Given $f(x) = \frac{2x-3}{x+1},$ find the value of $a$ such that $f(3a-2) = 1.$

$f(x)=\dfrac{2x-3}{x+1}\\~\\ f(3a-2)=\dfrac{2(3a-2)-3}{(3a-2)+1}\\~\\ \ \\~\\ f(3a-2)\ =\ 1\\~\\ \dfrac{2(3a-2)-3}{(3a-2)+1}\ =\ 1\\~\\ \ \\~\\ \dfrac{6a-4-3}{3a-2+1}\ =\ 1\\~\\ \dfrac{6a-7}{3a-1}\ =\ 1\\~\\ \begin{array}{} 6a-7\ =\ 1(3a-1)&\qquad\text{and}\quad3a-1\neq0\\~\\ 6a-7\ =\ 3a-1&\qquad\text{and}\quad3a-1\neq0\\~\\ 6a-3a\ =\ 7-1&\qquad\text{and}\quad3a-1\neq0\\~\\ 3a\ =\ 6&\qquad\text{and}\quad3a-1\neq0\\~\\ a\ =\ 2&\qquad\text{and}\quad a\neq\frac13 \end{array}\\~\\ a\ =\ 2$