Let f: R-{3} rightarrow R-{1} be defined by f | vedclass

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(C) Given $f(x) = \frac{x-2}{x-3}$. Let $y = \frac{x-2}{x-3}$.$y(x-3) = x-2 \implies xy - 3y = x - 2 \implies x(y-1) = 3y - 2 \implies x = \frac{3y-2}

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$5$

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(C) Given $f(x) = \frac{x-2}{x-3}$. Let $y = \frac{x-2}{x-3}$.$y(x-3) = x-2 \implies xy - 3y = x - 2 \implies x(y-1) = 3y - 2 \implies x = \frac{3y-2}{y-1}$.Thus,$f^{-1}(x) = \frac{3x-2}{x-1}$.Given $g(x) = 2x-3$. Let $y = 2x-3$.$y+3 = 2x \implies x = \frac{y+3}{2}$.Thus,$g^{-1}(x) = \frac{x+3}{2}$.We are given $f^{-1}(x) + g^{-1}(x) = \frac{13}{2}$.$\frac{3x-2}{x-1} + \frac{x+3}{2} = \frac{13}{2}$.Multiply by $2(x-1)$: $2(3x-2) + (x+3)(x-1) = 13(x-1)$.$6x - 4 + x^2 + 2x - 3 = 13x - 13$.$x^2 + 8x - 7 = 13x - 13$.$x^2 - 5x + 6 = 0$.$(x-2)(x-3) = 0$.The roots are $x = 2$ and $x = 3$. However,$f^{-1}(x)$ is defined for $x 
eq 1$. Since $x=3$ is in the domain of $f^{-1}$,we check the original function $f(x)$. The domain of $f$ is $R-\{3\}$,so $f^{-1}$ is defined on $R-\{1\}$. Both $x=2$ and $x=3$ are valid. The sum of the roots is $2+3 = 5$.

Let $f: R-\{3\} \rightarrow R-\{1\}$ be defined by $f(x) = \frac{x-2}{x-3}$. Let $g: R \rightarrow R$ 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

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