Let $$f : R - \{3\} \to R - \{1\}$$ be defined by $$f(x) = \frac{x-2}{x-3}$$. Let $$g : R \to R$$ be given as $$g(x) = 2x - 3$$. Then, the sum of all the values of $$x$$ for which $$f^{-1}(x) + g^{-1}(x) = \frac{13}{2}$$ is equal to

Given $$f(x) = \frac{x-2}{x-3}$$, we find $$f^{-1}$$. Setting $$y = \frac{x-2}{x-3}$$, we solve for $$x$$: $$y(x-3) = x - 2$$, so $$yx - 3y = x - 2$$, giving $$x(y-1) = 3y - 2$$, hence $$x = \frac{3y - 2}{y - 1}$$. Therefore $$f^{-1}(x) = \frac{3x - 2}{x - 1}$$.

Given $$g(x) = 2x - 3$$, we find $$g^{-1}(x) = \frac{x + 3}{2}$$.

The equation $$f^{-1}(x) + g^{-1}(x) = \frac{13}{2}$$ becomes $$\frac{3x-2}{x-1} + \frac{x+3}{2} = \frac{13}{2}$$. Multiplying through by $$2(x-1)$$: $$2(3x-2) + (x+3)(x-1) = 13(x-1)$$. Expanding: $$6x - 4 + x^2 + 2x - 3 = 13x - 13$$. This simplifies to $$x^2 + 8x - 7 = 13x - 13$$, so $$x^2 - 5x + 6 = 0$$, which factors as $$(x-2)(x-3) = 0$$.

Thus $$x = 2$$ or $$x = 3$$. Both values are in the domain of $$f^{-1}$$ (since $$x \neq 1$$). The sum of all values is $$2 + 3 = 5$$.