3}{2}
\\\\
y=-\dfrac{1}{2}x+\dfrac{3}{2}-2
\\\\
y=-\dfrac{1}{2}x-\dfrac{1}{2}
.\end{array}
In function notation, this is equivalent to $
f(x)=-\dfrac{1}{2}x-\dfrac{1}{2}
.$