$\begin{array}{l}\text{ Let }f:R-\left\{3\right\}\to R-\left\{1\right\}\text{ be defined by }f\left(x\right)=\frac{x-2}{x-3}\\ \text{ Let }g:R\to R\text{ be given as }g\left(x\right)=2x-3\text{ . Then, the sum of all the values of }x\text{ for }\end{array}$

$\text{ which }{f}^{-1}\left(x\right)+g^{-1}\left(x\right)=\frac{13}{2}\text{ is equal to. }$

$\begin{array}{c}\text{Given }f\left(x\right)=\frac{x-2}{x-3}=y\\ x-2=xy-3y\\ x\left(1-y\right)=-3y+2\\ x=\frac{3y-2}{y-1}\text{'}\\ f^{-1}\left(x\right)=\frac{3x-2}{x-1}\end{array}$

$\begin{array}{c}g\left(x\right)=2x-3=y\\ x=\frac{y+3}{2}\\ g^{-1}\left(x\right)=\frac{x+3}{2}\end{array}$

$\begin{array}{c}{f}^{-1}\left(x\right)+g^{-1}\left(x\right)=\frac{13}{2}\\ \frac{3x-2}{x-1}+\frac{x+3}{2}=\frac{13}{2}\\ 6x-4+x^2-x+3x-3=13\left(x-1\right)\\ x^2-5x+6=0\\ \text{solving this equation}\\ x=2,3\end{array}$

$sum of the values is 2+3=5$