Let f: R - 3 → R - 1 be defined by f(x) = (x - 2)/(x - 3) . Let g:...

$f\left(x\right) = y = \dfrac{x - 2}{x - 3}$

$∴ x = \dfrac{3y - 2}{y - 1}$

$∴ {f}^{- 1}\left(x\right) = \dfrac{3x - 2}{x - 1}$

$\text{ and } g ( x ) = y = 2x - 3$

$∴ x = \dfrac{y + 3}{2}$

$∴ {g}^{- 1}\left(x\right) = \dfrac{x + 3}{2}$

$∵ {f}^{- 1}\left(x\right) + {g}^{- 1}\left(x\right) = \dfrac{13}{2}$

$\dfrac{3x - 2}{x - 1} + \dfrac{x + 3}{2} = \dfrac{13}{2}$

$∴ {x}^{2} - 5x + 6 = 0$ has roots ${x}_{1}, {x}_{2}$

$∴$ sum of roots ${x}_{1} + {x}_{2} = 5$