Answer
$\text{a) }
f(g(x))=x-\dfrac{1}{10}
\\\\\text{b) }
g(f(x))=x+\dfrac{19}{10}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
With
\begin{array}{l}\require{cancel}
f(x)=
\dfrac{1}{2}x+\dfrac{1}{5}
\\g(x)=
2x-\dfrac{3}{5}
,\end{array}
replace $x$ with $g(x)$ in $f$ to find $f(g(x)).$ To find $g(f(x)),$ replace $x$ with $f(x)$ in $g.$
$\bf{\text{Solution Details:}}$
Replacing $x$ with $g(x)$ in $f,$ then
\begin{array}{l}\require{cancel}
f(g(x))=f\left( 2x-\dfrac{3}{5} \right)
\\\\
f(g(x))=\dfrac{1}{2}\left( 2x-\dfrac{3}{5} \right)+\dfrac{1}{5}
\\\\
f(g(x))=x-\dfrac{3}{10}+\dfrac{1}{5}
\\\\
f(g(x))=x-\dfrac{3}{10}+\dfrac{2}{10}
\\\\
f(g(x))=x-\dfrac{1}{10}
.\end{array}
Replacing $x$ with $f(x)$ in $g.$ Hence,
\begin{array}{l}\require{cancel}
g(f(x))=g\left( \dfrac{1}{2}x+\dfrac{1}{5} \right)
\\\\
g(f(x))=2\left( \dfrac{1}{2}x+\dfrac{1}{5} \right)-\dfrac{3}{5}
\\\\
g(f(x))=x+\dfrac{5}{2}-\dfrac{3}{5}
\\\\
g(f(x))=x+\dfrac{25}{10}-\dfrac{6}{10}
\\\\
g(f(x))=x+\dfrac{19}{10}
.\end{array}
Hence,
\begin{array}{l}\require{cancel}
\text{a) }
f(g(x))=x-\dfrac{1}{10}
\\\\\text{b) }
g(f(x))=x+\dfrac{19}{10}
.\end{array}
