Let f: R-{2} rightarrow R-{1} defined by f(x) | vedclass

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(A) Step $1$: Find $f^{-1}(x)$. Let $y = \frac{x-3}{x-2}$. Then $y(x-2) = x-3$,so $xy - 2y = x - 3$. Thus $x(y-1) = 2y-3$,which gives $x = \frac{2y-3}

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$\frac{5}{2}$

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(A) Step $1$: Find $f^{-1}(x)$. Let $y = \frac{x-3}{x-2}$. Then $y(x-2) = x-3$,so $xy - 2y = x - 3$. Thus $x(y-1) = 2y-3$,which gives $x = \frac{2y-3}{y-1}$. So,$f^{-1}(x) = \frac{2x-3}{x-1}$.Step $2$: Find $g^{-1}(x)$. Let $y = 3x-2$. Then $3x = y+2$,so $x = \frac{y+2}{3}$. Thus,$g^{-1}(x) = \frac{x+2}{3}$.Step $3$: Solve the equation $f^{-1}(x) + g^{-1}(x) = \frac{19}{6}$.$\frac{2x-3}{x-1} + \frac{x+2}{3} = \frac{19}{6}$.Multiply by $6(x-1)$ to clear denominators: $6(2x-3) + 2(x-1)(x+2) = 19(x-1)$.$12x - 18 + 2(x^2 + x - 2) = 19x - 19$.$12x - 18 + 2x^2 + 2x - 4 = 19x - 19$.$2x^2 + 14x - 22 = 19x - 19$.$2x^2 - 5x - 3 = 0$.Step $4$: Solve the quadratic equation $2x^2 - 5x - 3 = 0$.$(2x+1)(x-3) = 0$.The roots are $x = -\frac{1}{2}$ and $x = 3$.Step $5$: The sum of the values of $x$ is $-\frac{1}{2} + 3 = \frac{5}{2}$.

Let $f: R-\{2\} \rightarrow R-\{1\}$ defined by $f(x)=\frac{x-3}{x-2}$ and $g: R \rightarrow R$ defined by $g(x)=3x-2$. Then,the sum of all values of $x$ for which $f^{-1}(x)+g^{-1}(x)=\frac{19}{6}$ is

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