Find $g^{-1}(3)$ given $g(x) = \dfrac{3x + 1}{2x + g(x)}$
My Approach:
\begin{align*} y & = \frac{3x + 1}{2x + y} && \text{(does $g(x)$ become $y$ also?)}\\ x & = \frac{3y + 1}{2y + x} && \text{(I switched $x$ and $y$)}\\ \Rightarrow x(2y + x) & = 3y + 1\\ \Rightarrow 2xy + x^2 & = 3y + 1\\ \Rightarrow 2xy - 3y & = 1 - x^2\\ \Rightarrow y(2x - 3) & = 1 - x^2\\ \Rightarrow y & = \frac{1 - x^2}{2x - 3}\\ \Rightarrow g^{-1}(x) & = \frac{1 - x^2}{2x - 3}\\ g^{-1}(3) & = \frac{1 - 9}{6 - 3} = -\frac{8}{3} \end{align*}
Does this seem correct? If not any hints on what I'm doing wrong?
Any help would be appreciated, I'm not friendly with the website syntax/latex
Your work is correct.
Note that $g^{-1}(3) = -\dfrac{8}{3} \Longleftrightarrow g\left(-\dfrac{8}{3}\right) = 3$.
As a check,
\begin{align*} g\left(-\frac{8}{3}\right) & = \frac{3\left(-\frac{8}{3}\right) + 1}{2\left(-\frac{8}{3}\right) + g\left(-\frac{8}{3}\right)}\\ 3 & = \frac{-8 + 1}{-\frac{16}{3} + 3}\\ & = \frac{-24 + 3}{-16 + 9}\\ & = \frac{-21}{-7}\\ & = 3 \end{align*}